Question

Calculate emf of the following cell at 298 K :
\( \text{Sn} \mid \text{Sn}^{2+} (0.001 \text{ M}) \parallel \text{H}^+ (0.01 \text{ M}) \mid \text{H}_{2(g)} (1 \text{ bar}) \mid \text{Pt}_{(s)} \)
Given : \( E^\circ_{\text{Sn}^{2+}/\text{Sn}} = - 0.14 \text{ V}, \quad E^\circ_{\text{H}^+/\text{H}_2} = 0.00 \text{ V} \quad [\log 10 = 1] \)

Hint:

Don't forget the stoichiometric coefficient from the balanced equation becomes the power in the reaction quotient \( Q \). Here, \( 2\text{H}^+ \) means \( [\text{H}^+] \) must be squared!

Answer 1

Concept: The electromotive force (EMF) of a cell under non-standard conditions is calculated using the Nernst Equation: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q \] Where:

• \( E^\circ_{\text{cell}} \) is the standard cell potential.

• \( n \) is the number of moles of electrons transferred.

• \( Q \) is the reaction quotient.
Step 1: Writing the half-cell reactions and determining \(n\).

• Anode (Oxidation, left side): \( \text{Sn}_{(s)} \longrightarrow \text{Sn}^{2+}_{(aq)} + 2e^- \)

• Cathode (Reduction, right side): \( 2\text{H}^+_{(aq)} + 2e^- \longrightarrow \text{H}_{2(g)} \)
Overall cell reaction: \( \text{Sn}_{(s)} + 2\text{H}^+_{(aq)} \longrightarrow \text{Sn}^{2+}_{(aq)} + \text{H}_{2(g)} \) The number of electrons exchanged is \( n = 2 \).

Step 2: Calculating the Standard Cell Potential (\( E^\circ_{\text{cell}} \)).
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] \[ E^\circ_{\text{cell}} = E^\circ_{\text{H}^+/\text{H}_2} - E^\circ_{\text{Sn}^{2+}/\text{Sn}} \] \[ E^\circ_{\text{cell}} = 0.00 \text{ V} - (-0.14 \text{ V}) = +0.14 \text{ V} \]

Step 3: Applying the Nernst Equation to find \(E_{\text{cell}}\).
The reaction quotient \( Q = \frac{[\text{Sn}^{2+}][P_{\text{H}_2}]}{[\text{H}^+]^2} \). Since \( P_{\text{H}_2} = 1 \text{ bar} \): \[ E_{\text{cell}} = 0.14 - \frac{0.0591}{2} \log \left( \frac{[\text{Sn}^{2+}]}{[\text{H}^+]^2} \right) \] Substitute the given concentrations (\( [\text{Sn}^{2+}] = 0.001 \text{ M} = 10^{-3} \text{ M} \), \( [\text{H}^+] = 0.01 \text{ M} = 10^{-2} \text{ M} \)): \[ E_{\text{cell}} = 0.14 - 0.02955 \log \left( \frac{10^{-3}}{(10^{-2})^2} \right) \] \[ E_{\text{cell}} = 0.14 - 0.02955 \log \left( \frac{10^{-3}}{10^{-4}} \right) \] \[ E_{\text{cell}} = 0.14 - 0.02955 \log (10^1) \]

Step 4: Final Calculation.
Since \( \log 10 = 1 \): \[ E_{\text{cell}} = 0.14 - (0.02955 \times 1) \] \[ E_{\text{cell}} = 0.14 - 0.02955 = 0.11045 \text{ V} \] Rounding to appropriate significant figures, \( E_{\text{cell}} \approx 0.11 \text{ V} \). Final Answer: The emf of the given cell is \( 0.11 \text{ V} \).