Electrochemistry Class 12 NCERT Solutions

Step-by-step solution

1. 1Oxidation happens at the anode (Zn loses 2 electrons), reduction at the cathode (Ag⁺ gains electrons):
2. 2
   $$\text{Anode (oxidation): }Zn(s)\to Z{n}^{2+}(aq)+2e^{-}$$
3. 3
   $$\text{Cathode (reduction): }A{g}^{+}(aq)+e^{-}\to Ag(s)$$
4. 4Balanced overall with 2 electrons transferred. Cell notation, anode on the left:
5. 5
   $$Zn(s)|Z{n}^{2+}(aq)||A{g}^{+}(aq)|Ag(s)$$

Step-by-step solution

1. 1Standard cell potential = cathode potential − anode potential. Ag is the cathode, Zn the anode:
2. 2
   $$E_{\text{cell}}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }=0.80-(-0.76)$$
3. 3
   $$E_{\text{cell}}^{\circ }=1.56\text{V}$$
4. 4Positive E°cell confirms the cell reaction is spontaneous as written.

Step-by-step solution

1. 1Gibbs energy from cell potential (n = 2):
2. 2
   $$\Delta G^{\circ }=-nF{E}_{\text{cell}}^{\circ }=-2\times 96500\times 1.56$$
3. 3
   $$\Delta G^{\circ }=-3.01\times 10^5{\text{J mol}}^{-1}\approx -301{\text{kJ mol}}^{-1}$$
4. 4Equilibrium constant from $\Delta G^{\circ }=-RT\ln K$ (R = 8.314, T = 298 K):
5. 5
   $$\ln K=\frac{-\Delta G^{\circ }}{RT}=\frac{3.01\times 10^5}{8.314\times 298}\approx 121.5$$
6. 6
   $$K=e^{121.5}\approx 10^{52.8}$$

Step-by-step solution

1. 1Nernst equation at 298 K using the simplified log form:
2. 2
   $$E_{\text{cell}}=E_{\text{cell}}^{\circ }-\frac{0.0591}{n}\log Q$$
3. 3Reaction quotient for $Zn+2A{g}^{+}\to Z{n}^{2+}+2Ag$ is $Q=\frac{[Z{n}^{2+}]}{[A{g}^{+}{]}^2}$:
4. 4
   $$Q=\frac{0.01}{(0.1{)}^2}=\frac{0.01}{0.01}=1$$
5. 5
   $$E_{\text{cell}}=1.56-\frac{0.0591}{2}\log 1=1.56-0=1.56\text{V}$$

Step-by-step solution

1. 1Charge passed, then Faraday's first law. Cu²⁺ + 2e⁻ → Cu, so n = 2:
2. 2
   $$Q=It=2\times 965=1930\text{C}$$
3. 3
   $$m=\frac{QM}{nF}=\frac{1930\times 63.5}{2\times 96500}$$
4. 4
   $$m=\frac{122555}{193000}\approx 0.635\text{g}$$

Step-by-step solution

1. 1Molar conductivity relates conductivity to concentration (units checked):
2. 2
   $${\Lambda }_m=\frac{\kappa \times 1000}{c}$$
3. 3
   $${\Lambda }_m=\frac{1.2\times 10^{-3}\times 1000}{0.01}=\frac{1.2}{0.01}=120{\text{S cm}}^2{\text{mol}}^{-1}$$

Step-by-step solution

1. 1Fe³⁺ + 3e⁻ → Fe, so n = 3. First find the charge passed:
2. 2
   $$Q=2.68\times (30\times 60)=2.68\times 1800=4824\text{C}$$
3. 3
   $$m=\frac{QM}{nF}=\frac{4824\times 55.85}{3\times 96500}$$
4. 4
   $$m=\frac{269420}{289500}\approx 0.931\text{g}$$

Step-by-step solution

1. 1Cell constant = conductivity × resistance:
2. 2
   $$G^{\ast }=\kappa \times R=1.409\times 10^{-3}\times 712$$
3. 3
   $$G^{\ast }\approx 1.003{\text{cm}}^{-1}$$