Calculate the e.m.f. of the cell in which the redox reaction is :
`Mg(s)+2Ag+(aq) to Mg^(2+)(aq)+2Ag(s)` when `[Mg^(2+)]=0.130` M and `[Ag^(+)]=1.0xx10^(-4) M`. Given `E_(Mg^(2+)//Mg)^(@)=-2.37 V` and `E_(Ag^(+)//Ag)^(@)=+0.80 V`.

Explanation

To calculate the e.m.f. (electromotive force) of the cell for the given redox reaction:

**Step 1: Identify the half-reactions.**
- At the anode, magnesium (Mg) is oxidized to magnesium ions (Mg²⁺):
\[ \text{Mg(s)} \rightarrow \text{Mg}^{2+}(aq) + 2e^- \]
- At the cathode, silver ions (Ag⁺) are reduced to silver (Ag):
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