Question

Calculate emf of the following cell at 25°C:
Sn/Sn$^{2+}$(0.001 M) || H$^+$(0.01 M) | H$_2$(g)(1 bar) | Pt(s)
Given: E°(Sn$^{2+}$/Sn) = -0.14 V, E° H$^+$/H$_2$ = 0.00 V (log 10 = 1)

Hint:

Use the Nernst equation to calculate the emf of electrochemical cells, considering concentrations and standard electrode potentials.

Answer 1

We are given the following data:
- The cell consists of two half-reactions:
1. \( \text{Sn}^{2+} + 2e^- \rightarrow \text{Sn} \) with \( E^\circ = -0.14 \, \text{V} \)
2. \( 2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \) with \( E^\circ = 0.00 \, \text{V} \)
- Concentration of Sn$^{2+}$ = 0.001 M
- Concentration of H$^+$ = 0.01 M
- Standard pressure for H$_2$ = 1 bar
Step 1: Write the Nernst equation for the cell:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{n} \log Q \] Where:
- \( E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \)
- \( n \) is the number of electrons transferred (in this case, \( n = 2 \)).
- \( Q \) is the reaction quotient, which is calculated as:
\[ Q = \frac{[\text{Sn}^{2+}]}{[\text{H}^+]^2} \] Substitute the given values:
\[ Q = \frac{0.001}{(0.01)^2} = \frac{0.001}{0.0001} = 10 \] Step 2: Calculate the standard cell potential \( E^\circ_{\text{cell}} \):
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.00 \, \text{V} - (-0.14 \, \text{V}) = 0.14 \, \text{V} \] Step 3: Apply the values to the Nernst equation:
\[ E_{\text{cell}} = 0.14 \, \text{V} - \frac{0.0592}{2} \log 10 \] Since \( \log 10 = 1 \), we get:
\[ E_{\text{cell}} = 0.14 \, \text{V} - \frac{0.0592}{2} \times 1 \] \[ E_{\text{cell}} = 0.14 \, \text{V} - 0.0296 \, \text{V} = 0.1104 \, \text{V} \] Final Answer: The emf of the cell at 25°C is \( E_{\text{cell}} = 0.1104 \, \text{V} \).