NCERT Solutions for Class 12 Chemistry Chapter 3 - Electrochemistry
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1. Arrange the following metals in the order in which they displace each other from their salts.
Al, Cu, Fe, Mg and Zn
Ans. Mg > Al > Zn > Fe > Cu
It means that every metal on the left side can displace the next one from its salt solution.
2. Given the standard electrode potentials, K+/K = –2.93V, Ag+/Ag = 0.80V, Hg2+/Hg = 0.79V, Mg2+/Mg = –2.37V, Cr3+/Cr = 0.74V.
Arrange these metals in their increasing order of reducing power.
$$\textbf{Ans.}\space\text{Reducing power}\space\alpha\space\frac{1}{\text{Reduction potential}}$$
i.e., as the reduction potential decreases, reducing power increases. Thus, increasing order of reducing power of metal will be:
Ag < Hg < Cr < Mg < K
3. Depict the galvanic cell in which the reaction
$$\textbf{Zn(s) + 2Ag}^{\normalsize+}(aq)\xrightarrow{}\textbf{Zn}^{2\normalsize+}\textbf{(aq)}+2\textbf{Ag(s)}\space\textbf{takes place. Further show:}$$
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.
Ans. The set-up will be similar to as shown below.
The cell will be represented as:
Zn(s)| Zn2+(aq) || Ag+(aq)|Ag(s)
(i) Anode, i.e., zinc electrode will be negatively charged.
(ii) The carriers of the current in cell is electrons and ions.
$$\textbf{(iii) At anode:}\space\text{Zn(s)}\xrightarrow{}\text{Zn}^{2+}\text{(aq)}+2e^{\normalsize-}\text{(Reduction)}\\\textbf{At cathode:}\space\text{2Ag}^{+}(aq) + 2e^{\normalsize–}\xrightarrow{}\text{2Ag}(s)\text{(Oxidation)}$$
4. Calculate the standard cell potentials of the galvanic cells in which the following reactions take place.
(i) 2Cr(s) + 3Cd2+ (aq) → 2Cr3+ (aq) + 3Cd(s)
Given: E°Cr3+/Cr = –0.74V; E°Cd2+/Cd = –0.40V
(ii) Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)
Given: E°Ag+/Ag = 0.80V; E°Fe3+/Fe2+ = 0.77V
Calculate ΔrG° and equilibrium constant for the reaction.
Ans. (i) Calculation of E°cell,
E°cell = E°cathode – E°anode = 0.40 – (–0.74) = +0.34V
Calculation of ΔrG°
ΔrG° = nF E°cell = –(6 mol) × (96500 C mol–1) × (0.34 V)
= – 196860 CV = – 196860 J = – 196.86 kJ
Calculation of Equilibrium Constant (Kc)
ΔrG° = –2.303 RT log Kc
$$\text{log K}_c=\frac{(-)\Delta_rG°}{2.303\text{RT}}=(-)\frac{(-)196860}{2.303×8.314×298}=34.501$$
Kc = Antilog (34.501) = 3.17 × 1034
(ii) Calculation of E°cell,
E°cell = E°cathode – E°anode = (0.80 – 0.77) = 0.03V
Calculation of ΔG°,
ΔrG° = –nF E°cell = –(1 mol) × (96500 C mol–1) × (0.03 V)
= –2895 CV = –2895 J = – 2.895 kJ
Calculation of Equilibrium Constant (Kc)
ΔrG° = –2.303 RT log Kc
$$\text{log K}_c=(-)\frac{(-\Delta_rG\degree)}{2.303\text{RT}}=(-)\frac{(-)2895}{2.303×8.314×298}=0.5074$$
Kc = Anrilog (0.5074) = 3.22
5. Write the Nernst equation and emf of the following cells at 298 K:
(i) Mg(s) | Mg2+ (0.001 M) || Cu2+ (0.001 M) | Cu(s)
(ii) Fe(s) | Fe2+ (0.001 M) || H+ (1M) | H2(g) (1 bar) | Pt(s)
(iii) Sn(s) | Sn2+ (0.050 M) || H+ (0.020 M) | H2(g) (1 bar) | Pt(s)
(iv) Pt(s) | Br– (0.010 M) | Br2(1) || H+ (0.030 M) H2(g) (1 bar) | Pt(s)
Ans. (i) Cell reaction:
$$\text{Mg + Cu}^{2+}\xrightarrow{}\text{Mg}^{2+}+\text{Cu}(n=2)\\\text{Nernst equation:}\\\text{E}_\text{cell}=\text{E}\degree_\text{cell}-\frac{0.0591}{2}\text{log}\frac{[\text{Mg}^{2+}]}{[\text{Cu}^{2+}]}\\\therefore\text{E}_\text{cell}=0.34-(-2.37)-\frac{0.0591}{2}\text{log}\frac{10^{\normalsize-3}}{10^{\normalsize-4}}$$
= 2.71 – 0.02955 = 2.68 V
(ii) Cell reaction:
$$\text{Fe}+2\text{H}^{+}\xrightarrow{}\text{Fe}^{2+}+\text{H}_2(n=2)\\\text{Nernst equation:}\\\text{E}_\text{cell}=\text{E}\degree_\text{cell}-\frac{0.0591}{2}\text{log}\frac{[\text{Fe}^{2+}]}{[\text{H}^{+}]^{2}}\\\therefore\space\text{E}_\text{cell}=0-(-0.44)-\frac{0.0591}{2}\text{log}\frac{10^{\normalsize-3}}{(1)^{2}}\\=0.44-\frac{0.0591}{2}×(-3)$$
= 0.44 + 0.0887 = 0.5287V ≈ 0.53 V
(iii) Cell reaction:
$$\text{Sn+2H}^{+}\xrightarrow{}\text{Sn}^{2+}+\text{H}_2(n=2)\\\text{Nernst equation:}\\\text{E}_\text{cell}=\text{E}\degree_\text{cell}-\frac{0.0591}{2}\text{log}\frac{[\text{Sn}^{2+}]}{[\text{H}^{+}]^{2}}\\\text{E}_\text{cell}=\text{E}\degree_\text{cell}-\frac{0.0591}{2}\text{log}\frac{0.05}{(0.02)^{2}}\\=0-(-0.14)-\frac{0.0591}{2}\text{log}\frac{0.05}{(0.02)^{2}}\\=0.14-\frac{0.0591}{2}\text{log}125\\=0.14-\frac{0.0591}{2}(2.0969)=0.078\text{V}$$
(iv) Cell reaction:
$$\text{2Br}^{\normalsize-}+2\text{H}^{+}\xrightarrow{}\text{Br}_2+\text{H}_2(n=2)\\\text{Nernst equation:}\\\text{E}_\text{cell}=\text{E}\degree_\text{cell}-\frac{0.0591}{2}\text{log}\frac{1}{[\text{Br}^{\normalsize-}]^{2}[\text{H}^{+}]^{2}}\\=(0-1.08)-\frac{0.0591}{2}\text{log}\frac{1}{(0.01)^{2}(0.03)^{2}}\\=-1.08-\frac{0.0591}{2}\text{log}(1.111×107)\\=-1.08-\frac{0.0591}{2}(7.0457)$$
= 1.08 – 0.208 = –1.288V.
Hence, oxidation will occur at the hydrogen electrode and reduction will occur on Br2 electrode.
6. In the button cells widely used in watches and other devices the following reaction takes place:
$$\textbf{Zn(s) + Ag}_2\textbf{O(s) + H}_2\textbf{O(l)}\xrightarrow{}\textbf{Zn}^{2+}\textbf{(aq)+ 2\textbf{Ag(s)}+2\textbf{OH}}^{\normalsize-}\textbf{(aq)}\\\textbf{Determine} \Delta_r\textbf{G}^{\normalsize–} \textbf{and E}^{\normalsize–} \textbf{for the reaction:}\\\textbf{Given: \space Zn}\xrightarrow{}\textbf{Zn}^{2+}+2\textbf{e}^{\normalsize-},\textbf{E}\degree=0.76\textbf{V};\\\textbf{Ag}_2\textbf{O}+\textbf{H}_2\textbf{O+2e}^{\normalsize-}\xrightarrow{}\textbf{2Ag+2OH}^{\normalsize-},\textbf{E}\degree\textbf{=0.344 V}.$$
Ans. Zn is oxidised and Ag2O is reduced.
E°cell = E°Ag2O, Ag(reduction) – E°Zn/Zn2+ (oxidation)
= 0.344 – (–0.76) = 1.104 V
ΔG = –nFEcell = –2 × 96500 × 1.104 J
= –2.13 × 105 J.
7. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Ans. Conductivity: The reciprocal of resistivity is known as specific conductance or simply conductivity. It is denoted by κ (kappa). Thus, if k is the specific conductance and G is the conductance of the solution, then
$$\text{R}=\frac{1}{\text{G}}\space\text{and}\space\rho=\frac{1}{k}\\\frac{1}{\text{G}}=\frac{1}{\text{k}}×\frac{l}{\text{A}}\\k=\text{G}×\frac{l}{\text{A}}$$
Now, if l = 1 cm and A = 1 sq.cm, then k = G.
Hence, conductivity of a solution is defined as the conductance of a solution of 1 cm length and having
1 sq. cm as the area of cross-section. Alternatively, it may be defined as conductance of one centimetre cube of the solution of the electrolyte.
Molar conductivity: Molar conductivity of a solution at a dilution V is the conductance of all the ions produced from 1 mole of the electrolyte dissolved in V cm3 of the solution when the electrodes are one cm apart and the area of the electrodes is so large that the whole of the solution is contained between them. It is represented by Λm.
$$\Lambda_m=\frac{\text{kA}}{l}$$
Since, l = 1cm and A = V (volume containing 1 mole of electrolyte)
Λm = κV
Variation of conductivity and molar conductivity with concentration: Conductivity always decreases with decrease in concentration, for both week and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases on dilution.
Molar conductivity increases with decrease in concentration. This is because that total volume, V, of solution containing one mole of electrolyte also increases. It has been found that decrease in k on dilution of a solution is more than compensated by increase in its volume.
8. The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm–1. Calculate its molar conductivity.
$$\textbf{Ans.}\space\text{Molar conductivity Λ}_m =\frac{k×1000}{\text{Molarity}}=\frac{0.0248 \text{S cm}^{\normalsize-1}×1000 \text{cm}^{3}\text{L}^{\normalsize-1}}{\text{0.20\space\text{mol L}}^{\normalsize-1}}=124\space\text{S cm}^{2}\text{mol}^{\normalsize-1}.$$
9. The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω What is the cell constant if conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10–3 S cm–1.
Ans. Cell constant = Conductivity × Resistance
= 0.146 × 10–3 S cm–1 × 1500 W
= 0.219 cm–1.
10. The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:
| Concentration/M | 0.001 | 0.010 | 0.020 | 0.050 | 0.100 |
| 102 × k/S m–1 | 1.237 | 11.85 | 23.15 | 55.53 | 106.74 |
Calculate Λm for all concentrations and draw a plot between Λm and c1/2. Find the value of Λ°m.
$$\textbf{Ans.}\space\frac{1 \text{S cm}^{\normalsize-1}}{100\space\text{S m}^{\normalsize-1}}=1\text{(unit conversion factor)}$$
| Concentration(M) | k(S m–1) | k(S cm–1) | $$\Lambda_m=\frac{100×k}{\text{Molarity}}(\text{S cm}^{2}\text{mol}^{\normalsize-1})$$ | c1/2 (M1/2) |
| 10–3 | 1.237 × 10–2 | 1.237 × 10–4 | $$\frac{1000×1.237×10^{\normalsize-4}}{10^{\normalsize-3}}=123.7$$ | 0.0316 |
| 10–2 | 11.85×10-2 | 11.85×10-4 | $$\frac{1000×11.85×10^{\normalsize-4}}{10^{\normalsize-2}}=118.5$$ | 0.100 |
| 2×10-2 | 23.15×10-2 | 23.15×10-4 | $$\frac{1000×23.15×10^{\normalsize-4}}{2×10^{\normalsize-2}}=115.8$$ | 0.141 |
| 5 × 10–2 | 55.53×10-2 | 55.53 × 10–4 | $$\frac{1000×55.53×10^{\normalsize-4}}{5×10^{\normalsize-2}}=111.1$$ | 0.224 |
| 10–1 | 106.74×10-2 | 106.74 × 10–4 | $$\frac{1000×106.74×10^{\normalsize-4}}{10^{\normalsize-1}}=106.7$$ | 0.316 |
11. Conductivity of 0.00241 M acetic acid is 7.896 × 10–5 S cm–1. Calculate its molar conductivity. If Λ°m for acetic acid is 390.5 S cm2 mol–1, what is its dissociation constant?
Ans. Molar conductivity of acetic acid:
$$\Lambda_m^c=\frac{k}{\text{C}}=\frac{7.896×10^{\normalsize-5}\text{S cm}^{\normalsize-1}}{241×10^{\normalsize-8}\text{mol cm}^{\normalsize-3}}=32.76\space\text{S cm}^{2}\text{mol}^{\normalsize-1}\\\text{Dissociation constant}\space k \alpha=\frac{\text{ C}\alpha^{2}}{1-\alpha}\space\bigg[\alpha=\frac{\Lambda_{m}^{c}}{\Lambda_{m}^{\degree}} =\frac{32.76}{390.5}=0.084\bigg ]\\\frac{0.00241×(0.084)^{2}}{1-0.084}=1.85×10^{\normalsize-5}\text{mol L}^{\normalsize-1}$$
12. How much charge is required for the following reductions:
(i) 1 mol of Al3+ to Al?
(ii) 1 mol of Cu2+ to Cu?
(iii) 1 mol of MnO–4 to Mn2+?
Ans. (i) 1 mol of Al3+ to Al
$$\text{Electrode reaction:}\underset{1 \text{mol}}{\text{Al}^{3+}(aq)+3e^{\normalsize-}}\xrightarrow{}\underset{3 \text{mol}}{\text{Al(s)}}$$
Quantity of charge required to reduce 1 mol of Al3+ is = 3 faraday = 3 × 96500 C = 289500 C = 2.89 × 105 C
(ii) 1 mol Cu2+ to Cu
$$\text{Electrode reaction:}\space\underset{1 \text{mol}}{\text{Cu}^{2+}(aq)+2e^{\normalsize-}}\xrightarrow{}\underset{2 \text{mol}}{\text{Cu(s)}}$$
Quantity of charge required to reduce 1 mol of Cu2+ is = 2 faraday = 2 × 96500 C = 1.93 × 105 C
(iii) 1 mol of MnO–4 to Mn2+
$$\text{Electrode reaction:}\space\underset{1 \text{mol}}{\text{MnO}^{-}_4+5e^{\normalsize-}}\xrightarrow{}\underset{5 \text{mol}}{\text{Mn}}^{2\normalsize+}$$
Quantity of change require to reduce 1 mol MnO–4 is = 5 faraday = 5 × 96500 C = 4.83 × 105 C
13. How much electricity in term sof Faraday is required to produce:
(i) 20.0g of Ca from molten CaCl2?
(ii) 40.0g of Al from molten Al2O3?
$$\textbf{Ans.}\space\text{(i) Electrode reaction:}\space\underset{1\text{mol}}{\text{Ca}}^{2+}(aq)+2e^{-}\xrightarrow{}\underset{2 \text{mol}}{\text{Ca(s)}}$$
To produce 1 mol or 40gm of Ca from molten CaCl2, electricity required = 2 F
To produce 20gm of Ca from molten CaCl2, electricity required = 1 F
$$\text{(ii) Electrode reaction:}\underset{\space\text{1 mol}}{\text{Al}_2\text{O}_3(l)}+\underset{\space\text{6 mol}}{e^{\normalsize-}}\xrightarrow{}\text{2 Al+3O}^{2\normalsize-}\\\text{To produce 54g Al from molten Al}_2\text{O}_3, \text{electricity required = 6F}\\\text{To produce 40g Al from molten A}_2\text{O}_3,\text{electricity required =}\frac{6×40}{54}=4.44\text{F}$$
14. How much electricity is required in coulomb for the oxidation of:
(i) 1 mol of H2O to O2?
(ii) 1 mol of FeO to Fe2O3?
$$\textbf{Ans.}\space\text{(i)\space H}_2\text{O}\xrightarrow{}2\text{H}^{+}+\frac{1}{2}\text{O}_2+2e^{\normalsize-}\\\text{Electricity required to oxidise 1 mol H}_2\text{O to O}_2 = 2 \text{F}\\= 2 × 96500 \text{C} = 193000 \text{C} = 1.93 × 10^5 \text{C}\\\text{(ii)\space 2FeO}+\frac{1}{2}\text{O}_2\xrightarrow{}\text{Fe}_2\text{O}_3\\2\text{Fe}^{2+}\xrightarrow{}2\text{Fe}^{3+}+2e^{\normalsize-}$$
Electricity required to oxidise 2 moles FeO to Fe2O3 = 2 F
∴ Electricity required to oxidise 1 mole FeO to Fe2O3 = 1 F
= 1 × 96500 C = 9.65 × 104 C
15. A solution of Ni(NO3)2 is electrolysed between platinum electrodes used a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?
Ans. Charge = Current × time = 5 × 1200s = 6000 C
Chemical equation:
$$\text{Ni(NO}_3)_2 (aq) + 2\text{H}^ {\normalsize+} + 2e^{\normalsize –} \xrightarrow{} \text{Ni + 2HNO}_3$$
Charge required to deposite 1 mol nickel = 2 F
= 2 × 96500 = 1.93 × 105 C
So, 1 mol or 58.7g nickel produce charge of 1.93 × 105C
$$\text{Therefore, 6000 C change produced by =}\frac{58.7×6000}{1.93×10^{5}}=1.825\text{g nickel.}$$
Hence, the mass of Ni deposited at the cathode is 1.825g.
16. Three electrolytic cells A,B,C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A current of 1.5 amperes was passed through them unit 1.45g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?
$$\textbf{Ans.}\space\text{Ag}^{\normalsize+} (aq) + e^{\normalsize –}\xrightarrow{}\text{Ag(s)}\\\text{108g silver deposited by electricity = 1F = 96500 C}\\\text{1.45g silver deposited by electricity =}\frac{96500}{108}×1.45\\= 1295.6 \text{C}\\\text{Q = i × t}\\t=\frac{\text{Q}}{i}=\frac{1295.6}{1.5}=14\text{min}\space 39 s\\\text{Cu}^{2+}(aq)+2e^{\normalsize-}\xrightarrow{}\text{Cu(s)}\\\text{On passing electricity of 2 × 96500 C, Cu deposited = 63.5g}\\\text{On passing electricity of 1295.6 C, Cu deposited =}\frac{63.5×1295.6}{2×96500}=0.426\text{g}\\\text{Zn}^{2+}(aq)+2e^{\normalsize-}\xrightarrow{}\text{Zn(s)}\\\text{On passing electricity of 2 × 96500 C, Zn deposited = 65.3g}$$
$$\text{On passing electricity of 1295.6 C, Zn deposited =}\frac{63.5×1295.6}{2×96500}=0.438\text{g}$$
17. Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:
(i) Fe3+ (aq) and I– (aq)
(ii) Ag+ (aq) and Cu(s)
(iii) Fe3+ (aq) and B– (aq)
(iv) Ag(s) and Fe3+ (aq)
(v) Br2 (aq) and Fe2+ (aq)
Ans. The reaction is feasible, its E°cell value is positive.
$$\text{(i) Fe}^{3+}(aq)+\text{I}^{\normalsize-}(aq)\xrightarrow{}\text{Fe}^{2+}(aq)+\frac{1}{2}\text{I}_2\text{(g)}\\\text{E}\degree_{\text{cell}}=\text{E}\degree_{(\text{Fe}^{3+}/\text{Fe}^{2+})}-\text{E}\degree _{\bigg(\frac{1}{2I_2}/\text{I}^{\normalsize-}\bigg)}=0.77 – 0.54 = 0.23V\\\text{This reaction is feasible.}\\\text{(ii) Cu(s) + 2Ag}^+ (aq)\xrightarrow{}\text{2Ag(s) + Cu}^{2+} (aq)$$
E°cell = E°(Ag+/Ag) – E°(Cu2+/Cu) = 0.80 – 0.34 = 0.46V
This reaction is feasible
$$\text{(iii) Fe}^{3+} \text{(aq) + Br}^{–} \text{(aq)}\xrightarrow{}\text{Fe}^{2+}(aq)+\frac{1}{2}\text{Br}_2(aq)\\\text{E}\degree_{\text{cell}}=\text{E}\degree_{(\text{Fe}^{3+}/\text{Fe}^{2+})-}\text{E}\degree_{\bigg(\frac{1}{2}\text{Br}_2/\text{Br}^{\normalsize-}\bigg)}=0.77-1.09=-0.32\text{V}\\\text{This reaction is not feasible}\\\text{(iv) Ag(s) + Fe}^{3+} (aq)\xrightarrow{}\text{Ag}^{+}(aq)+\text{Fe}^{2+}(\text{aq})$$
E°cell = E°(Fe3+/Fe2+) – E°(Ag+/Ag) = 0.77 – 0.80 = – 0.03V
This reaction is not feasible.
$$\text{(V)}\frac{1}{2}\text{Br}_2(aq)+\text{Fe}^{2+}(aq)\xrightarrow{}\text{Br}^{\normalsize-}(aq)+\text{Fe}^{3+}(aq)\\\text{E}\degree_{\text{cell}}=\text{E}\degree_{\bigg(\frac{1}{2}\text{Br}_2/\text{Br}^{\normalsize-}\bigg)}-\text{E}\degree_{(\text{Fe}^{3+}/\text{Fe}^{2+})}=1.09-0.77=0.32\text{V}\\\text{This reaction is feasible.}$$
18. Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO3 with silver electrodes.
(ii) An aqueous solution o AgNO3 with platinum electrodes.
(iii) A dilute solution of H2SO4 with platinum electrodes.
(iv) An aqueous solution of CuCl2 with platinum electrodes.
Ans. (i) At Cathode:
The following reduction reactions compete to take place
$$\text{Ag}^{+}(aq)+e^{\normalsize-}\xrightarrow{}\text{Ag(s)}\\\text{H}^{+}(aq)+e^{\normalsize-}\xrightarrow{}\frac{1}{2}\text{H}_2$$
The reaction with a higher value of E° takes place at the cathode. Therefore, deposition of silver will takes place at the cathode.
At Anode:
The Ag anode is attacked by NO–3 ions. Therefore, the silver electrode at the anode dissolves in the solution to form Ag+.
(ii) At Cathode:
The following reduction reactions compete to take place
$$\text{Ag}^{+}(aq)+e^{\normalsize-}\xrightarrow{}\text{Ag(s)}\\\text{H}^{+}(aq)+e^{\normalsize-}\xrightarrow{}\frac{1}{2}\text{H}_2$$
The reaction with a higher value of E° takes place at the cathode. Therefore deposition of silver will takes place at the cathode.
At Anode:
Since Pt electrodes are inert, the anode is not attacked by NO3– ions. Therefore OH– or NO3– ions can be oxidized at the anode. But OH– ions having a lower discharge potential and get prefrence and decompose to liberate O2.
$$\text{OH}^{\normalsize-}\xrightarrow{}\text{OH}+e^{\normalsize-}\\\text{4OH}^{\normalsize-}\xrightarrow{}2\text{H}_2\text{O}+\text{O}_2\\\text{At Cathode:}\\\text{The following reduction reaction occurs to produce H}_2\space\text{gas}\\\text{H}^{\normalsize+}(aq)+e^{\normalsize-}\xrightarrow{}\frac{1}{2}\text{H}_2(\text{g})\\\text{(iii)}\space\text{At Anode:}\\2\text{H}_2\text{O(l)}\xrightarrow{}\text{O}_2\text{(g)}+4\text{H}^{\normalsize+}(aq)+4e^{\normalsize-}\space...(\text{i})\\\text{2SO}_4^{2\normalsize-}\xrightarrow{}\text{S}_2\text{O}_6^{2\normalsize-}+2e^{\normalsize-}\space...\text{(ii)}$$
For dilute sulphuric acid, reaction (i) is preferred to produce O2 gas. But for concentrated sulphuric acid, reaction (ii) occurs.
(iv) At Cathode:
The following reduction reactions compete to take place
$$\text{Cu}^{2+}(aq)+2e^{\normalsize-}\xrightarrow{}\text{Cu(s)}\\\text{H}^{+}(aq)+e^{\normalsize-}\xrightarrow{}\frac{1}{2}\text{H}_2(g)$$
The reaction with higher value of E° takes place at the cathode. Therefore, deposition of copper will takes place at cathode.
At Anode:
The following oxidation reactions are possible at the anode.
$$\text{Cl}^{\normalsize-}(aq)\xrightarrow{}\frac{1}{2}\text{Cl}_2(g)+e^{\normalsize-}\\2\text{H}_2\text{O(l)}\xrightarrow{}\text{O}_2\text{(g)}+4\text{H}^{\normalsize+}(aq)+4e^{\normalsize-}$$
At the anode, the reaction will lower the value of E° is preferred. But due to the over potential of O2, Cl– get oxidised at the anode to produce Cl2.
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