The EMF of the cell, `Zn|Zn^(2+) (0.1 M) || Cd^(2+) (M_(1)) |Cd` has been found to be 0.3305 V at 298 K . Calculate the value of `M_(1) [E_((Zn^(2+) |Zn))^(@) = - 0.76 V , E_((Cd^(2+) |Cd))^(Theta) = -0.40 V]`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the half-reactions
For the given cell, we need to identify the anodic (oxidation) and cathodic (reduction) reactions.

- **Anodic Reaction (Oxidation)**:
\[
\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^{-}
...