Standard Reduction Potentials and Cell Spontaneity for Metal Pairs

To determine which combinations of metal rods $\text{X}$ and $\text{Y}$ lead to the dissolution of metal $\text{X}$, we analyze the overall cell reaction and calculate the cell potential ($E_{\text{cell}}$) at $298\text{ K}$.

1. Cell Reactions and Nernst Equation

The dissolution of metal $\text{X}$ means that $\text{X}$ undergoes oxidation at the anode: $\text{Anode (Oxidation): } \text{X(s)} \rightarrow \text{X}^{2+}(\text{aq}) + 2\text{e}^-$

The metal $\text{Y}^{2+}$ ions undergo reduction at the cathode: $\text{Cathode (Reduction): } \text{Y}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Y(s)}$

Overall Cell Reaction: $\text{X(s)} + \text{Y}^{2+}(\text{aq}) \rightarrow \text{X}^{2+}(\text{aq}) + \text{Y(s)}$

The number of electrons transferred is $n = 2$. The reaction quotient $Q$ for the cell is: $Q = \frac{[\text{X}^{2+}]}{[\text{Y}^{2+}]} = \frac{0.001\text{ M}}{0.1\text{ M}} = \frac{10^{-3}}{10^{-1}} = 10^{-2}$

Using the Nernst equation at $298\text{ K}$: $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{2.303 R T}{n F} \log_{10} Q$

Given values: $\frac{2.303 R T}{F} = \frac{2.303 \times 8.314 \times 298}{96500} \approx 0.0591\text{ V}$

Substituting $n = 2$ and $Q = 10^{-2}$: $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{2} \log_{10}(10^{-2})$ $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{2} (-2) = E^\circ_{\text{cell}} + 0.0591\text{ V}$

Where the standard cell potential is given by: $E^\circ_{\text{cell}} = E^\circ_{\text{Y}^{2+}/\text{Y}} - E^\circ_{\text{X}^{2+}/\text{X}}$

For the dissolution of metal $\text{X}$ to be spontaneous, we require: $E_{\text{cell}} > 0 \implies E^\circ_{\text{cell}} + 0.0591\text{ V} > 0 \implies E^\circ_{\text{cell}} > -0.0591\text{ V}$

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2. Evaluation of Options

Given standard reduction potentials ($E^\circ$):

• $E^\circ_{\text{Pb}^{2+}/\text{Pb}} = -0.13\text{ V}$
• $E^\circ_{\text{Ni}^{2+}/\text{Ni}} = -0.24\text{ V}$
• $E^\circ_{\text{Cd}^{2+}/\text{Cd}} = -0.40\text{ V}$
• $E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44\text{ V}$

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Option A: $\text{X} = \text{Cd}$ and $\text{Y} = \text{Ni}$

$E^\circ_{\text{cell}} = E^\circ_{\text{Ni}^{2+}/\text{Ni}} - E^\circ_{\text{Cd}^{2+}/\text{Cd}} = -0.24\text{ V} - (-0.40\text{ V}) = +0.16\text{ V}$ $E_{\text{cell}} = +0.16\text{ V} + 0.0591\text{ V} = +0.2191\text{ V} > 0$ Since $E_{\text{cell}} > 0$, the reaction is spontaneous and metal $\text{Cd}$ dissolves. (Valid)

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Option B: $\text{X} = \text{Cd}$ and $\text{Y} = \text{Fe}$

$E^\circ_{\text{cell}} = E^\circ_{\text{Fe}^{2+}/\text{Fe}} - E^\circ_{\text{Cd}^{2+}/\text{Cd}} = -0.44\text{ V} - (-0.40\text{ V}) = -0.04\text{ V}$ $E_{\text{cell}} = -0.04\text{ V} + 0.0591\text{ V} = +0.0191\text{ V} > 0$ Since $E_{\text{cell}} > 0$, the reaction is spontaneous and metal $\text{Cd}$ dissolves. (Valid)

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Option C: $\text{X} = \text{Ni}$ and $\text{Y} = \text{Pb}$

$E^\circ_{\text{cell}} = E^\circ_{\text{Pb}^{2+}/\text{Pb}} - E^\circ_{\text{Ni}^{2+}/\text{Ni}} = -0.13\text{ V} - (-0.24\text{ V}) = +0.11\text{ V}$ $E_{\text{cell}} = +0.11\text{ V} + 0.0591\text{ V} = +0.1691\text{ V} > 0$ Since $E_{\text{cell}} > 0$, the reaction is spontaneous and metal $\text{Ni}$ dissolves. (Valid)

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Option D: $\text{X} = \text{Ni}$ and $\text{Y} = \text{Fe}$

$E^\circ_{\text{cell}} = E^\circ_{\text{Fe}^{2+}/\text{Fe}} - E^\circ_{\text{Ni}^{2+}/\text{Ni}} = -0.44\text{ V} - (-0.24\text{ V}) = -0.20\text{ V}$ $E_{\text{cell}} = -0.20\text{ V} + 0.0591\text{ V} = -0.1409\text{ V} < 0$ Since $E_{\text{cell}} < 0$, the reaction is non-spontaneous. (Invalid)

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Conclusion

The valid combinations for $\text{X}$ and $\text{Y}$ are (A), (B), and (C).