Calculate the emf for the following cell at 298 K.
`Cd//Cd^(2+)(0.1 M) || Ag^(+)(0.1 M)//Ag`
`"Given"E_(Cd^(2+)//Cd)^(@)=-0.40" V ",E_(Ag^(+)//Ag)^(@)=0.80" V "`

Explanation

To calculate the electromotive force (emf) for the cell represented as `Cd//Cd^(2+)(0.1 M) || Ag^(+)(0.1 M)//Ag`, we will follow these steps:

### Step 1: Identify the half-reactions and their standard electrode potentials.
The half-reactions for the cell are:
1. For cadmium:
\[ \text{Cd}^{2+} + 2e^- \rightarrow \text{Cd} \quad E^\circ_{\text{Cd}^{2+}/\text{Cd}} = -0.40 \, \text{V} \]
2. For silver:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \quad E^\circ_{\text{Ag}^+/\text{Ag}} = 0.80 \, \text{V} \]

### Step 2: Calculate the standard cell potential \(E^\circ_{\text{cell}}\).
The standard cell potential can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Here, silver acts as the cathode (reduction occurs) and cadmium acts as the anode (oxidation occurs):
\[ E^\circ_{\text{cell}} = E^\circ_{\text{Ag}^+/\text{Ag}} - E^\circ_{\text{Cd}^{2+}/\text{Cd}} = 0.80 \, \text{V} - (-0.40 \, \text{V}) = 0.80 \, \text{V} + 0.40 \, \text{V} = 1.20 \, \text{V} \]

### Step 3: Use the Nernst equation to calculate the cell potential under non-standard conditions.
The Nernst equation is given by:
\[ E = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q \]
where:
- \(n\) is the number of moles of electrons transferred (2 for this reaction),
- \(Q\) is the reaction quotient.

For the reaction:
\[ \text{Cd} + 2\text{Ag}^+ \rightarrow \text{Cd}^{2+} + 2\text{Ag} \]
The reaction quotient \(Q\) is given by:
\[ Q = \frac{[\text{Cd}^{2+}]}{[\text{Ag}^+]^2} = \frac{0.1}{(0.1)^2} = \frac{0.1}{0.01} = 10 \]

### Step 4: Substitute the values into the Nernst equation.
Now substituting the values into the Nernst equation:
\[ E = 1.20 \, \text{V} - \frac{0.0591}{2} \log(10) \]
Since \(\log(10) = 1\):
\[ E = 1.20 \, \text{V} - \frac{0.0591}{2} \cdot 1 \]
\[ E = 1.20 \, \text{V} - 0.02955 \, \text{V} \]
\[ E \approx 1.17045 \, \text{V} \]

### Step 5: Round the final answer.
Rounding to two decimal places, we get:
\[ E \approx 1.17 \, \text{V} \]

### Final Answer:
The emf of the cell is approximately **1.17 V**.

---