Question

Calculate the electromotive force of the following cell: \[ \text{Cu} | \text{Cu}^{2+} (1M) || \text{Ag}^{+} (1M) | \text{Ag} \] Given \( E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34V \), \( E^\circ_{\text{Ag}^{+}/\text{Ag}} = +0.80V \)

Hint:

When the concentrations of the solutions are 1 M, the electromotive force equals the standard cell potential.

Answer 1

Step 1: Write the Nernst equation
The Nernst equation for a cell is given by: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{n} \log Q \] where: - \( E_{\text{cell}} \) is the electromotive force (EMF) of the cell, - \( E^\circ_{\text{cell}} \) is the standard cell potential, - \( n \) is the number of electrons involved, - \( Q \) is the reaction quotient.
Step 2: Standard Cell Potential
The standard cell potential is: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] Since the reduction half-reaction occurs at the cathode and the oxidation half-reaction occurs at the anode: - \( E^\circ_{\text{cell}} = E^\circ_{\text{Ag}^{+}/\text{Ag}} - E^\circ_{\text{Cu}^{2+}/\text{Cu}} \) \[ E^\circ_{\text{cell}} = 0.80V - 0.34V = 0.46V \]
Step 3: Calculate the Electromotive Force
Since both solutions are 1M, the reaction quotient \( Q = 1 \), and hence the log term becomes zero. Therefore, the electromotive force is equal to the standard cell potential: \[ E_{\text{cell}} = 0.46V \] Final Answer: The electromotive force of the cell is \( 0.46V \). Correct Answer: 0.46V