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

The cell reaction is written as:
\[ \text{Sn}(s) \mid \text{Sn}^{2+} (\text{0.004 M}) \parallel \text{H}^+ (\text{0.02 M}) \mid \text{H}_2 (\text{1 Bar}) \mid \text{Pt}(s) \] The cell potential \(E_{\text{cell}}\) can be calculated using the Nernst equation: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \frac{[\text{products}]}{[\text{reactants}]} \]
Where:
- \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\),
- \(E^\circ_{\text{cathode}} = E^\circ_{\text{H}^+/\text{H}_2} = 0.00 \, \text{V}\),
- \(E^\circ_{\text{anode}} = E^\circ_{\text{Sn}^{2+}/\text{Sn}} = -0.14 \, \text{V}\),
- \(n = 2\) (since the reaction involves the transfer of 2 electrons).
Now, calculate the cell potential: \[ E_{\text{cell}} = 0.00 - (-0.14) - \frac{0.0591}{2} \log \frac{[\text{H}^+]^2}{[\text{Sn}^{2+}]} \]
Substitute the given concentrations: \[ E_{\text{cell}} = 0.14 - \frac{0.0591}{2} \log \frac{(0.02)^2}{0.004} \] Simplifying the logarithmic term: \[ E_{\text{cell}} = 0.14 - \frac{0.0591}{2} \log \frac{0.0004}{0.004} \] \[ E_{\text{cell}} = 0.14 - \frac{0.0591}{2} \log 0.1 \] Since \(\log 0.1 = -1\): \[ E_{\text{cell}} = 0.14 - \frac{0.0591}{2} (-1) \] \[ E_{\text{cell}} = 0.14 + 0.02955 \] \[ E_{\text{cell}} = 0.16955 \, \text{V} \] Thus, the e.m.f. of the cell is 0.170 V.