Consider the reaction `2Ag^(+) + Cd to 2Ag + Cd^(2+)`. The standard reduction potentials of `Ag^(+)//Ag and Cd^(2+)//Cd are + 0*80 " volt and " 0*40` volt, respectively.
What will be the emf of the cell if concentration of `Cd^(2+) is 0*1 M and Ag^(+) is 0*2 M`?

Explanation

To solve the problem, we need to determine the emf (electromotive force) of the cell based on the given reaction and the concentrations of the ions involved. Here’s a step-by-step solution:

### Step 1: Identify the half-reactions and their standard reduction potentials
The given reaction is:
\[ 2Ag^+ + Cd \rightarrow 2Ag + Cd^{2+} \]

The half-reactions are:
1. Reduction of silver ions:
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