Question

Calculate the e.m.f. of the following cell writing the cell reaction: \[ \text{Cu} | \text{Cu}^{2+}(1M) || \text{Ag}^+(1M) | \text{Ag} \] Given: \[ E^0_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \text{ volt}, \quad E^0_{\text{Ag}^+/\text{Ag}} = +0.80 \text{ volt} \]

Hint:

The standard e.m.f. of a galvanic cell is calculated using \( E_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}} \).

Answer 1

Step 1: The cell reaction is: \[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad (E^0 = +0.34V) \] \[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \quad (E^0 = +0.80V) \] Step 2: The overall cell reaction: \[ \text{Cu} + 2\text{Ag}^+ \rightarrow \text{Cu}^{2+} + 2\text{Ag} \] Step 3: The e.m.f. of the cell is calculated as: \[ E_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}} \] \[ E_{\text{cell}} = 0.80 - 0.34 = 0.46V \] Thus, the e.m.f. of the cell is **0.46V**.