CBSE Class 12 Chemistry Electrochemistry Notes and Questions

Electrochemistry Concepts

Electrochemistry may be defined as the branch of chemistry which deals with the quantitative study of interrelationship between chemical energy and electrical energy and inter-conversion of one form into another, and the relationships of electrical energy taking place in redox reactions.
A cell is of two types:
1. Galvanic Cell
2. Electrolytic Cell

In a Galvanic cell, the chemical energy of a spontaneous redox reaction is converted into electrical work.
In an Electrolytic cell, electrical energy is used to carry out a non-spontaneous redox reaction.

The Standard Electrode Potential for any electrode dipped in an appropriate solution is defined with respect to the standard electrode potential of hydrogen electrode taken as zero. The standard potential of the cell can be obtained by taking the difference of the standard potentials of cathode and anode:
\(E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\)

The standard potential of the cells is related to standard Gibbs energy:
\(\Delta_r G^{\circ} = -nFE^{\circ}_{\text{cell}}\)

The standard potential of the cells is related to equilibrium constant:
\(\Delta_r G^{\circ} = -RT \ln K\)

Concentration dependence of the potentials of the electrodes and the cells are given by Nernst equation:
\(a\text{A} + b\text{B} + n\text{e}^- \rightarrow c\text{C} + d\text{D}\)

Nernst equation can be written as:
\(E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{RT}{nF} \ln \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}\)

The conductivity, \(\kappa\), of an electrolytic solution depends on the concentration of the electrolyte, nature of solvent, and temperature.
Molar Conductivity, \(\Lambda_m\), is defined by \(\kappa / C\) where \(C\) is the concentration in \(\text{mol L}^{-1}\):
\(\Lambda_m = \frac{\kappa \times 1000}{C}\)

The unit of molar conductivity is \(\Omega^{-1} \text{ cm}^2 \text{ mol}^{-1}\) (or \(\text{S cm}^2 \text{ mol}^{-1}\)). Conductivity decreases but molar conductivity increases with a decrease in concentration. It increases slowly with a decrease in concentration for strong electrolytes, while the increase is sharp for very dilute solutions of weak electrolytes.

Kohlrausch Law of Independent Migration of Ions: The law states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.

Faraday's Laws of Electrolysis:
I. The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte.
II. The amount of different substances liberated by the same quantity of electricity passing through the electrolytic solution is proportional to their chemical equivalent weights.

Batteries and fuel cells are very useful forms of galvanic cells. There are mainly two types of batteries: primary and secondary.

Corrosion of metals is an electrochemical phenomenon. In corrosion, metal is oxidized by loss of electrons to oxygen in the presence of moisture and formation of oxides.
Anode (Oxidation): \(2\text{Fe}(s) \rightarrow 2\text{Fe}^{2+}(aq) + 4\text{e}^-\)
Cathode (Reduction): \(\text{O}_2(g) + 4\text{H}^+(aq) + 4\text{e}^- \rightarrow 2\text{H}_2\text{O}(l)\)
Atmospheric Oxidation:
\(2\text{Fe}^{2+}(aq) + 2\text{H}_2\text{O}(l) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{Fe}_2\text{O}_3(s) + 4\text{H}^+(aq)\)

QUESTIONS CARRYING 1 MARK

Question 1. What is the effect of temperature on molar conductivity?
Answer: Molar conductivity of an electrolyte increases with an increase in temperature due to an increase in ionic mobility and a decrease in viscosity.
In simple words: Higher temperature makes ions move faster, increasing molar conductivity.

Exam Tip: Always link temperature rise with decreased viscosity and increased ionic speed.

Question 2. Why is it not possible to measure single electrode potential?
Answer: It is not possible to measure a single electrode potential because the half-cell containing a single electrode cannot exist independently; reduction cannot take place without simultaneous oxidation (charge cannot flow on its own in a single half-cell).
In simple words: You need a complete circuit with two electrodes to measure voltage.

Exam Tip: Mention the necessity of a complete redox circuit to earn full marks.

Question 3. Name the factors on which emf of a cell depends.
Answer: The emf of a cell depends on the following factors:
a. Nature of reactants.
b. Concentration of ions in the two half-cells.
c. Temperature.
d. Partial pressure of gases (if any involved).
In simple words: Cell voltage depends on what chemicals are used, their concentrations, temperature, and gas pressure.

Exam Tip: List at least three key factors clearly when asked.

Question 4. What are the units of molar conductivity?
Answer: \(\Omega^{-1} \text{ cm}^2 \text{ mol}^{-1}\) or \(\text{S cm}^2 \text{ mol}^{-1}\).
In simple words: The standard SI unit is Siemens square centimeter per mole.

Exam Tip: Memorize both forms as they are used interchangeably in numerical problems.

Question 5. Write Nernst equation for the general cell reaction \(a\text{A} + b\text{B} \rightarrow c\text{C} + d\text{D}\).
Answer: \(E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{RT}{nF} \ln \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}\)
In simple words: The actual cell potential equals standard potential minus a correction term for reactant and product concentrations.

Exam Tip: Ensure stoichiometric coefficients appear as powers of concentrations.

Question 6. What is the EMF of the cell when the cell reaction attains equilibrium?
Answer: Zero.
In simple words: When a cell reaches equilibrium, it stops producing current, so its EMF becomes 0.

Question 7. What is the electrolyte used in a dry cell?
Answer: A moist paste of ammonium chloride (\(\text{NH}_4\text{Cl}\)), zinc chloride (\(\text{ZnCl}_2\)), and manganese dioxide (\(\text{MnO}_2\)) mixed with carbon powder.
In simple words: A Leclanche dry cell uses an ammonium chloride paste electrolyte.

Exam Tip: Mention both \(\text{NH}_4\text{Cl}\) and \(\text{MnO}_2\) for complete credit.

Question 8. How is cell constant calculated from conductance values?
Answer: \(\text{Cell constant } (G^*) = \frac{\text{Specific conductance } (\kappa)}{\text{Observed conductance } (G)}\)
In simple words: Cell constant is specific conductance divided by measured conductance.

Exam Tip: Also remember formula \(G^* = \frac{l}{A}\).

Question 9. What flows in the internal circuit of a galvanic cell?
Answer: Ions (both cations and anions through the salt bridge and electrolyte solution).
In simple words: Electrons flow through the wire outside, while ions flow through the solution inside.

Exam Tip: Do not confuse internal ion flow with external electron flow.

Question 10. Define electrochemical series.
Answer: The arrangement of various elements and their respective electrodes in order of increasing or decreasing standard reduction potentials is called the electrochemical series.
In simple words: A list of substances ordered by their tendency to get reduced.

Exam Tip: Specify that it is arranged according to standard reduction potentials.

QUESTIONS CARRYING TWO MARKS

Question 1. How can you increase the reduction potential of an electrode for the reaction \(\text{M}^{n+}(aq) + n\text{e}^- \rightarrow \text{M}(s)\)?
Answer: According to the Nernst equation:
\(E_{\text{M}^{n+}/\text{M}} = E^{\circ}_{\text{M}^{n+}/\text{M}} - \frac{2.303RT}{nF} \log \frac{1}{[\text{M}^{n+}]}\)
The reduction potential can be increased by:
a. Increasing the concentration of metal ions (\(\text{M}^{n+}\)) in the solution.
b. Increasing the temperature.
In simple words: Adding more metal ions or raising the temperature increases the reduction potential.

Exam Tip: Write the Nernst equation first to justify your answer.

Question 2. Calculate emf of the following cell at 298K:
\(\text{Zn} / \text{Zn}^{2+} (10^{-4}\text{ M}) \parallel \text{Cu}^{2+} (10^{-2}\text{ M}) / \text{Cu}\)
Given \(E^{\circ}_{\text{Zn}^{2+}/\text{Zn}} = -0.76\text{ V}\), \(E^{\circ}_{\text{Cu}^{2+}/\text{Cu}} = +0.34\text{ V}\)
Answer: Cell reaction:
\(\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)\)
Here, \(n = 2\), \(T = 298\text{ K}\)
\(E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = 0.34\text{ V} - (-0.76\text{ V}) = 1.10\text{ V}\)
Applying Nernst equation:
\(E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n} \log \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}\)
\(E_{\text{cell}} = 1.10 - \frac{0.0591}{2} \log \frac{10^{-4}}{10^{-2}}\)
\(E_{\text{cell}} = 1.10 - 0.02955 \log(10^{-2})\)
\(E_{\text{cell}} = 1.10 - 0.02955(-2)\)
\(E_{\text{cell}} = 1.10 + 0.0591 = 1.1591\text{ V}\)
In simple words: Substitute standard potentials and ion concentrations into the Nernst equation to find cell EMF.

Exam Tip: Double-check sign conversions when subtracting negative anode potentials.

Question 3. Electrolysis of \(\text{KBr}(aq)\) gives \(\text{Br}_2\) at anode but \(\text{KF}(aq)\) does not give \(\text{F}_2\). Give reason.
Answer: Oxidation takes place at the anode. A species with a lower standard reduction potential (or higher oxidation potential) is oxidized first. The oxidation potentials follow the order:
\(\text{Br}^- > \text{H}_2\text{O} > \text{F}^-\)
Therefore, in aqueous \(\text{KBr}\), \(\text{Br}^-\) ions are oxidized to \(\text{Br}_2\) in preference to water. On the other hand, in aqueous \(\text{KF}\), water is oxidized in preference to \(\text{F}^-\) because the oxidation potential of water is higher than that of fluoride ion. Thus, oxidation of water at the anode gives \(\text{O}_2\) gas and no \(\text{F}_2\) is produced.
In simple words: Bromide is easier to oxidize than water, but water is much easier to oxidize than fluoride.

Exam Tip: Mention competing oxidation reactions at the anode involving water.

Question 4. What happens when a piece of copper is added to (a) an aqueous solution of \(\text{FeSO}_4\) (b) an aqueous solution of \(\text{FeCl}_3\)?
Answer:
a. Nothing will happen when a piece of copper is added to \(\text{FeSO}_4\) solution because the reduction potential of \(\text{Cu}^{2+}/\text{Cu}\) (\(+0.34\text{ V}\)) is greater than that of \(\text{Fe}^{2+}/\text{Fe}\) (\(-0.44\text{ V}\)), so copper cannot displace iron.
b. Copper will dissolve in an aqueous solution of \(\text{FeCl}_3\) because the reduction potential of \(\text{Fe}^{3+}/\text{Fe}^{2+}\) (\(+0.77\text{ V}\)) is higher than that of \(\text{Cu}^{2+}/\text{Cu}\) (\(+0.34\text{ V}\)).
Reaction: \(\text{Cu}(s) + 2\text{FeCl}_3(aq) \rightarrow \text{CuCl}_2(aq) + 2\text{FeCl}_2(aq)\)
In simple words: Copper cannot reduce iron(II), but it reduces iron(III) to iron(II) while dissolving.

Exam Tip: Compare standard reduction potentials to predict metal displacement feasibility.

Question 5. Define corrosion. Write chemical formula of rust.
Answer: Corrosion is the slow degradation and deterioration of a metal as a result of its electrochemical reaction with moisture, oxygen, and other gases present in the atmosphere, leading to the formation of oxides, carbonates, sulphides, or hydroxides.
Chemical formula of rust: \(\text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O}\)
In simple words: Corrosion is metal rusting or decaying via air and water attack.

Exam Tip: Include water of crystallization (\(x\text{H}_2\text{O}\)) in the formula for rust.

Question 6. Write short notes on reduction and oxidation potentials.
Answer:
- Oxidation Potential: The tendency of an electrode to lose electrons and undergo oxidation when dipped in its own ion solution.
- Reduction Potential: The tendency of an electrode to gain electrons and undergo reduction when dipped in its own ion solution. By IUPAC convention, standard reduction potential is taken as the standard electrode potential.
In simple words: Oxidation potential measures how easily a substance oxidizes, while reduction potential measures how easily it reduces.

Exam Tip: Note that reduction potential is the exact negative of oxidation potential for the same half-cell.

Question 7. How are standard electrode potentials measured?
Answer: Standard electrode potentials are measured by setting up a galvanic cell coupling the given electrode (at \(1\text{ M}\) concentration, \(298\text{ K}\), and \(1\text{ atm}\) pressure for gases) with a Standard Hydrogen Electrode (SHE), whose potential is arbitrarily fixed at zero volts. The EMF of this cell gives the standard electrode potential of the test half-cell.
In simple words: Connect the test electrode to a Standard Hydrogen Electrode and measure the total cell voltage.

Exam Tip: Mention SHE and standard conditions (\(1\text{ M}, 298\text{ K}, 1\text{ atm}\)).

Question 8. What is cell constant? How is it determined?
Answer: Cell constant is defined as the ratio of the distance between the electrodes (\(l\)) to the cross-sectional area of the electrodes (\(A\)):
\(G^* = \frac{l}{A}\)
It is determined experimentally by measuring the conductance of a standard solution (usually \(\text{KCl}\)) of known specific conductivity (\(\kappa\)) using the relation: \(G^* = \frac{\kappa}{G}\).
In simple words: Cell constant characterizes the geometry of a conductivity cell and is found using standard KCl solution.

Exam Tip: State both formulas (\(l/A\) and \(\kappa/G\)) for complete clarity.

Question 9. What is conductivity water?
Answer: Conductivity water is extremely pure water that has a very low electrical conductivity, prepared by repeated distillation of ordinary distilled water with a small amount of potassium permanganate and sodium hydroxide to remove volatile impurities.
In simple words: Ultra-pure water used as a solvent in conductivity measurements so impurities do not affect readings.

Question 10. Why is it necessary to platinize the electrodes of a conductivity cell before it is used for conductance measurement?
Answer: Platinization (coating with finely divided platinum black) increases the effective surface area of the electrodes enormously, which minimizes polarization effects and increases the sharpness and accuracy of the conductance measurements.
In simple words: Platinizing electrodes increases their surface area to give precise electrical readings.

Exam Tip: Mention "platinum black" and "increased surface area" as key terms.

Question 11. Why does a mercury cell give a constant voltage?
Answer: A mercury cell gives a constant voltage throughout its life because the overall cell reaction does not involve any ions in solution whose concentration can change during its operation.
In simple words: No ions change concentration during discharge, keeping the voltage steady.

Exam Tip: Emphasize that ion concentrations do not alter overall in solution.

Question 12. What is a fuel cell? Write the reaction involved in \(\text{H}_2-\text{O}_2\) fuel cell.
Answer: A fuel cell is a galvanic cell designed to convert the chemical energy of combustion fuels (like hydrogen, methane, methanol) directly into electrical energy continuously.
Reactions in \(\text{H}_2-\text{O}_2\) fuel cell:
Anode: \(2\text{H}_2(g) + 4\text{OH}^-(aq) \rightarrow 4\text{H}_2\text{O}(l) + 4\text{e}^-\)
Cathode: \(\text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4\text{e}^- \rightarrow 4\text{OH}^-(aq)\)
Overall: \(2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l)\)
In simple words: Fuel cells continuously turn fuel and oxygen into electricity and water.

Exam Tip: Clearly write anode and cathode half-reactions along with the overall equation.

QUESTIONS CARRYING THREE MARKS

Question 1. Write any three differences between potential difference and e.m.f.
Answer:

In simple words: EMF is open-circuit voltage, while potential difference is closed-circuit voltage under load.

Exam Tip: Tabulate differences clearly for full marks.

Question 2. Why does an electrochemical cell stop working after sometime?
Answer: As the electrochemical cell works, the concentration of reactants in the anodic compartment increases while the concentration of reactant metal ions in the cathodic compartment decreases. According to Le Chatelier's principle, this change shifts the equilibrium, causing the electrode potentials to change. Specifically, the cathode potential decreases and the anode potential increases, making the cell EMF gradually drop until \(E_{\text{cell}} = 0\), at which point the cell stops working.
In simple words: As reactants get used up, cell voltage drops to zero and the cell dies.

Exam Tip: Reference Le Chatelier's principle and Nernst equation concentration dependence.

Question 3. For the standard cell \(\text{Cu}(s) / \text{Cu}^{2+}(aq) \parallel \text{Ag}^+(aq) / \text{Ag}(s)\), given \(E^{\circ}_{\text{Cu}^{2+}/\text{Cu}} = +0.34\text{ V}\) and \(E^{\circ}_{\text{Ag}^+/\text{Ag}} = +0.80\text{ V}\):
i. Identify the cathode and the anode as the current is drawn from the cell.
ii. Write the reaction taking place at the electrodes.
iii. Calculate the standard cell potential.
Answer:
i. Anode: Copper electrode (\(\text{Cu}/\text{Cu}^{2+}\))
Cathode: Silver electrode (\(\text{Ag}^+/\text{Ag}\))
ii. Electrode Reactions:
At Anode (Oxidation): \(\text{Cu}(s) \rightarrow \text{Cu}^{2+}(aq) + 2\text{e}^-\)
At Cathode (Reduction): \(2\text{Ag}^+(aq) + 2\text{e}^- \rightarrow 2\text{Ag}(s)\)
iii. Standard Cell Potential:
\(E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\)
\(E^{\circ}_{\text{cell}} = E^{\circ}_{\text{Ag}^+/\text{Ag}} - E^{\circ}_{\text{Cu}^{2+}/\text{Cu}}\)
\(E^{\circ}_{\text{cell}} = +0.80\text{ V} - (+0.34\text{ V}) = +0.46\text{ V}\)
In simple words: Silver acts as cathode and copper as anode; subtract anode potential from cathode potential to get cell EMF.

Exam Tip: Always write balanced half-cell equations with correct electron stoichiometry.

Question 4. Can we store copper sulphate in (i) Zinc vessel (ii) Silver vessel? Give reasons. Given \(E^{\circ}_{\text{Cu}^{2+}/\text{Cu}} = +0.34\text{ V}\), \(E^{\circ}_{\text{Zn}^{2+}/\text{Zn}} = -0.76\text{ V}\), \(E^{\circ}_{\text{Ag}^+/\text{Ag}} = +0.80\text{ V}\).
Answer: A metal with a lower standard reduction potential can displace a metal with a higher reduction potential from its salt solution.
i. Zinc vessel: Since the standard reduction potential of zinc (\(-0.76\text{ V}\)) is much lower than that of copper (\(+0.34\text{ V}\)), zinc will displace copper from \(\text{CuSO}_4\) solution and react with it. Therefore, \(\text{CuSO}_4\) cannot be stored in a zinc vessel.
ii. Silver vessel: Since the standard reduction potential of silver (\(+0.80\text{ V}\)) is higher than that of copper (\(+0.34\text{ V}\)), silver cannot displace copper and will not react with \(\text{CuSO}_4\). Therefore, \(\text{CuSO}_4\) can be stored in a silver vessel.
In simple words: Zinc reacts with copper sulfate and destroys the vessel, but silver does not react.

Question 5. How many grams of chlorine can be produced by the electrolysis of molten NaCl with a current of \(1.02\text{ A}\) for \(15\text{ min}\)?
Answer:
Electrolysis reaction at anode:
\(2\text{Cl}^-(l) \rightarrow \text{Cl}_2(g) + 2\text{e}^-\)
\(2\text{ moles}\) of electrons (\(2F\)) produce \(1\text{ mole}\) of \(\text{Cl}_2\).
Total charge passed (\(Q\)) = \(I \times t\)
\(I = 1.02\text{ A}\)
\(t = 15 \times 60 = 900\text{ s}\)
\(Q = 1.02 \times 900 = 918\text{ C}\)
Since \(2 \times 96500\text{ C}\) produce \(1\text{ mole of }\text{Cl}_2\) (\(71\text{ g}\)):
Mass of chlorine produced = \(\frac{71 \times 918}{2 \times 96500} = \frac{65178}{193000} = 0.3377\text{ g}\)
In simple words: Use Faraday's laws (\(m = \frac{M \cdot I \cdot t}{n \cdot F}\)) to calculate mass deposited.

Exam Tip: Always convert minutes into seconds before calculating charge (\(Q = It\)).

Question 6. What is understood by a normal hydrogen electrode? Give its significance.
Answer: A Standard Hydrogen Electrode (SHE) consists of a platinum electrode coated with platinum black dipped in an acidic solution of \(1\text{ M}\) \(\text{H}^+\) ion concentration, around which pure hydrogen gas at \(1\text{ atm}\) pressure is bubbled at \(298\text{ K}\).
Significance: Its potential is arbitrarily fixed at \(0.00\text{ V}\), serving as a universal reference standard to measure the electrode potentials of all other half-cells.
In simple words: SHE is the universal zero-volt reference electrode used to measure standard potentials.

Question 7. Define electrode potential. Why absolute value of reduction potential of an electrode cannot be determined?
Answer: Electrode potential is the measure of the tendency of an electrode to lose or gain electrons when it is in contact with a solution of its own ions.
Absolute value of a single electrode potential cannot be determined experimentally because a half-cell reaction cannot occur independently; oxidation and reduction must take place simultaneously in a complete cell circuit.
In simple words: You cannot measure a single electrode's voltage alone because a complete circuit with two electrodes is required for charge flow.

Exam Tip: Emphasize the requirement of a closed circuit involving both oxidation and reduction.

Question 8. Write the equation showing the effect of concentration on the electrode potential.
Answer: For a general electrode reaction \(\text{M}^{n+}(aq) + n\text{e}^- \rightarrow \text{M}(s)\), the Nernst equation is:
\(E_{\text{M}^{n+}/\text{M}} = E^{\circ}_{\text{M}^{n+}/\text{M}} - \frac{RT}{nF} \ln \frac{1}{[\text{M}^{n+}]}\)
or in base-10 log form at \(298\text{ K}\):
\(E_{\text{M}^{n+}/\text{M}} = E^{\circ}_{\text{M}^{n+}/\text{M}} - \frac{0.0591}{n} \log \frac{1}{[\text{M}^{n+}]}\)
In simple words: Nernst equation relates electrode potential directly to ion concentration.

Exam Tip: Remember to substitute constant values at \(298\text{ K}\) for numerical applications.

Question 9. Derive the relationship between Gibb's free energy change and the cell potential.
Answer: The maximum electrical work obtained from a galvanic cell is equal to the decrease in Gibbs free energy (\(-\Delta G\)).
Electrical work = Total charge passed \(\times\) EMF of the cell
\(-\Delta G = nFE_{\text{cell}}\)
Under standard conditions:
\(\Delta^{\circ} G = -nFE^{\circ}_{\text{cell}}\)
In simple words: Free energy change equals negative total electrical work done by the cell.

Question 10. How can Nernst equation be applied in the calculation of equilibrium constant of any cell reaction?
Answer: At chemical equilibrium, the cell potential (\(E_{\text{cell}}\)) becomes zero, and the reaction quotient (\(Q\) or concentration ratio) equals the equilibrium constant (\(K_c\)).
From Nernst equation:
\(E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{RT}{nF} \ln K_c = 0\)
\(E^{\circ}_{\text{cell}} = \frac{2.303RT}{nF} \log K_c\)
At \(298\text{ K}\):
\(E^{\circ}_{\text{cell}} = \frac{0.0591}{n} \log K_c\)
In simple words: Set \(E_{\text{cell}} = 0\) in the Nernst equation to find the equilibrium constant \(K_c\).

Exam Tip: This formula is extremely important for board exam numerical problems.

Question 11. The cell reaction as written is spontaneous if the overall EMF of the cell is positive. Comment on this statement.
Answer: The statement is correct. According to thermodynamics, the Gibbs free energy change is given by \(\Delta G = -nFE_{\text{cell}}\). For a reaction to be spontaneous, \(\Delta G\) must be negative. Since \(n, F\) are positive constants, \(\Delta G\) will be negative only when \(E_{\text{cell}}\) is positive.
In simple words: Positive cell EMF means negative free energy change, making the reaction spontaneous.

Question 1. Explain the term electrolysis. Discuss briefly the electrolysis of (i) molten NaCl (ii) aqueous sodium chloride solution (iii) molten lead bromide (iv) water.
Answer: Electrolysis: The process of decomposition of an electrolyte solution or molten electrolyte by passing direct electric current through it, converting electrical energy into chemical energy.
i. Molten NaCl: Sodium ions migrate to cathode and gain electrons to form liquid sodium metal (\(\text{Na}^+ + \text{e}^- \rightarrow \text{Na}\)), while chloride ions migrate to anode and lose electrons to form chlorine gas (\(2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-\)).
ii. Aqueous NaCl solution: At cathode, reduction of water occurs in preference to sodium ions, releasing hydrogen gas (\(2\text{H}_2\text{O} + 2\text{e}^- \rightarrow \text{H}_2 + 2\text{OH}^-\)). At anode, chloride ions are oxidized to chlorine gas (\(\text{Cl}^-\)). NaOH is left in solution.
iii. Molten Lead Bromide (\(\text{PbBr}_2\)): Lead is deposited at the cathode (\(\text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb}\)) and reddish-brown bromine vapors are liberated at the anode (\(2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-\)).
iv. Water (acidified): Hydrogen gas is liberated at the cathode and oxygen gas is liberated at the anode due to oxidation of water molecules.
In simple words: Electrolysis uses electricity to break down compounds into their elements at electrodes.

Exam Tip: Clearly write anodic and cathodic half-reactions for each case.

Question 2. State and explain Faraday's laws of electrolysis. What is Electrochemical equivalent?
Answer:
- First Law: The amount of any substance (\(m\)) deposited or liberated at any electrode is directly proportional to the quantity of electricity (\(Q\) or \(It\)) passed through the electrolyte: \(m \propto Q\) or \(m = Z \cdot I \cdot t\).
- Second Law: When the same quantity of electricity is passed through different electrolytic solutions connected in series, the masses of substances liberated at the electrodes are directly proportional to their chemical equivalent weights: \(\frac{m_1}{m_2} = \frac{E_1}{E_2}\).
- Electrochemical Equivalent (ECE, \(Z\)): It is defined as the mass of a substance deposited when a current of \(1\text{ ampere}\) is passed for \(1\text{ second}\) (i.e., a charge of \(1\text{ Coulomb}\)): \(Z = \frac{m}{It}\).
In simple words: Faraday's laws relate deposited mass to electrical charge passed.

Exam Tip: State both laws mathematically and define ECE precisely.

Question 3. What do you understand by 'electrolytic conduction'? What are the factors on which electrolyte conduction depends? What is the effect of temperature on electrolytic conduction?
Answer: Electrolytic conduction: The flow of electric current through an electrolytic solution due to the movement of ions.
Factors affecting electrolytic conduction:
1. Nature of electrolyte (strong or weak).
2. Concentration of the solution.
3. Temperature.
4. Nature and viscosity of the solvent.
Effect of temperature: Electrolytic conduction increases with an increase in temperature because viscosity decreases and ionic mobility increases.
In simple words: Electrolytic conduction is current carried by moving ions, and it increases with temperature.

Exam Tip: Contrast electrolytic conduction with metallic conduction where resistance increases with temperature.

Question 4. How is electrolytic conductance measured experimentally?
Answer: The resistance of an electrolytic solution is measured using a Wheatstone bridge circuit. An alternating current (AC) source is used to prevent polarization and composition changes at the electrodes. The conductivity cell is filled with the solution, and variable resistance is adjusted until null point is obtained on the detector. The solution resistance (\(R\)) is calculated, from which conductance (\(G = 1/R\)), specific conductance (\(\kappa = G \times G^*\)), and molar conductivity are determined.
In simple words: Use a Wheatstone bridge with an AC source to measure solution resistance without chemical polarization.

Question 5. Describe normal hydrogen electrode and its applications.
Answer: Normal Hydrogen Electrode (NHE / SHE): Comprises a platinum wire sealed in a glass tube with a platinum foil coated with platinum black at the end. It is immersed in a \(1\text{ M}\) acidic solution (\(\text{H}^+\)) and pure hydrogen gas at \(1\text{ atm}\) and \(298\text{ K}\) is bubbled around it. Its potential is set to zero.
Applications:
1. Determination of standard electrode potentials of other half-cells.
2. Determination of pH of solutions.
3. Determination of dissociation constants of weak acids.
In simple words: SHE is the zero-volt reference electrode used to measure unknown electrode potentials and pH.

Exam Tip: List at least two key electrochemical applications.

HOT QUESTIONS (1 MARK)

Question 1. Why in a concentrated solution, a strong electrolyte shows deviations from Debye-Huckel-Onsager equation?
Answer: Because interionic forces of attraction and repulsion become very large in concentrated solutions, restricting free ionic mobility.
In simple words: Crowded ions attract each other strongly at high concentrations, causing deviations.

Exam Tip: Mention strong interionic forces at high concentrations.

Question 2. What is the use of Platinum foil in the hydrogen electrode?
Answer: It is used for the inflow and outflow of electrons and provides an inert catalytic surface for the redox equilibrium of hydrogen gas and protons.
In simple words: Platinum conducts electrons and acts as an inert surface for the hydrogen reaction.

Question 3. Corrosion of motor cars is a greater problem in winter when salts are spread on roads to melt ice and snow. Why?
Answer: Salt increases the ionic conductivity of moisture on the car body, accelerating the electrochemical corrosion process.
In simple words: Road salt makes water more conductive, speeding up rusting.

Exam Tip: Connect road salt with enhanced electrolytic conductivity in moisture films.

Question 4. Is it safe to stir \(\text{AgNO}_3\) solution with a copper spoon? Given \(E^{\circ}_{\text{Ag}^+/\text{Ag}} = 0.80\text{ V}\), \(E^{\circ}_{\text{Cu}^{2+}/\text{Cu}} = 0.34\text{ V}\).
Answer: No, it is not safe because copper will react with \(\text{AgNO}_3\) solution (cell EMF will be positive, meaning spontaneous displacement of silver by copper).
In simple words: Copper displaces silver from solution, so the copper spoon will get damaged.

Exam Tip: Verify that \(E^{\circ}_{\text{cell}} > 0\) for the displacement reaction.

Question 5. Why is it necessary to use a salt bridge in a galvanic cell?
Answer: To complete the inner electrical circuit and maintain electrical neutrality in the two half-cell solutions.
In simple words: Salt bridge completes the circuit and stops charge buildup in half-cells.

Exam Tip: Mention both completing the circuit and maintaining electrical neutrality.

HOT QUESTIONS (2 MARKS)

Question 1. Why is Li the best reducing agent while Fluorine is the best oxidizing agent?
Answer: Lithium has the most negative standard reduction potential (\(-3.05\text{ V}\)), meaning it has the highest tendency to lose electrons and get oxidized (best reducing agent). Fluorine has the most positive standard reduction potential (\(+2.87\text{ V}\)), meaning it has the highest tendency to accept electrons and get reduced (best oxidizing agent).
In simple words: Lithium loses electrons most easily, while fluorine accepts electrons most easily.

Exam Tip: Refer to standard reduction potential values to explain oxidizing and reducing strengths.

Question 2. Equilibrium constant is related to \(E^{\circ}_{\text{cell}}\) but not to \(E_{\text{cell}}\). Explain.
Answer: Equilibrium constant is a unique characteristic constant of a reaction at a given temperature. It is related to standard cell potential \(E^{\circ}_{\text{cell}}\) because at equilibrium, \(E_{\text{cell}} = 0\) and reactant concentrations are at standard or equilibrium states. In contrast, \(E_{\text{cell}}\) changes continuously as the cell operates and becomes zero at equilibrium, so it cannot define a fixed constant.
In simple words: Cell potential changes as the reaction proceeds, but standard potential and equilibrium constant are fixed.

Exam Tip: State that \(E_{\text{cell}}\) becomes zero at equilibrium, making it unusable for calculating \(K\).

Question 3. Why is sodium metal not obtained at cathode when aqueous NaCl is electrolysed with Pt electrodes, but obtained when molten NaCl is electrolysed?
Answer: In aqueous NaCl, water molecules have a much higher reduction potential (\(-0.83\text{ V}\)) than sodium ions (\(-2.71\text{ V}\)), so water is reduced to hydrogen gas at the cathode instead of \(\text{Na}^+\). In molten NaCl, water is absent, so \(\text{Na}^+\) ions are directly reduced to sodium metal at the cathode.
In simple words: Water reduces more easily than sodium ions in water, but molten salt has no water so sodium deposits.

Question 4. A Zn rod weighing \(25\text{ g}\) was kept in \(100\text{ mL}\) of \(1\text{ M}\) copper sulphate solution. After certain time interval, the molarity of \(\text{Cu}^{2+}\) was found to be \(0.8\text{ M}\). What is the molarity of \(\text{SO}_4^{2-}\) in the resulting solution and what should be the mass of Zn rod after cleaning and drying?
Answer:
- Molarity of \(\text{SO}_4^{2-}\) ions remains constant at \(1\text{ M}\) because sulfate ions are spectator ions and do not participate in the redox reaction.
- Initial moles of \(\text{Cu}^{2+} = \frac{1 \times 100}{1000} = 0.1\text{ mol}\)
- Final moles of \(\text{Cu}^{2+} = \frac{0.8 \times 100}{1000} = 0.08\text{ mol}\)
- Moles of \(\text{Cu}^{2+}\) reacted = \(0.1 - 0.08 = 0.02\text{ mol}\)
Since 1 mole of Zn reacts with 1 mole of \(\text{Cu}^{2+}\), moles of Zn consumed = \(0.02\text{ mol}\)
Mass of Zn consumed = \(0.02\text{ mol} \times 65.38\text{ g/mol} = 1.308\text{ g}\)
Remaining mass of Zn rod = \(25\text{ g} - 1.308\text{ g} = 23.692\text{ g}\)
In simple words: Sulfate ions don't change concentration; calculate zinc reacted using moles of copper consumed.

Exam Tip: Remember spectator ions like sulfate do not change in concentration during metal displacement.

Question 5. Which will have greater molar conductivity and why? Sol A: 1 mol KCl dissolved in \(200\text{ cc}\) of solution or Sol B: 1 mol KCl dissolved in \(500\text{ cc}\) of solution.
Answer: Solution B will have greater molar conductivity. Molar conductivity increases with dilution. Since Solution B is more dilute (\(500\text{ cc}\) contains 1 mol compared to \(200\text{ cc}\)), ionic dissociation is greater and interionic interactions are weaker, resulting in higher molar conductivity.
In simple words: More dilute solutions have higher molar conductivity due to better dissociation of ions.

Exam Tip: Link higher molar conductivity directly with greater dilution (\(\Lambda_m = \kappa \times V\)).

HOT QUESTIONS (3 / 5 MARKS)

Question 1. What do you mean by (i) negative standard electrode potential and (ii) positive standard electrode potential?
Answer:
i. Negative Standard Electrode Potential: Means that the reduction of the given electrode is non-spontaneous compared to SHE, and the metal has a greater tendency to lose electrons (get oxidized) than hydrogen.
ii. Positive Standard Electrode Potential: Means that the reduction of the given electrode is spontaneous compared to SHE, and the species has a greater tendency to gain electrons (get reduced) than hydrogen.
In simple words: Negative potential means it oxidizes easier than hydrogen; positive means it reduces easier than hydrogen.

Question 2. Which cell is generally used in hearing aids? Name the material of the anode, cathode, and the electrolyte. Write the reactions involved.
Answer: Zinc-mercury button cell (Mercury cell) is generally used in hearing aids.
- Anode: Zinc-mercury amalgam (\(\text{Zn(Hg)}\))
- Cathode: Paste of mercuric oxide (\(\text{HgO}\)) and carbon
- Electrolyte: Moist paste of potassium hydroxide (\(\text{KOH}\)) and zinc oxide (\(\text{ZnO}\))
Reactions:
Anode: \(\text{Zn(Hg)} + 2\text{OH}^- \rightarrow \text{ZnO}(s) + \text{H}_2\text{O} + 2\text{e}^-\)
Cathode: \(\text{HgO}(s) + \text{H}_2\text{O} + 2\text{e}^- \rightarrow \text{Hg}(l) + 2\text{OH}^-\)
Overall: \(\text{Zn(Hg)} + \text{HgO}(s) \rightarrow \text{ZnO}(s) + \text{Hg}(l)\)
In simple words: Hearing aids use mercury button cells with zinc anode and HgO cathode.

Exam Tip: Clearly state anode, cathode, electrolyte, and half-cell reactions.

Question 3. Iron does not rust even if Zinc coating is broken in a galvanized iron pipe, but rusting occurs much faster if tin coating over iron is broken. Explain.
Answer: Zinc has a lower standard reduction potential (\(-0.76\text{ V}\)) than iron (\(-0.44\text{ V}\)), so zinc acts as the anode and sacrifices itself to protect iron cathodically (sacrificial protection). Conversely, tin has a higher standard reduction potential than iron, so when the tin coating is broken, iron acts as the anode and rusts much faster due to galvanic action.
In simple words: Zinc protects iron by corroding first, but tin accelerates iron rusting when scratched.

Exam Tip: Use reduction potentials to explain sacrificial protection by zinc versus galvanic corrosion by tin.

Question 4. "Corrosion is an electrochemical phenomenon", explain.
Answer: Corrosion involves the formation of miniature electrochemical cells on the metal surface. At anodic regions, the metal undergoes oxidation (losing electrons to form metal ions). At cathodic regions, oxygen in the presence of moisture and hydrogen ions gains electrons and undergoes reduction. The electrons flow through the metal from anode to cathode, while ions flow through the moisture layer acting as an electrolyte, producing rust/oxides.
In simple words: Rusting happens because tiny electrochemical galvanic cells form on the metal surface.

Exam Tip: Mention anodic oxidation of metal and cathodic reduction of oxygen.

Question 5. Calculate the pH of the following cell: \(\text{Pt, H}_2(1\text{ atm}) / \text{H}_2\text{SO}_4\), if its electrode potential is \(0.03\text{ V}\).
Answer:
For hydrogen electrode: \(\text{H}^+(aq) + \text{e}^- \rightarrow \frac{1}{2}\text{H}_2(g)\)
Nernst equation: \(E_{\text{H}^+/\text{H}_2} = E^{\circ}_{\text{H}^+/\text{H}_2} - \frac{0.0591}{1} \log \frac{1}{[\text{H}^+]}\)
Given \(E_{\text{H}^+/\text{H}_2} = -0.03\text{ V}\) (oxidation potential is \(+0.03\text{ V}\), so reduction potential is \(-0.03\text{ V}\)), \(E^{\circ} = 0\):
\(-0.03 = 0 - 0.0591(-\log[\text{H}^+])\)
\(-0.03 = -0.0591(\text{pH})\)
\(\text{pH} = \frac{0.03}{0.0591} \approx 0.508\)
In simple words: Substitute electrode potential into the Nernst equation to find hydrogen ion concentration and pH.

Question 6. A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of \(10^{-5}\text{ M}\) \(\text{H}^+\) ions. The emf of the cell is \(0.118\text{ V}\) at \(298\text{ K}\). Calculate the concentration of the \(\text{H}^+\) ions at the positive electrode.
Answer:
Cell reaction: \(\text{H}_2(\text{anode}) + 2\text{H}^+(\text{cathode}) \rightarrow 2\text{H}^+(\text{anode}) + \text{H}_2(\text{cathode})\)
Applying Nernst equation at \(298\text{ K}\):
\(E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{2} \log \frac{[\text{H}^+]_{\text{anode}}^2}{[\text{H}^+]_{\text{cathode}}^2}\)
Since \(E^{\circ}_{\text{cell}} = 0\):
\(0.118 = -0.0591 \log \frac{10^{-5}}{[\text{H}^+]_{\text{cathode}}}\)
\(\frac{0.118}{-0.0591} = \log \left(\frac{10^{-5}}{[\text{H}^+]_{\text{cathode}}}\right)\)
\(-2 = \log \left(\frac{10^{-5}}{[\text{H}^+]_{\text{cathode}}}\right)\)
\(10^{-2} = \frac{10^{-5}}{[\text{H}^+]_{\text{cathode}}}\)
\([\text{H}^+]_{\text{cathode}} = \frac{10^{-5}}{10^{-2}} = 10^{-3}\text{ M}\)
In simple words: Use concentration cell Nernst equation to find unknown hydrogen ion concentration.

Question 7. Crude copper containing Fe and Ag as contaminations was subjected to electro-refining by using a current of \(175\text{ A}\) for \(6.434\text{ min}\). The mass of anode was found to decrease by \(22.260\text{ g}\), while that of cathode was increased by \(22.011\text{ g}\). Estimate the percentage of copper, iron, and silver in crude copper.
Answer:
Total charge passed (\(Q\)) = \(I \times t = 175\text{ A} \times 6.434 \times 60\text{ s} = 67557\text{ C}\)
Mass of copper deposited at cathode according to Faraday's law:
\(m_{\text{Cu}} = \frac{M \cdot Q}{nF} = \frac{63.5 \times 67557}{2 \times 96500} = 22.22\text{ g}\)
Assuming total anode loss minus copper dissolved gives iron dissolved, and silver settles as anode mud.
Calculated percentages based on total mass dissolved: Cu % \(\approx 98.9\%\), Fe % \(\approx 1.1\%\), Ag % in mud.
In simple words: Use Faraday's laws to find copper mass, then calculate percentages of impurities.

Exam Tip: Silver forms anode mud as it is less reactive than copper.

Question 8. Zinc electrode is constituted at \(298\text{ K}\) by placing Zn rod in \(0.1\text{ M}\) aq solution of zinc sulphate which is \(95\%\) dissociated at this concentration. What will be the electrode potential of the electrode given that \(E^{\circ}_{\text{Zn}^{2+}/\text{Zn}} = -0.76\text{ V}\)?
Answer:
Concentration of \(\text{Zn}^{2+}\) ions = \(c \times \alpha = 0.1 \times \frac{95}{100} = 0.095\text{ M}\)
Applying Nernst equation:
\(E_{\text{Zn}^{2+}/\text{Zn}} = E^{\circ}_{\text{Zn}^{2+}/\text{Zn}} - \frac{0.0591}{2} \log \frac{1}{[\text{Zn}^{2+}]}\)
\(E_{\text{Zn}^{2+}/\text{Zn}} = -0.76 - \frac{0.0591}{2} \log \left(\frac{1}{0.095}\right)\)
\(E_{\text{Zn}^{2+}/\text{Zn}} = -0.76 - 0.02955 \times \log(10.526)\)
\(E_{\text{Zn}^{2+}/\text{Zn}} = -0.76 - 0.02955 \times 1.022 = -0.76 - 0.0302 = -0.7902\text{ V}\)
In simple words: Calculate actual zinc ion concentration using percentage dissociation, then plug into Nernst equation.

Exam Tip: Always multiply molarity by the degree of dissociation \(\alpha\) to get active ion concentration.

Question 9. At what pH will hydrogen electrode at \(298\text{ K}\) show an electrode potential of \(-0.118\text{ V}\), when hydrogen gas is bubbled at \(1\text{ atm}\) pressure?
Answer:
Nernst equation for hydrogen electrode:
\(E_{\text{H}^+/\text{H}_2} = 0 - \frac{0.0591}{1} \log \frac{1}{[\text{H}^+]}\)
Given \(E_{\text{H}^+/\text{H}_2} = -0.118\text{ V}\):
\(-0.118 = -0.0591(\text{pH})\)
\(\text{pH} = \frac{-0.118}{-0.0591} \approx 2.0\)
In simple words: Divide the given electrode potential by \(0.0591\) to find the pH.

Question 10. Electrolysis of the solution of \(\text{MnSO}_4\) in aq sulphuric acid is a method for the preparation of \(\text{MnO}_2\) as per the chemical reaction:
\(\text{Mn}^{2+} + 2\text{H}_2\text{O} \rightarrow \text{MnO}_2 + 2\text{H}^+ + \text{H}_2\)
Passing a current of \(27\text{ A}\) for \(24\text{ hrs}\) gives \(1\text{ kg}\) of \(\text{MnO}_2\). What is the current efficiency? What are the reactions occurring at anode and cathode?
Answer:
- Total charge (\(Q\)) = \(27\text{ A} \times 24 \times 3600\text{ s} = 2,332,800\text{ C}\)
Molar mass of \(\text{MnO}_2 = 54.94 + 32 = 86.94\text{ g/mol}\)
For the reaction, 2 moles of electrons (\(2F\) or \(2 \times 96500\text{ C}\)) produce 1 mole (\(86.94\text{ g}\)) of \(\text{MnO}_2\).
Theoretical mass of \(\text{MnO}_2\) expected = \(\frac{86.94 \times 2332800}{2 \times 96500} = 1050.5\text{ g} = 1.0505\text{ kg}\)
Actual mass produced = \(1\text{ kg} = 1000\text{ g}\)
Current efficiency = \(\frac{\text{Actual mass}}{\text{Theoretical mass}} \times 100 = \frac{1000}{1050.5} \times 100 = 95.19\%\)
Reactions:
Anode: \(\text{Mn}^{2+} + 2\text{H}_2\text{O} \rightarrow \text{MnO}_2 + 4\text{H}^+ + 2\text{e}^-\)
Cathode: \(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2(g)\)
In simple words: Compare actual yield with theoretical Faraday yield to find current efficiency.

Exam Tip: Current efficiency is always calculated as \((\text{Actual Yield} / \text{Theoretical Yield}) \times 100\).

Additional Electrochemistry Problems

Question 1. What do you mean by Kohlrausch's law? From the following molar conductivities at infinite dilution:
\(\Lambda_m^{\infty} \text{ Ba(OH)}_2 = 457.6\text{ }\Omega^{-1}\text{ cm}^2\text{ mol}^{-1}\)
\(\Lambda_m^{\infty} \text{ BaCl}_2 = 240.6\text{ }\Omega^{-1}\text{ cm}^2\text{ mol}^{-1}\)
\(\Lambda_m^{\infty} \text{ NH}_4\text{Cl} = 129.8\text{ }\Omega^{-1}\text{ cm}^2\text{ mol}^{-1}\)
Calculate \(\Lambda_m^{\infty}\) for \(\text{NH}_4\text{OH}\).
Answer:
According to Kohlrausch's law, limiting molar conductivity of an electrolyte is the sum of ionic conductivities of its constituent cations and anions.
We need \(\Lambda_m^{\infty}(\text{NH}_4\text{OH}) = \lambda^{\infty}_{\text{NH}_4^+} + \lambda^{\infty}_{\text{OH}^-}\)
Using given values:
\(\Lambda_m^{\infty}(\text{Ba(OH)}_2) = \lambda^{\infty}_{\text{Ba}^{2+}} + 2\lambda^{\infty}_{\text{OH}^-} = 457.6\)-(1)
\(\Lambda_m^{\infty}(\text{BaCl}_2) = \lambda^{\infty}_{\text{Ba}^{2+}} + 2\lambda^{\infty}_{\text{Cl}^-} = 240.6\)-(2)
\(\Lambda_m^{\infty}(\text{NH}_4\text{Cl}) = \lambda^{\infty}_{\text{NH}_4^+} + \lambda^{\infty}_{\text{Cl}^-} = 129.8\)-(3)
Calculation: \(\frac{1}{2}(\text{Eq 1}) + \text{Eq 3} - \frac{1}{2}(\text{Eq 2})\):
\(\Lambda_m^{\infty}(\text{NH}_4\text{OH}) = \frac{457.6}{2} + 129.8 - \frac{240.6}{2}\)
\(= 228.8 + 129.8 - 120.3 = 238.3\text{ }\Omega^{-1}\text{ cm}^2\text{ mol}^{-1}\)
In simple words: Add and subtract limiting molar conductivities of strong electrolytes to get \(\Lambda_m^{\infty}\) for \(\text{NH}_4\text{OH}\).

Exam Tip: Set up simultaneous ionic equations correctly to find target molar conductivity.

Question 2. Calculate the equilibrium constant for the reaction:
\(\text{Zn} + \text{Cd}^{2+} \rightleftharpoons \text{Zn}^{2+} + \text{Cd}\)
Given \(E^{\circ}_{\text{Cd}^{2+}/\text{Cd}} = -0.403\text{ V}\), \(E^{\circ}_{\text{Zn}^{2+}/\text{Zn}} = -0.763\text{ V}\)
Antilog \(12.1827 = 1.52 \times 10^{12}\)
Answer:
\(E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = -0.403\text{ V} - (-0.763\text{ V}) = +0.360\text{ V}\)
Number of electrons transferred (\(n\)) = 2
Using relation between \(E^{\circ}_{\text{cell}}\) and \(K_c\):
\(\log K_c = \frac{n \cdot E^{\circ}_{\text{cell}}}{0.0591} = \frac{2 \times 0.360}{0.0591} = \frac{0.720}{0.0591} = 12.1827\)
\(K_c = \text{antilog}(12.1827) = 1.52 \times 10^{12}\)
In simple words: Calculate standard cell potential and substitute into the Nernst equilibrium formula to find \(K_c\).

Exam Tip: Use \(\log K_c = \frac{nE^{\circ}}{0.0591}\) for quick calculations at \(298\text{ K}\).

Question 3. Predict the products of electrolysis of the following:
(a) a dil. solution of \(\text{H}_2\text{SO}_4\) with Pt electrodes.
(b) an aqueous solution of \(\text{AgNO}_3\) with silver electrode.
Answer:
(a) Dilute \(\text{H}_2\text{SO}_4\) with Pt electrodes: Hydrogen gas (\(\text{H}_2\)) is liberated at the cathode, and oxygen gas (\(\text{O}_2\)) is liberated at the anode due to the electrolysis of water.
(b) Aqueous \(\text{AgNO}_3\) with silver electrodes: Silver is deposited at the cathode (\(\text{Ag}^+ + \text{e}^- \rightarrow \text{Ag}\)), and the silver anode dissolves into the solution as silver ions (\(\text{Ag} \rightarrow \text{Ag}^+ + \text{e}^-\)).
In simple words: Dilute acid electrolyzes water into hydrogen and oxygen; silver nitrate with silver electrodes plates silver and dissolves the anode.

Exam Tip: Active metal anodes like silver dissolve during electrolysis rather than liberating oxygen.