Question

The emf of the cell \(\text{Zn(s)} \,|\, \text{Zn}^{2+}(0.1\,\text{M}) \,||\, \text{Cd}^{2+}(M_1) \,|\, \text{Cd(s)}\) has been found to be \(0.3305\ \text{V}\) at \(298\,\text{K}\). Calculate the value of \(M_1\). Given: \(E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\ \text{V}\) and \(E^\circ_{\text{Cd}^{2+}/\text{Cd}} = -0.40\ \text{V}\).

Hint:

For electrochemical concentration cells, the Nernst equation relates emf directly to ion concentrations through the reaction quotient.

Answer 1

Step 1: Write the cell reaction: \[ \text{Zn(s)} + \text{Cd}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cd(s)} \]
Step 2: Calculate the standard emf of the cell: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] \[ E^\circ_{\text{cell}} = (-0.40) - (-0.76) = 0.36\ \text{V} \]
Step 3: Apply the Nernst equation at \(298\,\text{K}\): \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n}\log Q \] Here, \(n = 2\).
Step 4: Write the reaction quotient: \[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Cd}^{2+}]} = \frac{0.1}{M_1} \]
Step 5: Substitute the given values: \[ 0.3305 = 0.36 - \frac{0.0591}{2}\log\left(\frac{0.1}{M_1}\right) \]
Step 6: Simplify: \[ 0.36 - 0.3305 = 0.02955 \log\left(\frac{0.1}{M_1}\right) \] \[ 0.0295 = 0.02955 \log\left(\frac{0.1}{M_1}\right) \]
Step 7: Solve: \[ \log\left(\frac{0.1}{M_1}\right) = 1 \] \[ \frac{0.1}{M_1} = 10 \] \[ M_1 = 0.01 \times 10 = 1.0\ \text{M} \]