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 need to calculate the cell e.m.f. (electromotive force) and the change in Gibbs free energy (ΔG) for the given electrochemical cell reaction at 298 K.

### Step 1: Identify the half-reactions and their standard electrode potentials
The cell consists of two half-reactions:
1. Anode (oxidation): \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \) with \( E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.763 \, \text{V} \)
2. Cathode (reduction): \( \text{Cd}^{2+} + 2e^- \rightarrow \text{Cd} \) with \( E^\circ_{\text{Cd}^{2+}/\text{Cd}} = -0.403 \, \text{V} \)

### Step 2: Write the overall cell reaction
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