Question

\( E^\circ_{\text{Cell}} = 1.1 \, \text{V}. \, \text{Find} \, E_{\text{Cell}} \, \text{for the reaction:} \, \text{Zn} + \text{Cu}^{2+} \, (0.1M) \rightleftharpoons \text{Zn}^{2+} \, (0.001M) + \text{Cu} \)

• A. 1.05 V
• B. 1.2 V
• C. 1.1 V
• D. 1.0 V

Hint:

Use the Nernst equation to calculate the cell potential at non-standard conditions. Remember that the logarithmic term adjusts the potential based on ion concentrations.

Answer 1

The Correct Option is A

We are given the standard cell potential \( E^\circ_{\text{Cell}} = 1.1 \, \text{V} \), and the concentrations of \( \text{Zn}^{2+} \) and \( \text{Cu}^{2+} \) are 0.001 M and 0.1 M, respectively. We can use the **Nernst equation** to calculate the cell potential at these concentrations: \[ E_{\text{Cell}} = E^\circ_{\text{Cell}} - \frac{0.0592}{n} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \right) \] Where: - \( E^\circ_{\text{Cell}} = 1.1 \, \text{V} \), - \( n = 2 \) (since the change in oxidation state for both Zn and Cu is 2), - \( [\text{Zn}^{2+}] = 0.001 \, \text{M} \), - \( [\text{Cu}^{2+}] = 0.1 \, \text{M} \). ### Step 1: Apply the Nernst Equation Substitute the values into the Nernst equation: \[ E_{\text{Cell}} = 1.1 - \frac{0.0592}{2} \log \left( \frac{0.001}{0.1} \right) \] \[ E_{\text{Cell}} = 1.1 - \frac{0.0592}{2} \log(0.01) \] \[ E_{\text{Cell}} = 1.1 - \frac{0.0592}{2} \times (-2) \] \[ E_{\text{Cell}} = 1.1 + 0.0592 \] \[ E_{\text{Cell}} = 1.05 \, \text{V} \] Thus, the cell potential at the given concentrations is: \[ \boxed{(A) \, 1.05 \, \text{V}} \]