Question

Calculate the emf of the following cell at 25°C: 
\[ \text{Zn(s)} | \text{Zn}^{2+}(0.1M) || \text{Cd}^{2+}(0.01M) | \text{Cd(s)} \] Given: \[ E^\circ_{\text{Cd}^{2+}/\text{Cd}} = -0.40 \, V, \, E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \, V \] \[ [\log 10 = 1] \]

Hint:

The Nernst equation allows you to calculate the emf of a cell under non-standard conditions by considering the concentrations of the ions involved in the cell reaction.

Answer 1

The emf of the cell can be calculated using the Nernst equation: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{n} \log Q \] Where: - \( E^\circ_{\text{cell}} \) is the standard cell potential - \( n \) is the number of moles of electrons exchanged (here, \(n = 2\) because 2 electrons are involved) - \( Q \) is the reaction quotient, which is the ratio of concentrations of products to reactants First, calculate the standard cell potential (\( E^\circ_{\text{cell}} \)) using the formula: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] In this case: - The cathode is the Cd/Cd\(^{2+}\) half-reaction (\(E^\circ_{\text{Cd}^{2+}/\text{Cd}} = -0.40 \, V\)) - The anode is the Zn/Zn\(^{2+}\) half-reaction (\(E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \, V\)) Thus: \[ E^\circ_{\text{cell}} = -0.40 - (-0.76) = +0.36 \, V \] Now, calculate the reaction quotient \( Q \): \[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Cd}^{2+}]} = \frac{0.1}{0.01} = 10 \] Substitute the values into the Nernst equation: \[ E_{\text{cell}} = 0.36 - \frac{0.0592}{2} \log(10) \] Since \( \log(10) = 1 \), we have: \[ E_{\text{cell}} = 0.36 - \frac{0.0592}{2} \times 1 = 0.36 - 0.0296 = 0.3304 \, V \] Thus, the emf of the cell at 25°C is 0.3304 V.