Question

Calculate \( E_{\text{cell}} \) of a galvanic cell in which the following reaction takes place at 25°C:
\[ \text{Zn(s)} + \text{Pb}^{2+}(0.02M) \longrightarrow \text{Zn}^{2+}(0.1M) + \text{Pb(s)} \quad \text{[Given: } E^0_{\text{Zn}^{2+}/\text{Zn}} = -0.76V, \; E^0_{\text{Pb}^{2+}/\text{Pb}} = -0.13V, \; \log 2 = 0.3010, \; \log 4 = 0.6021, \; \log 5 = 0.6990 \text{]} \]

Hint:

Remember to use the Nernst equation when solving for cell potentials at non-standard conditions, and ensure you substitute the correct values for concentration and cell configuration.

Answer 1

Using the Nernst equation for the cell potential: \[ E_{\text{cell}} = E^0_{\text{cell}} - \frac{0.0592}{n} \log Q \] Here, \( E^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}} = (-0.13V) - (-0.76V) = +0.63V \)
For this reaction, \( n = 2 \) (since 2 electrons are involved in the reaction), and the reaction quotient \( Q \) is given by: \[ Q = \frac{[\text{Zn}^{2+}]_{\text{product}} \cdot [\text{Pb}]_{\text{product}}}{[\text{Zn}]_{\text{reactant}} \cdot [\text{Pb}^{2+}]_{\text{reactant}}} = \frac{0.1}{0.02} = 5 \] Thus, we can substitute the values into the Nernst equation: \[ E_{\text{cell}} = 0.63V - \frac{0.0592}{2} \log 5 \] \[ E_{\text{cell}} = 0.63V - \frac{0.0592}{2} \times 0.6990 \] \[ E_{\text{cell}} = 0.63V - 0.0207V = 0.6093V \] Thus, the cell potential \( E_{\text{cell}} \) is 0.6093V.