Calculate the e.m.f. of the cell in which the following reaction takes place :
`Ni(s) +2Ag^(+)(0.002 M)to Ni^(2+)(0.160 M)+2Ag(s)`
Given `E_(cell)^(@)`=1.05 v

Explanation

To calculate the e.m.f. (electromotive force) of the cell for the given reaction:
\[ \text{Ni(s) + 2Ag}^+ \text{(0.002 M)} \rightarrow \text{Ni}^{2+} \text{(0.160 M) + 2Ag(s)} \]
we will use the Nernst equation. The standard cell potential (\( E^\circ_{\text{cell}} \)) is given as 1.05 V.

### Step-by-Step Solution:

1. **Identify the half-cell reactions:**
- **Anode (oxidation):**
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