Calculate the e.m.f. of the following cell at 298 K
`Co//Co^(2+)(0.1M)"||"Ag^(+) (0.1M)//Ag` Given
`E_(Co^(2+)//Co)^(Theta)= -0.28 V, E_(Ag^(+)//Ag)^(Theta)=0.8V`

`R = 8.31JK^(-1)"mol"^(-), F = 96500` coulombs.

Explanation

To calculate the e.m.f. (electromotive force) of the given electrochemical cell at 298 K, we will follow these steps:

### Step 1: Identify the Anode and Cathode
In the given cell notation `Co//Co^(2+)(0.1M) || Ag^(+) (0.1M)//Ag`, the left side represents the anode and the right side represents the cathode.

- **Anode (oxidation)**: Cobalt (Co) is oxidized to Co²⁺.
- **Cathode (reduction)**: Silver ions (Ag⁺) are reduced to silver (Ag).

### Step 2: Write the Half-Reactions
1. **Oxidation at Anode**:
\[ \text{Co} \rightarrow \text{Co}^{2+} + 2e^{-} \quad \text{(Equation 1)} \]

2. **Reduction at Cathode**:
\[ \text{Ag}^{+} + e^{-} \rightarrow \text{Ag} \quad \text{(Equation 2)} \]

### Step 3: Balance the Overall Reaction
To balance the overall cell reaction, we need to ensure that the number of electrons lost in oxidation equals the number of electrons gained in reduction. Since the oxidation of Co produces 2 electrons, we multiply the reduction reaction by 2:

\[ 2 \text{Ag}^{+} + 2e^{-} \rightarrow 2 \text{Ag} \quad \text{(Equation 2 multiplied by 2)} \]

### Step 4: Write the Overall Cell Reaction
Combining the half-reactions gives:
\[ \text{Co} + 2 \text{Ag}^{+} \rightarrow \text{Co}^{2+} + 2 \text{Ag} \]

### Step 5: Calculate the Standard Cell Potential (E°cell)
Using the standard electrode potentials provided:
- \( E^{\circ}_{\text{Co}^{2+}/\text{Co}} = -0.28 \, \text{V} \)
- \( E^{\circ}_{\text{Ag}^{+}/\text{Ag}} = 0.80 \, \text{V} \)

The standard cell potential is calculated as:
\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = 0.80 \, \text{V} - (-0.28 \, \text{V}) = 0.80 + 0.28 = 1.08 \, \text{V} \]

### Step 6: Apply the Nernst Equation
The Nernst equation is given by:
\[ E = E^{\circ}_{\text{cell}} - \frac{RT}{nF} \ln Q \]
Where:
- \( R = 8.31 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( T = 298 \, \text{K} \)
- \( n = 2 \) (number of electrons transferred)
- \( F = 96500 \, \text{C/mol} \)
- \( Q = \frac{[\text{Co}^{2+}]}{[\text{Ag}^{+}]^2} = \frac{0.1}{(0.1)^2} = 10 \)

Substituting the values into the Nernst equation:
\[ E = 1.08 - \frac{(8.31)(298)}{(2)(96500)} \ln(10) \]

Calculating the second term:
\[ \frac{(8.31)(298)}{(2)(96500)} = \frac{2477.58}{193000} \approx 0.01283 \]

Since \( \ln(10) \approx 2.303 \):
\[ E = 1.08 - 0.01283 \times 2.303 \approx 1.08 - 0.0295 \approx 1.0505 \, \text{V} \]

### Step 7: Final Answer
Thus, the e.m.f. of the cell at 298 K is approximately:
\[ \boxed{1.05 \, \text{V}} \]