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 reaction:

\[ \text{Ni(s)} + 2\text{Ag}^+(0.002 \, \text{M}) \rightarrow \text{Ni}^{2+}(0.160 \, \text{M}) + 2\text{Ag(s)} \]

Given that \( E^\circ_{\text{cell}} = 1.05 \, \text{V} \), we will use the Nernst equation:

### Step 1: Identify the half-reactions
The half-reactions for this cell can be written as:
...