CBSE Class 12 Chemistry Electrochemistry Worksheet Set 02

Important Questions for NCERT Class 12 Chemistry Electrochemistry 

Question. Given that the standard reduction potentials for M+/M and N+/N electrodes at 298 K are 0.52 V and 0.25 V respectively. Which of the following is correct in respect of the following electrochemical cell ? M/M⁺ | | N⁺/N 
(a) The overall cell reaction is a spontaneous reaction.
(b) The standard EMF of the cell is – 0.27 V.
(c) The standard EMF of the cell is 0.77 V.
(d) The standard EMF of the cell is – 0.77 V. 

Question. The specific conductance of a 0.1 N KCl solution at 23°C is 0.012 ohm–1 cm–1. The resistance of cell containing the solution at the same temperature was found to be 55 ohm. The cell constant will be
(a) 0.918 cm^–1
(b) 0.66 cm^–1
(c) 1.142 cm^–1
(d) 1.12 cm^–1  

Question. On heating one end of a piece of a metal, the other end becomes hot because of
(a) energised electrons moving to the other end
(b) minor perturbation in the energy of atoms
(c) resistance of the metal
(d) mobility of atoms in the metal. 

Question. On electrolysis of dil. sulphuric acid using platinum (Pt) electrode, the product obtained at anode will be
(a) hydrogen gas
(b) oxygen gas

(c) H₂S gas
(d) SO₂ gas.  

Commercial cells

Question. Match the following:

Answer:
The correct matched pairs are:

In simple words: This table matches different types of chemical batteries and cells with the specific conducting liquids or pastes they contain.

Exam Tip: Memorize the active electrode materials and electrolytes for all major commercial batteries, as this is a frequent topic in short-answer board exam questions.

Question. State True or False: Dry cell does not provide a constant voltage throughout life.

Answer: True. The dry cell does not maintain a steady potential because the overall cell reaction involves ions whose concentrations change during discharge, causing a gradual decline in voltage.
In simple words: True. A dry cell's power weakens slowly as it runs out of its active chemical compounds.

Exam Tip: State clearly in your explanations that the accumulation of ionic reaction products around the electrodes changes the internal cell potential over time.

Question. State True or False: A Zn/Hg Cell is superior to dry cell.

Answer: True. A zinc-mercury cell is superior to a dry cell because it delivers a highly stable operating potential of approximately \( 1.35\text{ V} \) throughout its entire useful life.
In simple words: True. A mercury cell is better because it gives a steady, unvarying voltage until the battery completely runs out.

Exam Tip: Note that mercury cells maintain a constant potential because the overall reaction does not involve any ions in solution whose concentration can change.

Question. State True or False: A Fuel Cell has about 70% chemical efficiency.

Answer: True. Fuel cells convert chemical energy directly into electrical energy with an outstanding efficiency of about 70%, which is significantly higher than conventional thermal power plants.
In simple words: True. Fuel cells are highly efficient, converting about 70% of the fuel's chemical energy directly into usable electrical power.

Exam Tip: Compare the efficiency of fuel cells (~70%) with conventional steam turbines (~40%) to gain extra points in descriptive questions.

Question. State True or False: The lead storage battery is an example of primary cell.

Answer: False. The lead storage battery is a secondary cell because the electrochemical cell reactions can be fully reversed by passing an external electric current, allowing it to be recharged.
In simple words: False. A lead storage battery is a secondary cell because it is rechargeable, unlike one-time-use primary cells.

Exam Tip: Always categorize batteries based on rechargeability: non-rechargeable are primary cells, while rechargeable are secondary cells.

VSQ Related with Dry Cell

Question. For the Leclanche cell, write the chemical reactions involved at cathode.

Answer: At the cathode, manganese dioxide is reduced in the presence of ammonium ions to manganese oxyhydroxide: \[ \text{MnO}_2 + \text{NH}_4^+ + \text{e}^- \rightarrow \text{Mn(OH)O} + \text{NH}_3 \]
In simple words: At the cathode, manganese dioxide gains an electron and hydrogen from ammonium to produce manganese oxyhydroxide and release ammonia gas.

Exam Tip: Pay special attention to the formulas: manganese is reduced to \( \text{Mn(OH)O} \), which can also be represented as \( \text{MnO(OH)} \).

Question. For the Leclanche cell, write the change in oxidation state of Mn.

Answer: During the cathode reduction reaction, the oxidation state of manganese decreases from \( +4 \) in \( \text{MnO}_2 \) to \( +3 \) in \( \text{Mn(OH)O} \).
In simple words: The oxidation state of manganese decreases from +4 to +3 during the cell's discharge.

Exam Tip: Show the calculation of the oxidation state for both compounds to prove your answer clearly.

Question. For the Leclanche cell, write the complex entity formed between Zn2+(aq) and NH3(g).

Answer: The zinc ions generated at the anode react with the ammonia gas released at the cathode to form a stable coordination complex: \[ [\text{Zn}(\text{NH}_3)_4]^{2+} \]
In simple words: Zinc ions bind with the released ammonia gas to form a complex ion, preventing pressure buildup from gas inside the cell.

Exam Tip: Write the coordination complex with its correct formula and a \( 2+ \) superscript charge to secure full marks.

Concept 2: Units of conductivity, molar conductivity etc.

Question. Match the following:

Answer:
The correct matched properties and units are:

In simple words: This table matches different physical terms in electrochemistry to their standard units. Conductance is measured in Siemens, while cell constant depends on length.

Exam Tip: Be comfortable with both SI units (such as \( \text{S m}^{-1} \)) and common units (such as \( \text{S cm}^{-1} \)) for conductivity calculations.

Concept 3: Products of electrolysis

Question. Predict the products of electrolysis for an aqueous solution of AgNO3 using Ag electrodes.

Answer: When \( \text{AgNO}_3 \) is electrolyzed using active silver electrodes:
- **At Cathode:** Silver ions have a higher reduction potential than water, so silver metal is deposited: \[ \text{Ag}^+(\text{aq}) + \text{e}^- \rightarrow \text{Ag}(\text{s}) \] - **At Anode:** Since the silver anode is active, it oxidizes preferentially over water or nitrate ions: \[ \text{Ag}(\text{s}) \rightarrow \text{Ag}^+(\text{aq}) + \text{e}^- \]
In simple words: Silver dissolves from the anode and gets deposited as pure metal on the cathode.

Exam Tip: When active electrodes are used, the anode material itself dissolves instead of water undergoing oxidation.

Question. Predict the products of electrolysis for an aqueous solution of CuSO4 using Cu electrodes.

Answer: When \( \text{CuSO}_4 \) is electrolyzed using active copper electrodes:
- **At Cathode:** Copper ions have a higher reduction potential than water and are reduced to deposit copper metal: \[ \text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s}) \] - **At Anode:** The active copper anode oxidized preferentially, dissolving copper into the solution: \[ \text{Cu}(\text{s}) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \]
In simple words: Copper metal dissolves from the positive electrode and deposits onto the negative electrode in its pure form.

Exam Tip: This active-electrode mechanism is the fundamental principle used in the industrial electrolytic refining of copper.

Question. Predict the products of electrolysis for an aqueous solution of AgNO3 using Pt electrodes.

Answer: When \( \text{AgNO}_3 \) is electrolyzed using inert platinum electrodes:
- **At Cathode:** Silver ions are reduced in preference to water, depositing silver metal: \[ \text{Ag}^+(\text{aq}) + \text{e}^- \rightarrow \text{Ag}(\text{s}) \] - **At Anode:** Since platinum is inert, water is oxidized in preference to nitrate ions, releasing oxygen gas: \[ 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{O}_2(\text{g}) + 4\text{H}^+(\text{aq}) + 4\text{e}^- \]
In simple words: Pure silver is deposited on the negative cathode, and bubbles of oxygen gas are released at the positive anode.

Exam Tip: Nitrate ions (\( \text{NO}_3^- \)) have a much higher oxidation potential than water, so oxygen gas is always liberated at an inert anode in nitrate solutions.

Question. Predict the products of electrolysis for an aqueous solution of NaCl using Pt electrodes.

Answer: When aqueous \( \text{NaCl} \) (brine) is electrolyzed using inert platinum electrodes:
- **At Cathode:** Water has a higher reduction potential than sodium ions, so water is reduced to liberate hydrogen gas: \[ 2\text{H}_2\text{O}(\text{l}) + 2\text{e}^- \rightarrow \text{H}_2(\text{g}) + 2\text{OH}^-(\text{aq}) \] - **At Anode:** Chloride ions are oxidized to chlorine gas in preference to water due to the overvoltage of oxygen: \[ 2\text{Cl}^-(\text{aq}) \rightarrow \text{Cl}_2(\text{g}) + 2\text{e}^- \] The remaining ions (\( \text{Na}^+ \) and \( \text{OH}^- \)) leave a solution of sodium hydroxide (\( \text{NaOH} \)).
In simple words: Hydrogen gas bubbles out at the negative electrode, chlorine gas bubbles out at the positive electrode, and the solution turns basic as sodium hydroxide is formed.

Exam Tip: Clearly explain that chloride ions oxidize preferentially over water due to the kinetic barrier of oxygen evolution, commonly known as overpotential.