Calculate the E.M.F of the following cell at 298 K:
`Mg//Mg^(2+) (0.130 M)"||" Ag^(+) (1.0 xx 10^(-4)M) "|"Ag`
Given that `E_((Mg^(2+)"|"Mg))^(Theta)= -2.37V, E_((Ag^(+)"|"Ag))^(@)=0.80V`

Explanation

To calculate the E.M.F (Electromotive Force) of the given cell at 298 K, we can follow these steps:

### Step 1: Identify the half-reactions
The cell consists of two half-reactions:
1. **Anode (oxidation)**:
\[ \text{Mg} \rightarrow \text{Mg}^{2+} + 2e^- \]
2. **Cathode (reduction)**:
\[ \text{Ag}^{+} + e^- \rightarrow \text{Ag} \]

### Step 2: Write the overall cell reaction
To balance the electrons, we multiply the cathode reaction by 2:
\[ 2 \text{Ag}^{+} + 2e^- \rightarrow 2 \text{Ag} \]
Now, adding the two half-reactions gives the overall cell reaction:
\[ \text{Mg} + 2 \text{Ag}^{+} \rightarrow \text{Mg}^{2+} + 2 \text{Ag} \]

### Step 3: Calculate the standard cell potential (E°cell)
Using the standard reduction potentials provided:
- For the cathode (Ag): \(E^\circ_{\text{Ag}^{+}/\text{Ag}} = 0.80 \, \text{V}\)
- For the anode (Mg): \(E^\circ_{\text{Mg}^{2+}/\text{Mg}} = -2.37 \, \text{V}\)

The standard cell potential is calculated as:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.80 - (-2.37) = 0.80 + 2.37 = 3.17 \, \text{V} \]

### Step 4: Apply the Nernst equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]
Where \(n\) is the number of moles of electrons transferred (which is 2 in this case).

### Step 5: Substitute concentrations into the Nernst equation
The concentrations are:
- \([\text{Mg}^{2+}] = 0.130 \, \text{M}\)
- \([\text{Ag}^{+}] = 1.0 \times 10^{-4} \, \text{M}\)

Substituting into the Nernst equation:
\[ E_{\text{cell}} = 3.17 - \frac{0.0591}{2} \log \left( \frac{0.130^2}{(1.0 \times 10^{-4})^2} \right) \]

### Step 6: Calculate the logarithm
Calculating the logarithm:
\[ \log \left( \frac{0.130^2}{(1.0 \times 10^{-4})^2} \right) = \log \left( \frac{0.0169}{1.0 \times 10^{-8}} \right) = \log(1.69 \times 10^{6}) \approx 6 + \log(1.69) \approx 6 + 0.227 = 6.227 \]

### Step 7: Substitute back into the Nernst equation
Now substituting back:
\[ E_{\text{cell}} = 3.17 - 0.02955 \times 6.227 \]
Calculating this gives:
\[ E_{\text{cell}} = 3.17 - 0.183 \]
\[ E_{\text{cell}} \approx 2.987 \, \text{V} \approx 2.96 \, \text{V} \]

### Final Answer
The E.M.F of the cell is approximately **2.96 V**.

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