Question

Calculate emf of the following cell at 25°C:
\(\text{Zn (s)} | \text{Zn}^{2+} (0.001M) || \text{Cd}^{2+} (0.1M) | \text{Cd (s)}\) Given: \( E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \, \text{V}, E^\circ_{\text{Cd}^{2+}/\text{Cd}} = -0.40 \, \text{V} \)
\([ \log 10 = 1 ]\)

Hint:

The Nernst equation is essential for calculating the emf of a cell under non-standard conditions. Ensure to substitute the correct values of concentrations and standard electrode potentials.

Answer 1

The emf of the cell can be calculated using the Nernst equation: \[ E = E^\circ - \frac{0.0591}{n} \log Q \] Where:
- \( E^\circ = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \),
- \( n \) is the number of moles of electrons transferred (here, \(n = 2\) for the zinc and cadmium half-reactions),
- \( Q \) is the reaction quotient, which for the given cell is: \[ Q = \frac{[\text{Zn}^{2+}][\text{Cd}^{2+}]}{[\text{Cd}^{2+}][\text{Zn}^{2+}]} \] Given concentrations are \( [\text{Zn}^{2+}] = 0.001M \) and \( [\text{Cd}^{2+}] = 0.1M \). Therefore, \[ Q = \frac{0.001 \times 0.1}{1} = 0.0001 \] Now, substituting the values into the Nernst equation: \[ E = (-0.40 - (-0.76)) - \frac{0.0591}{2} \log (0.0001) \] \[ E = 0.36 - \frac{0.0591}{2} \times (-4) \] \[ E = 0.36 + 0.1182 \] \[ E = 0.4782 \, \text{V} \] So, the emf of the cell is \( 0.4782 \, \text{V} \).