Following cell is set up between copper and silver electrodes `Cu//Cu^(2+)(aq)"||"Ag^(+)//Ag`. If two half-cells work under standard conditions, calculate the e.m.f. of the cell. [Given `E^(Theta)Cu^(2+)//Cu (E^(Theta)" reduction")= +0.34"volt. "E^(Theta)Ag^(+)//Ag(E^(Theta)" reduction")= +0.80" volt"`]

Explanation

To calculate the electromotive force (e.m.f.) of the cell set up between copper and silver electrodes, we can follow these steps:

### Step 1: Identify the half-reactions and their standard reduction potentials.
- The half-reaction for copper is:
\[ Cu^{2+} + 2e^- \rightarrow Cu \quad (E^\circ_{Cu^{2+}/Cu} = +0.34 \, \text{V}) \]
- The half-reaction for silver is:
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