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. (electromotive force) of the given electrochemical cell at 298 K, we will follow these steps:

### Step 1: Identify the half-reactions and standard electrode potentials
The cell notation provided is:
\[ \text{Cu(s)} | \text{Cu}^{2+}(0.130M) || \text{Ag}^{+}(1.00 \times 10^{-4}M) | \text{Ag(s)} \]

From the information given:
- The half-reaction for copper is:
...