The EMF of the following cell is found to be 0.20 V at 298 K
`Cd|Cd^(2+)(?)||Ni^(2+)(2.0M)|Ni`
What is the molar concentration of `Cd^(2+)` ions in the solution?
`(E_(Cd^(2+)//Cd)^(@)=-0.40V,E_(Ni^(2+)//Ni)^(@)=-0.25V)`

Explanation

To find the molar concentration of \( \text{Cd}^{2+} \) ions in the solution, we can use the Nernst equation, which relates the cell potential (EMF) to the concentrations of the reactants and products.

### Step-by-Step Solution:

1. **Identify the Cell Reaction:**
The cell reaction can be represented as:
\[
\text{Cd}^{2+} + 2 \text{e}^- \rightarrow \text{Cd} \quad (E^\circ = -0.40 \, \text{V})
...