Calculate the cell e.m.f. and `DeltaG` for the cell reaction at 298K for the cell.
`Zn(s) | Zn^(2+) (0.0004M) ||Cd^(2+) (0.2M)|Cd(s)`
Given, `E_(Zn^(2+)//Zn)^(@) =- 0.763 V, E_(Cd^(+2)//Cd)^(@) = - 0.403 V` at `298K`.
`F = 96500 C mol^(-1)`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the Cell Reaction
The cell consists of a zinc electrode and a cadmium electrode. The half-reactions are:
- Oxidation (at anode): \( \text{Zn(s)} \rightarrow \text{Zn}^{2+}(aq) + 2e^- \)
- Reduction (at cathode): \( \text{Cd}^{2+}(aq) + 2e^- \rightarrow \text{Cd(s)} \)

Combining these half-reactions, the overall cell reaction is:
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