Question

Calculate the e.m.f. of the following cell: \[ \text{Cu(s)} | \text{Cu}^{2+} (1M) || \text{Ag}^{+} (0.01M) | \text{Ag(s)} \] Given: \[ E^0_{\text{Cu}^{2+}/\text{Cu}} = +0.34V, \quad E^0_{\text{Ag}^{+}/\text{Ag}} = +0.80V \]

Hint:

The Nernst equation helps calculate cell potential under non-standard conditions.

Answer 1

Step 1: Cell Reaction: \[ \text{Cu} + 2\text{Ag}^{+} \rightarrow \text{Cu}^{2+} + 2\text{Ag} \] Step 2: Standard E.M.F. Calculation: \[ E_{\text{cell}}^0 = E_{\text{cathode}}^0 - E_{\text{anode}}^0 \] \[ = 0.80 - 0.34 = 0.46V \] Thus, the e.m.f. of the cell is 0.46V.