Related Formula
Ecell = E°cell - (0.059)/(n) Q$$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.059}{n} \log Q$$
Core Logic
First, write the complete balanced cell reaction.
At Anode (Oxidation): Ag(s) + Cl^-(aq) arrow AgCl(s) + e^-$Ag_{(s)} + Cl^-_{(aq)} \rightarrow AgCl_{(s)} + e^-$
At Cathode (Reduction): Fe³⁺(aq) + e^- arrow Fe²⁺(aq)$Fe^{3+}_{(aq)} + e^- \rightarrow Fe^{2+}_{(aq)}$
Overall cell reaction:
Ag(s) + Cl^-(aq) + Fe³⁺(aq) arrow AgCl(s) + Fe²⁺(aq)$$Ag_{(s)} + Cl^-_{(aq)} + Fe^{3+}_{(aq)} \rightarrow AgCl_{(s)} + Fe^{2+}_{(aq)}$$
Step 1: Applying the Nernst Equation
Applying the Nernst equation at 298 K$298 \text{ K}$ with n = 1$n = 1$:
Ecell = E°cell - (0.059)/(1) [Fe²⁺][Cl^-][Fe³⁺]$$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.059}{1} \log \frac{[Fe^{2+}]}{[Cl^-][Fe^{3+}]}$$
Step 2: Analysis of Variables
To increase Ecell$E_{\text{cell}}$, the value of the logarithmic term [Fe²⁺][Cl^-][Fe³⁺]$\log \frac{[Fe^{2+}]}{[Cl^-][Fe^{3+}]}$ must decrease.
This happens if the numerator decreases or the denominator increases.
- Decreasing [Fe²⁺]$[Fe^{2+}]$ (Statement C)
- Increasing [Fe³⁺]$[Fe^{3+}]$ (Statement D)
- Increasing [Cl^-]$[Cl^-]$ (Statement E)
Thus, statements C, D, and E will increase the cell potential.
Pattern Recognition
Nernst equation trick: Ecell$E_{\text{cell}}$ increases when product concentrations decrease or reactant concentrations increase (Le Chatelier's perspective of pushing the forward reaction).
Chapter Mix
Class 12 Chemistry: Electrochemistry