Question

At $298 \text{ K}$ the equilibrium constant for the reaction, $\text{M}(\text{s})+2\text{Ag}^+(\text{aq}) \rightleftharpoons \text{M}^{2+}(\text{aq})+2\text{Ag}(\text{s})$ is $10^{15}$. What is the $E_{\text{cell}}^{\circ}$ (in $\text{V}$) for this reaction? ($\frac{2.303 \text{ RT}}{F} = 0.06 \text{ V}$)

• A. 0.45
• B. 0.90
• C. 0.225
• D. 1.10

Hint:

The relation $E_{\text{cell}}^{\circ} = \frac{2.303 \text{ RT}}{nF} \log K$ is a key equation in electrochemistry. In many exam problems, the factor $\frac{2.303 \text{ RT}}{F}$ is approximated as $0.0592 \text{ V}$ or $0.06 \text{ V}$ at $298 \text{ K}$ for simpler calculation.

Answer 1

The Correct Option is A

Step 1: State the relationship between the standard cell potential and the equilibrium constant.
The standard cell potential ($E_{\text{cell}}^{\circ}$) is related to the equilibrium constant ($K$) by the Nernst equation at equilibrium: \[ E_{\text{cell}}^{\circ} = \frac{2.303 \text{ RT}}{nF} \log K. \] The problem provides the value of $\frac{2.303 \text{ RT}}{F} = 0.06 \text{ V}$ (which is an approximation of the value at $298 \text{ K}$).

Step 2: Determine the number of electrons transferred ($n$).
The reaction is $\text{M}(\text{s})+2\text{Ag}^+(\text{aq}) \rightleftharpoons \text{M}^{2+}(\text{aq})+2\text{Ag}(\text{s})$. The oxidation half-reaction is $\text{M} \to \text{M}^{2+} + 2e^-$. The reduction half-reaction is $2\text{Ag}^+ + 2e^- \to 2\text{Ag}$. The total number of electrons transferred is $n=2$.

Step 3: Substitute the known values into the equation.
We are given $K = 10^{15}$ and $n=2$. \[ E_{\text{cell}}^{\circ} = \frac{1}{n} \left(\frac{2.303 \text{ RT}}{F}\right) \log K. \] \[ E_{\text{cell}}^{\circ} = \frac{1}{2} (0.06 \text{ V}) \log(10^{15}). \]

Step 4: Calculate the standard cell potential.
Using the logarithm property $\log(10^x) = x$: \[ E_{\text{cell}}^{\circ} = \frac{1}{2} (0.06) \times 15 \text{ V}. \] \[ E_{\text{cell}}^{\circ} = 0.03 \times 15 \text{ V} = 0.45 \text{ V}. \]