CBSE Class 12 Chemistry Electrochemistry Important Questions Set 08

Very Short Answer Questions (1 Mark)

Question 1. Can we store copper sulphate in iron vessel ? Why ?
Answer:
No, because iron is more reactive than copper.
(Iron would displace copper: Fe + CuSO₄ → FeSO₄ + Cu.)

Teacher's Note:
1. E°(Fe²⁺/Fe) = −0.44 V is lower than E°(Cu²⁺/Cu) = +0.34 V, so the displacement reaction has positive E°_cell.

Question 2. Why does zinc react with dilute sulphuric acid but copper does not ? (Jharkhand Board 2011)
Answer:
Zinc lies above hydrogen whereas copper lies below hydrogen in the electrochemical series. Hence, zinc is more reactive than H⁺ whereas copper is less reactive. Therefore, zinc reacts with dilute H₂SO₄ but copper cannot liberate H₂ gas by reaction with dilute H₂SO₄.

Teacher's Note:
1. Metals with negative E° (reduction potential) displace H₂ from dilute acids; those with positive E° do not.

Question 3. Write Nernst equation to calculate the cell potential of Mg(s) | Mg²⁺(aq) || Ag⁺(aq) | Ag
Answer:
The cell reaction is Mg(s) + 2Ag⁺(aq) → Mg²⁺(aq) + 2Ag(s), n = 2
∴ Nernst equation is :
\( E_{cell} = E^\circ_{cell} - \frac{2.303\,RT}{2F}\log\frac{[Mg^{2+}]}{[Ag^+]^2} \)

Teacher's Note:
1. Balance the electrons first: two Ag⁺ ions are reduced for one Mg oxidised, so [Ag⁺] is squared and n = 2.

Question 4. Write Nernst equation for the reaction
(i) 2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd(s)
(ii) 2Cr + 3Fe²⁺ → 2Cr³⁺ + 3Fe
Answer:
(i) \( E_{cell} = E^\circ_{cell} - \frac{RT}{6F}\ln\frac{[Cr^{3+}]^2}{[Cd^{2+}]^3} \) (∵ n = 6)
(ii) \( E_{cell} = E^\circ_{cell} - \frac{RT}{6F}\ln\frac{[Cr^{3+}]^2}{[Fe^{2+}]^3} \) (∵ n = 6)

Teacher's Note:
1. 2 Cr atoms each lose 3 electrons, so 6 electrons are transferred; with log₁₀ the factor becomes 2.303RT/6F.

Question 5. Why a galvanic cell stops working after some time ?
Answer:
With time, concentrations of the electrolytic solutions change. Hence, their electrode potentials change. When the electrode potentials of the two half-cells become equal, the cell stops working.

Teacher's Note:
1. At this stage E_cell = 0 and the cell reaction has reached equilibrium (ΔG = 0).

Question 6. What is a concentration cell ?
Answer:
A cell in which both the electrodes are of the same metal dipping in solutions of its ions of different concentrations (or the same gas electrode at different pressures) is called a concentration cell. Its emf arises only due to the difference in concentrations.
For example, Zn | Zn²⁺(C₁) || Zn²⁺(C₂) | Zn ; E_cell = \( \frac{0.059}{2}\log\frac{C_2}{C_1} \) at 298 K (C₂ > C₁).

Teacher's Note:
1. For a concentration cell E°_cell = 0, because both electrodes are identical.

Question 7. How is free energy change of a cell reaction related to (i) its emf (ii) equilibrium constant of the cell reaction ?
Answer:
(i) −Δ_rG = nFE_cell where n is number of electrons involved in the cell reaction.
(ii) −ΔG° = RT ln K where K is equilibrium constant.

Teacher's Note:
1. Combining the two gives E°_cell = (2.303RT/nF) log K_c, which is used to find K_c from cell data.

Question 8. Is free energy change of a cell reaction an intensive property or extensive property ?
Answer:
Extensive property.

Teacher's Note:
1. ΔG = −nFE depends on n, i.e. on how the equation is written, but E_cell itself is intensive.

Question 9. Write the product obtained at anode on electrolysis of concentrated sulphuric acid using platinum electrodes.
Answer:
S₂O₈²⁻(aq)
(2SO₄²⁻ → S₂O₈²⁻ + 2e⁻)

Teacher's Note:
1. With dilute H₂SO₄ the anode product is O₂; only with concentrated acid is peroxodisulphate formed.

Question 10. Name the electrolyte used in (i) dry cell (ii) mercury cell.
Answer:
(i) Paste of NH₄Cl and ZnCl₂ in the zinc cylinder and powdered MnO₂ + C surrounding the graphite rod.
(ii) Paste of KOH and ZnO.

Teacher's Note:
1. In the mercury cell, HgO mixed with carbon is the cathode paste and the zinc-mercury amalgam is the anode; the electrolyte is the KOH-ZnO paste.

Question 11. Give an example of a secondary cell. (HP Board 2013)
Answer:
Lead storage battery.

Teacher's Note:
1. Nickel-cadmium cell is another common example of a secondary (rechargeable) cell.

Question 12. Write the name of a cell used in small watches. (Raj. Board 2011)
Answer:
Mercury cell (Ruben-Mallory cell).

Teacher's Note:
1. The mercury cell is preferred in watches and hearing aids because its voltage (about 1.35 V) stays constant during its life.

Short Answer Questions (2 Marks)

Question 13. What is understood by a normal hydrogen electrode ? Give its significance.
Answer:
A normal (standard) hydrogen electrode consists of a platinum foil coated with platinum black, dipping in a solution of 1 M H⁺ ions (e.g., 1 M HCl), through which pure hydrogen gas at 1 bar pressure is bubbled at 298 K. It is represented as Pt, H₂(g, 1 bar) | H⁺(aq, 1 M). Its electrode potential is arbitrarily taken as zero.
Significance : Since the absolute potential of a single electrode cannot be measured, the SHE is used as the reference electrode. The potential of any other electrode is measured by coupling it with SHE; the emf of the cell gives the potential of that electrode.

Teacher's Note:
1. E° = 0 for SHE is a convention, not a measured value; all tabulated E° values are relative to it.

Question 14. What do you understand by the following ?
(i) negative standard electrode potential
(ii) positive standard electrode potential.
Answer:
(i) A negative standard (reduction) electrode potential means that the electrode has a smaller tendency to get reduced than the hydrogen electrode. When coupled with SHE, it acts as the anode (oxidation occurs on it) and H⁺ ions are reduced to H₂. Such a metal is a stronger reducing agent than H₂, e.g., E°(Zn²⁺/Zn) = −0.76 V.
(ii) A positive standard electrode potential means that the electrode has a greater tendency to get reduced than the hydrogen electrode. When coupled with SHE, it acts as the cathode, and H₂ is oxidised to H⁺ ions. Its ions are stronger oxidising agents than H⁺ ions, e.g., E°(Cu²⁺/Cu) = +0.34 V.

Teacher's Note:
1. Metals with negative E° liberate H₂ from dilute acids; metals with positive E° do not.

Question 15. What is meant by reduction electrode potentials of zinc and copper being −0.76 V and +0.34 V respectively ? Can an aqueous solution of CuSO₄ be stored in a zinc vessel ? Answer with reason.
Answer:
E°(Zn²⁺/Zn) = −0.76 V means that zinc has a smaller tendency to get reduced (or a greater tendency to get oxidised) than hydrogen; when coupled with SHE, zinc acts as anode. E°(Cu²⁺/Cu) = +0.34 V means that Cu²⁺ ions have a greater tendency to get reduced than H⁺ ions; when coupled with SHE, copper acts as cathode.
No, CuSO₄ solution cannot be stored in a zinc vessel. Zinc has lower reduction potential than copper, so zinc will displace copper from CuSO₄ solution : Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) ; E°_cell = 0.34 − (−0.76) = 1.10 V (positive, so the reaction is spontaneous).

Teacher's Note:
1. A positive E°_cell for the vessel-metal-reduces-solution-ion reaction means the vessel will be eaten away.

Question 16. Define electrode potential, oxidation potential and reduction potential. Why is it not possible to determine the absolute value of electrode potential ? (Karnataka Board 2012)
Answer:
Electrode potential : The potential difference set up between the metal and the solution of its ions when the metal is placed in the solution is called electrode potential.
Oxidation potential : The tendency of an electrode to lose electrons (or get oxidised) is called its oxidation potential.
Reduction potential : The tendency of an electrode to gain electrons (or get reduced) is called its reduction potential. (Reduction potential = − Oxidation potential.)
The absolute value cannot be determined because oxidation or reduction cannot take place alone; a half-cell cannot work by itself. Its potential can be measured only relative to another (reference) electrode.

Teacher's Note:
1. By the IUPAC convention, the term “electrode potential” now means reduction potential unless stated otherwise.

Question 17. Explain with example standard electrode potential. (UP Board 2012)
Answer:
The electrode potential of an electrode when the concentrations of all the species involved in the half-cell are unity (1 M), gases are at 1 bar pressure and temperature is 298 K is called standard electrode potential (E°). It is measured with respect to the standard hydrogen electrode, whose potential is taken as zero.
For example, the cell Pt, H₂(1 bar) | H⁺(1 M) || Zn²⁺(1 M) | Zn has emf −0.76 V. Hence E°(Zn²⁺/Zn) = −0.76 V. Similarly, E°(Cu²⁺/Cu) = +0.34 V.

Teacher's Note:
1. The negative sign for zinc shows it is oxidised more readily than hydrogen; zinc becomes the anode when paired with SHE.

Teacher's Note:
1. E.m.f. exceeds the potential difference by the voltage dropped across the internal resistance of the cell.

Question 19. What is an electrochemical series ? How does it help us in predicting whether a redox reaction is feasible in a given direction or not ?
Answer:
The arrangement of various electrodes in the decreasing (or increasing) order of their standard reduction potentials is called electrochemical series.
A redox reaction is feasible if the species which is to be oxidised lies lower in the series (has lower reduction potential) than the species which is to be reduced, i.e., if E°_cell = E°_cathode − E°_anode comes out positive. For example, Zn + Cu²⁺ → Zn²⁺ + Cu is feasible (E°_cell = 0.34 − (−0.76) = +1.10 V) but the reverse reaction Cu + Zn²⁺ → Cu²⁺ + Zn is not feasible (E°_cell = −1.10 V).

Teacher's Note:
1. Positive E°_cell means negative ΔG°, which is the real test of feasibility.

Question 20. What is an electrochemical series ? How does it help in calculating the e.m.f. of a standard cell ?
Answer:
The arrangement of various electrodes in the decreasing (or increasing) order of their standard reduction potentials is called electrochemical series.
The e.m.f. of a standard cell is calculated from the standard reduction potentials taken from the series : E°_cell = E°_cathode − E°_anode = E°_right − E°_left, where the electrode with higher reduction potential acts as cathode.
For example, for Zn | Zn²⁺(1 M) || Cu²⁺(1 M) | Cu, E°_cell = +0.34 − (−0.76) = 1.10 V.

Teacher's Note:
1. Never multiply E° values by the stoichiometric coefficients; electrode potential is an intensive property.

Question 21. Give the reason why blue colour of copper sulphate solution is discharged slowly when an iron rod is dipped in it. Given E°_Cu²⁺/Cu = 0.34 V, E°_Fe²⁺/Fe = −0.44 V
Answer:
Iron has lower reduction potential than copper, so iron displaces copper from CuSO₄ solution :
Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s) ; E°_cell = 0.34 − (−0.44) = +0.78 V
As the reaction proceeds, the blue Cu²⁺ ions are removed from the solution (copper is deposited on the iron rod) and are replaced by almost colourless (pale green) Fe²⁺ ions. Hence, the blue colour fades gradually.

Teacher's Note:
1. The reaction happens only at the surface of the rod, and the copper deposited on it slows further contact, so the colour fades slowly.

Question 22. 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. The free energy change of a cell reaction is related to its emf by Δ_rG = −nFE_cell.
If E_cell is positive, Δ_rG is negative and the reaction is spontaneous in the direction written. If E_cell is negative, Δ_rG is positive and the reaction is non-spontaneous as written (the reverse reaction is spontaneous).

Teacher's Note:
1. If E_cell = 0, ΔG = 0 and the reaction is at equilibrium; the cell gives no current.

Question 23. What is the effect of change in (a) concentration (b) temperature on the electrode potential of a given half-cell ?
Answer:
For the half-cell Mⁿ⁺ + ne⁻ → M, \( E = E^\circ - \frac{2.303\,RT}{nF}\log\frac{1}{[M^{n+}]} \)
(a) On increasing the concentration of Mⁿ⁺ ions, log(1/[Mⁿ⁺]) decreases and hence the reduction potential increases. On decreasing the concentration, the reduction potential decreases.
(b) The electrode potential changes with temperature because the term 2.303RT/nF depends upon T (and E° also varies with temperature). For [Mⁿ⁺] < 1 M, increase of temperature decreases the reduction potential.

Teacher's Note:
1. At 298 K a tenfold change in ion concentration changes E by only 0.059/n volt.

Question 24. Write Nernst equation for the general electrochemical change of the following type at 25°C.
aA + bB \( \xrightarrow{ne^-} \) cC + dD (AP Board 2012)
Answer:
\( E_{cell} = E^\circ_{cell} - \frac{2.303\,RT}{nF}\log\frac{[C]^c[D]^d}{[A]^a[B]^b} \)
At 25°C (298 K), putting R = 8.314 J K⁻¹ mol⁻¹, F = 96500 C mol⁻¹ :
\( E_{cell} = E^\circ_{cell} - \frac{0.059}{n}\log\frac{[C]^c[D]^d}{[A]^a[B]^b} \)

Teacher's Note:
1. Pure solids and liquids do not appear in the concentration quotient; gases enter as partial pressures in bar.

Question 25. What is a concentration cell ? Give one example. How is the emf of such a cell calculated ?
Answer:
A cell in which both the electrodes are made of the same material dipping in solutions of the same electrolyte but of different concentrations is called a concentration cell.
Example : Cu | Cu²⁺(C₁) || Cu²⁺(C₂) | Cu, where C₂ > C₁.
Since both electrodes are the same, E°_cell = 0. Applying Nernst equation, E_cell = \( \frac{2.303\,RT}{nF}\log\frac{C_2}{C_1} \) = \( \frac{0.059}{n}\log\frac{C_2}{C_1} \) at 298 K.

Teacher's Note:
1. The electrode in the more dilute solution is the anode; the cell works until the two concentrations become equal.

Short Answer Questions (3 Marks)

Teacher's Note:
1. In both cells oxidation takes place at the anode and reduction at the cathode; only the signs of the electrodes differ.

Question 27. What do you understand by Normal Hydrogen Electrode ? Give its construction and working.
Answer:
A normal hydrogen electrode (NHE or SHE) is a reference electrode whose electrode potential is taken as zero at all temperatures.
Construction : It consists of a platinum foil coated with platinum black, attached to a platinum wire sealed in a glass tube. The foil is placed in a beaker containing 1 M HCl (1 M H⁺ ions). Pure hydrogen gas at 1 bar pressure is bubbled continuously over the foil at 298 K. It is represented as Pt, H₂(g, 1 bar) | H⁺(aq, 1 M).
Working : It can act as anode or cathode depending upon the other electrode with which it is coupled.
As anode (oxidation) : H₂(g) → 2H⁺(aq) + 2e⁻
As cathode (reduction) : 2H⁺(aq) + 2e⁻ → H₂(g)
Since E° of SHE = 0, the emf of the cell formed with any other electrode directly gives the standard electrode potential of that electrode.

Teacher's Note:
1. Platinum black adsorbs H₂ and speeds up the equilibrium H₂ ⇌ 2H⁺ + 2e⁻; Pt itself does not react.

Question 28. Daniell cell is a galvanic cell made of zinc and copper electrodes
(i) Write anode and cathode reactions in Daniell cell
(ii) Nernst equation for the electrode reaction, Mⁿ⁺ + ne⁻ → M is
\( E_{M^{n+}/M} = E^\circ_{M^{n+}/M} - \frac{2.303\,RT}{nF}\log\frac{1}{[M^{n+}]} \)
Derive Nernst equation for Daniell cell. (Kerala Board 2012)
Answer:
(i) At anode : Zn(s) → Zn²⁺(aq) + 2e⁻
At cathode : Cu²⁺(aq) + 2e⁻ → Cu(s)
(ii) For the cathode : \( E_{Cu^{2+}/Cu} = E^\circ_{Cu^{2+}/Cu} - \frac{2.303\,RT}{2F}\log\frac{1}{[Cu^{2+}]} \)
For the anode (written as reduction) : \( E_{Zn^{2+}/Zn} = E^\circ_{Zn^{2+}/Zn} - \frac{2.303\,RT}{2F}\log\frac{1}{[Zn^{2+}]} \)
E_cell = E_Cu²⁺/Cu − E_Zn²⁺/Zn
= (E°_Cu²⁺/Cu − E°_Zn²⁺/Zn) − \( \frac{2.303\,RT}{2F}\left[\log\frac{1}{[Cu^{2+}]} - \log\frac{1}{[Zn^{2+}]}\right] \)
\( E_{cell} = E^\circ_{cell} - \frac{2.303\,RT}{2F}\log\frac{[Zn^{2+}]}{[Cu^{2+}]} \)
At 298 K : E_cell = E°_cell − \( \frac{0.059}{2}\log\frac{[Zn^{2+}]}{[Cu^{2+}]} \)

Teacher's Note:
1. The log term always has products over reactants of the cell reaction; increasing [Cu²⁺] therefore raises E_cell.

Question 29. What is free energy and free energy change ? Explain how is free energy related to cell potential ? Also explain its significance in predicting the feasibility of the cell reaction.
Answer:
Free energy : The Gibbs free energy (G) of a system is the energy available for doing useful work. It is defined as G = H − TS.
Free energy change : The change in Gibbs free energy during a process, ΔG = ΔH − TΔS, is equal to the maximum useful work that can be obtained from the process, i.e., −ΔG = W_max.
Relation with cell potential : In a cell, if n moles of electrons flow under emf E_cell, the electrical work = nFE_cell. Hence Δ_rG = −nFE_cell and Δ_rG° = −nFE°_cell.
Significance : If E_cell is positive, ΔG is negative and the cell reaction is feasible (spontaneous). If E_cell is negative, ΔG is positive and the reaction is not feasible. If E_cell = 0, ΔG = 0 and the reaction is at equilibrium.

Teacher's Note:
1. Keep the sign convention straight: positive E_cell and negative ΔG both mean the reaction goes.

Question 30. Explain why electrolysis of an aqueous solution of NaCl gives H₂ at cathode and Cl₂ at anode ? Write overall reaction
(E°_Na⁺/Na = −2.71 V ; E°_H2O/H2 = −0.83 V ; E°_Cl2/2Cl⁻ = +1.36 V ; E°_H⁺/O2/H2O = 1.23 V)
Answer:
At cathode : Two reductions are possible :
Na⁺(aq) + e⁻ → Na(s) ; E° = −2.71 V
2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) ; E° = −0.83 V
The reaction with higher reduction potential takes place. Hence, water is reduced and H₂ is liberated at the cathode.
At anode : Two oxidations are possible :
2Cl⁻(aq) → Cl₂(g) + 2e⁻ ; E°_red = +1.36 V
2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ ; E°_red = +1.23 V
On the basis of E° values, oxidation of water (lower reduction potential) should be preferred. But oxidation of water needs a large overvoltage. Hence, Cl⁻ ions are oxidised and Cl₂ is liberated at the anode.
Overall reaction : 2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + H₂(g) + Cl₂(g)

Teacher's Note:
1. Overvoltage of oxygen is the key word examiners look for in explaining Cl₂ at the anode.

Question 31. Draw a well labelled diagram of “Dry cell”. Give the reactions taking place at the anode and the cathode.

Answer:
(Diagram above)
Anode : Zinc container ; Cathode : Graphite (carbon) rod surrounded by powdered MnO₂ and carbon ; Electrolyte : moist paste of NH₄Cl and ZnCl₂.
At anode : Zn(s) → Zn²⁺(aq) + 2e⁻
At cathode : MnO₂(s) + NH₄⁺(aq) + e⁻ → MnO(OH)(s) + NH₃(g)
The NH₃ formed combines with Zn²⁺ to form the complex [Zn(NH₃)₄]²⁺. The cell gives a potential of about 1.5 V.

Teacher's Note:
1. In the cathode reaction manganese is reduced from +4 (MnO₂) to +3 (MnO(OH)).

Question 32. Which cell is generally used in the hearing aids ? Name the material of the anode, cathode and the electrolyte. Write the reactions involved.
Answer:
Mercury cell is generally used in hearing aids.
Anode : zinc-mercury amalgam ; Cathode : paste of HgO and carbon ; Electrolyte : paste of KOH and ZnO.
At anode : Zn(Hg) + 2OH⁻ → ZnO(s) + H₂O + 2e⁻
At cathode : HgO(s) + H₂O + 2e⁻ → Hg(l) + 2OH⁻
Overall : Zn(Hg) + HgO(s) → ZnO(s) + Hg(l)

Teacher's Note:
1. No ion appears in the overall reaction, so the cell voltage (about 1.35 V) stays constant throughout its life.

Question 33. What type of battery is lead storage battery ? Write the anode and the cathode reactions and the overall reactions occurring in a lead storage battery. (CBSE 2011)
Answer:
Lead storage battery is a secondary battery (it can be recharged).
At anode : Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻
At cathode : PbO₂(s) + SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l)
Overall : Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l)

Teacher's Note:
1. Sulphuric acid is consumed during discharge, so measuring the density of the acid tells how far the battery is discharged.

Question 34. Give following information about ‘Nickel-Cadmium storage cell’ :
(i) Material of the cathode (ii) Material of the anode (iii) Electrolyte used
(iv) Reactions involved at the anode and cathode (v) Approximate voltage of the cell.
Answer:
(i) Cathode : metal grid containing nickel(III) hydroxide [Ni(OH)₃, also written as NiO(OH)].
(ii) Anode : cadmium.
(iii) Electrolyte : KOH solution (moist paste).
(iv) At anode : Cd(s) + 2OH⁻(aq) → CdO(s) + H₂O(l) + 2e⁻
At cathode : 2Ni(OH)₃(s) + 2e⁻ → 2Ni(OH)₂(s) + 2OH⁻(aq)
Overall : Cd(s) + 2Ni(OH)₃(s) → CdO(s) + 2Ni(OH)₂(s) + H₂O(l)
(v) Voltage : about 1.2 V (1.2–1.4 V).

Teacher's Note:
1. The Ni-Cd cell is rechargeable and has a longer life than the lead storage cell, but it is more expensive.

Question 35. What is mercury cell ? Give the electrode reactions.
Answer:
Mercury cell is a small primary cell used in watches, hearing aids and other low-current devices. It consists of zinc-mercury amalgam as anode, a paste of HgO and carbon as cathode and a paste of KOH and ZnO as electrolyte.
At anode : Zn(Hg) + 2OH⁻ → ZnO(s) + H₂O + 2e⁻
At cathode : HgO(s) + H₂O + 2e⁻ → Hg(l) + 2OH⁻
Overall : Zn(Hg) + HgO(s) → ZnO(s) + Hg(l). The cell potential is about 1.35 V and remains constant during its life.

Teacher's Note:
1. The constant voltage follows from the overall equation containing no ions whose concentration could change.

Long Answer Questions (5 Marks)

Question 36. (a) The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if the conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10⁻³ S cm⁻¹ ?
(b) Predict the products of electrolysis of the solution of H₂SO₄ with platinum electrodes. (CBSE 2008)
Answer:
(a) Cell constant = Conductivity × Resistance = 0.146 × 10⁻³ S cm⁻¹ × 1500 Ω = 0.219 cm⁻¹
(b) At cathode : 2H⁺(aq) + 2e⁻ → H₂(g). Hence, H₂ gas is liberated at the cathode.
At anode : with dilute H₂SO₄, water is oxidised : 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻. Hence, O₂ gas is liberated at the anode.
With concentrated H₂SO₄, SO₄²⁻ ions are oxidised at the anode : 2SO₄²⁻(aq) → S₂O₈²⁻(aq) + 2e⁻ (peroxodisulphate ions).

Teacher's Note:
1. Cell constant = κ × R; S × Ω cancel, leaving cm⁻¹.

Question 37. (a) Define molar conductivity of a substance and describe how for weak and strong electrolytes, molar conductivity changes with concentration of solute. How is such change obtained ?
(b) A voltaic cell is set up at 25°C with the following half cells :
Ag⁺ (0.001 M) | Ag and Cu²⁺ (0.10 M) | Cu
What would be the voltage of this cell ? (E°_cell = 0.46 V) (CBSE 2009)

Answer:
(a) Molar conductivity (Λ_m) is the conductance of the volume of solution containing one mole of the electrolyte placed between two electrodes of large area at unit distance apart. Λ_m = \( \frac{\kappa\times 1000}{M} \) S cm² mol⁻¹.
Molar conductivity increases on dilution for both types of electrolytes. For a strong electrolyte, Λ_m increases slowly and linearly with √C according to Λ_m = Λ°_m − A√C, because inter-ionic attractions decrease on dilution. For a weak electrolyte, Λ_m increases sharply at low concentrations because its degree of dissociation increases on dilution.
The change is obtained by measuring the conductivity (κ) of solutions of different concentrations with a conductivity cell, calculating Λ_m = κ × 1000/M for each and plotting Λ_m against √C (graph above).
(b) Cell : Cu | Cu²⁺(0.10 M) || Ag⁺(0.001 M) | Ag
Cell reaction : Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s), n = 2
E_cell = E°_cell − \( \frac{0.0591}{2}\log\frac{[Cu^{2+}]}{[Ag^+]^2} \) = 0.46 − \( \frac{0.0591}{2}\log\frac{0.10}{(0.001)^2} \)
= 0.46 − 0.02955 × log 10⁵ = 0.46 − 0.02955 × 5 = 0.46 − 0.148 = 0.312 V

Teacher's Note:
1. In the Nernst equation [Ag⁺] is squared because two Ag⁺ ions take part in the balanced cell reaction.

Question 38. (a) Corrosion is essentially an electrochemical phenomenon. Explain the reactions occurring during corrosion of iron kept in an open atmosphere.
(b) One half-cell in a voltaic cell is constructed from a silver wire dipped in silver nitrate solution of unknown concentration. Its other half-cell consists of a zinc electrode dipping in 1.0 M solution of Zn(NO₃)₂. A voltage of 1.48 V is measured for this cell. Use this information to calculate the concentration of silver nitrate solution used.
(E°_Zn²⁺|Zn = −0.76 V, E°_Ag⁺|Ag = +0.80 V). (CBSE 2009)
Answer:
(a) In open atmosphere, small electrochemical cells are set up on the surface of iron, with water containing dissolved CO₂ and O₂ as electrolyte.
At anode : 2Fe(s) → 2Fe²⁺(aq) + 4e⁻ ; E°_Fe²⁺/Fe = −0.44 V
At cathode : O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l) ; E° = 1.23 V
Overall : 2Fe(s) + O₂(g) + 4H⁺(aq) → 2Fe²⁺(aq) + 2H₂O(l) ; E°_cell = 1.67 V
Fe²⁺ ions are further oxidised by atmospheric oxygen to Fe³⁺, which form hydrated ferric oxide (rust), Fe₂O₃.xH₂O.
(b) Cell : Zn | Zn²⁺(1.0 M) || Ag⁺(C) | Ag ; Cell reaction : Zn + 2Ag⁺ → Zn²⁺ + 2Ag, n = 2
E°_cell = 0.80 − (−0.76) = 1.56 V
E_cell = E°_cell − \( \frac{0.0591}{2}\log\frac{[Zn^{2+}]}{[Ag^+]^2} \)
1.48 = 1.56 − \( \frac{0.0591}{2}\log\frac{1.0}{[Ag^+]^2} \) or \( \log\frac{1}{[Ag^+]^2} = \frac{0.08\times 2}{0.0591} = 2.707 \)
or −2 log [Ag⁺] = 2.707 or log [Ag⁺] = −1.3535 = \( \bar{2} \).6465
[Ag⁺] = 4.4 × 10⁻² M

Teacher's Note:
1. A measured E_cell below E°_cell tells you at once that [Ag⁺] is less than 1 M; use this as a check on your answer.

Question 39. (a) Write the anode and cathode reactions and the overall reaction occurring in a lead storage battery.
(b) A copper-silver cell is set up. The copper ion concentration is 0.10 M. The concentration of silver ions is not known. The cell potential when measured was 0.422 V. Determine the concentration of silver ions in the cell. (Given : E°_Ag⁺/Ag = +0.80 V, E°_Cu²⁺/Cu = +0.34 V). (CBSE 2010)
Answer:
(a) At anode : Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻
At cathode : PbO₂(s) + SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l)
Overall : Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l)
(b) Cell : Cu | Cu²⁺(0.10 M) || Ag⁺(C) | Ag ; Cu + 2Ag⁺ → Cu²⁺ + 2Ag, n = 2
E°_cell = 0.80 − 0.34 = 0.46 V
0.422 = 0.46 − \( \frac{0.0591}{2}\log\frac{0.10}{[Ag^+]^2} \) or \( \log\frac{0.10}{[Ag^+]^2} = \frac{0.038\times 2}{0.0591} = 1.286 \)
\( \frac{0.10}{[Ag^+]^2} \) = 19.32 or [Ag⁺]² = 5.18 × 10⁻³
[Ag⁺] = 7.1 × 10⁻² M

Teacher's Note:
1. Copper has the lower E°, so it is the anode; Cu²⁺ goes in the numerator of the log term.

Question 40. (a) What is a fuel cell ? What is its main advantage ?
(b) What are the reactions occurring at the cathode and anode of a Lechlanche cell ?
(c) In the button cell widely used for watches and other devices, the following reaction takes place :
Zn(s) + Ag₂O(s) + H₂O(l) → Zn²⁺(aq) + 2Ag(s) + 2OH⁻(aq)
Give the cell representation and determine the value of K_c for the above reaction using the following data :
Ag₂O(s) + H₂O(l) + 2e⁻ → 2Ag(s) + 2OH⁻(aq) (E° = 0.344 V)
Zn²⁺(aq) + 2e⁻ → Zn(s) (E° = −0.76 V)
Answer:
(a) A fuel cell is a galvanic cell in which the chemical energy of combustion of a fuel (like H₂, CH₄, CH₃OH) is directly converted into electrical energy. Its main advantage is that it is pollution free and has high efficiency (about 60–70%).
(b) At anode : Zn(s) → Zn²⁺(aq) + 2e⁻
At cathode : MnO₂(s) + NH₄⁺(aq) + e⁻ → MnO(OH)(s) + NH₃(g)
(c) Zn(s) | Zn²⁺(aq) || Ag₂O(s) | OH⁻(aq)
E° = E°_Ag2O/Ag − E°_Zn²⁺/Zn = 0.344 − (−0.76) = 1.104 V
Zn(s) + Ag₂O(s) + H₂O(l) → Zn²⁺(aq) + 2Ag(s) + 2OH⁻(aq)
n = 2, Apply E° = \( \frac{0.0591}{n} \) log K_c, log K_c = \( \frac{2\times 1.104}{0.0591} \) = 37.36
K_c = 2.29 × 10³⁷

Teacher's Note:
1. When the antilog of a large log value is taken, carry at least two decimal places in the log, since a change of 0.01 in log changes K_c by about 2%.