Question

Write the cell reaction and calculate the e.m.f. of the following cell at 298 K:
\[ \text{Sn}(s) \mid \text{Sn}^{2+} (\text{0.004 M}) \parallel \text{H}^+ (\text{0.02 M}) \mid \text{H}_2 (\text{1 Bar}) \mid \text{Pt}(s) \]
(Given: \( E^\circ_{\text{Sn}^{2+}/\text{Sn}} = -0.14 \, \text{V}, E^\circ_{\text{H}^+/\text{H}_2} = 0.00 \, \text{V} \))

Hint:

The Nernst equation allows for the calculation of the cell potential by taking into account the concentrations of the ions involved in the electrochemical reaction.

Answer 1

To solve the problem, we need to write the cell reaction and calculate the e.m.f. of the given cell at 298 K.

1. Understanding the Cell Setup:
The given cell is represented as:
\[ \text{Sn}(s) \mid \text{Sn}^{2+} (\text{0.004 M}) \parallel \text{H}^+ (\text{0.02 M}) \mid \text{H}_2 (\text{1 Bar}) \mid \text{Pt}(s) \]

2. Identifying the Half-Reactions:
The cell consists of two half-cells: - The first half-cell involves the reduction of tin (Sn) to tin ions (\( \text{Sn}^{2+} \)) at the anode:
\[ \text{Sn}(s) \rightarrow \text{Sn}^{2+} (\text{aq}) + 2e^- \] - The second half-cell involves the reduction of hydrogen ions (\( \text{H}^+ \)) to hydrogen gas (\( \text{H}_2 \)) at the cathode:
\[ 2e^- + 2\text{H}^+ (\text{aq}) \rightarrow \text{H}_2 (g) \]

3. Standard Cell Potential (E°):
The standard electrode potential for the Sn\(^{2+}\)/Sn half-reaction is given as \( E^\circ_{\text{Sn}^{2+}/\text{Sn}} = -0.14 \, \text{V} \), and for the H\(^+\)/H\(_2\) half-reaction, \( E^\circ_{\text{H}^+/\text{H}_2} = 0.00 \, \text{V} \).
Therefore, the standard cell potential \( E^\circ_{\text{cell}} \) is: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.00 \, \text{V} - (-0.14 \, \text{V}) = +0.14 \, \text{V} \]

4. Nernst Equation:
The Nernst equation is used to calculate the cell potential under non-standard conditions: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q \] Where: - \( E^\circ_{\text{cell}} = 0.14 \, \text{V} \) - \( n = 2 \) (the number of electrons transferred) - \( Q \) is the reaction quotient, which is given by: \[ Q = \frac{[\text{Sn}^{2+}]_{\text{anode}} [\text{H}_2]_{\text{cathode}}}{[\text{H}^+]_{\text{anode}}} = \frac{[\text{Sn}^{2+}]}{[\text{H}^+]^2} \] Substitute the given concentrations and pressures: \[ Q = \frac{0.004}{(0.02)^2} = \frac{0.004}{0.0004} = 10 \]

5. Calculating the Cell Potential:
Now, substitute the values into the Nernst equation: \[ E_{\text{cell}} = 0.14 \, \text{V} - \frac{0.0591}{2} \log 10 \] Since \( \log 10 = 1 \), we get: \[ E_{\text{cell}} = 0.14 \, \text{V} - \frac{0.0591}{2} \times 1 = 0.14 \, \text{V} - 0.02955 \, \text{V} = 0.11045 \, \text{V} \]

6. Final Answer:
The cell reaction is:
\[ \text{Sn}(s) + 2\text{H}^+ (\text{aq}) \rightarrow \text{Sn}^{2+} (\text{aq}) + \text{H}_2 (g) \]
The e.m.f. of the cell at 298 K is 0.11045 V.