Write the nearest equation and calculate the e.m.f. of the following cell at 298 K
`Cu(s)|Cu^(2+)(0.130M)||Ag^(+)(1.00xx10^(-4)M)|Ag(s)`
Given `:E_(Cu^(2+)//Cu)^(@)=0.34V` and `E_(Ag^(+)//Ag)^(@)=+0.80V`

Explanation

To calculate the e.m.f. of the given electrochemical cell at 298 K, we will follow these steps:

### Step 1: Write the half-reactions
For the given cell `Cu(s)|Cu^(2+)(0.130M)||Ag^(+)(1.00x10^(-4)M)|Ag(s)`, we need to identify the oxidation and reduction half-reactions.

**Oxidation half-reaction (at anode):**
\[ \text{Cu(s)} \rightarrow \text{Cu}^{2+} + 2e^- \]

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