Calculate the e.m.f. of the cell,
`Mg//Mg^(2+)(0.1 M)||Ag^(+)(1.0xx10^(-3)M)//Ag`
The values of `E_(Mg^(2+)//Mg)^(@)` and `E_(Ag^(+)//Ag)^(@)` are -2.37`" V"` and +0.80`" V "` respectively.

Explanation

To calculate the e.m.f. (electromotive force) of the cell represented by the notation `Mg//Mg^(2+)(0.1 M)||Ag^(+)(1.0x10^(-3)M)//Ag`, we will follow these steps:

### Step 1: Identify the standard reduction potentials
We are given the standard reduction potentials:
- \( E^\circ_{Mg^{2+}/Mg} = -2.37 \, V \)
- \( E^\circ_{Ag^{+}/Ag} = +0.80 \, V \)

### Step 2: Calculate the standard cell potential \( E^\circ_{cell} \)
The standard cell potential can be calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
In this case:
- The cathode (where reduction occurs) is silver (Ag), so \( E^\circ_{cathode} = +0.80 \, V \).
- The anode (where oxidation occurs) is magnesium (Mg), so \( E^\circ_{anode} = -2.37 \, V \).

Thus, we have:
\[ E^\circ_{cell} = 0.80 \, V - (-2.37 \, V) = 0.80 \, V + 2.37 \, V = 3.17 \, V \]

### Step 3: Determine the reaction quotient \( Q \)
The reaction quotient \( Q \) is given by the concentrations of the products over the reactants, raised to the power of their coefficients in the balanced equation. For the cell reaction:
\[ Mg + 2Ag^{+} \rightarrow Mg^{2+} + 2Ag \]
The expression for \( Q \) is:
\[ Q = \frac{[Mg^{2+}]}{[Ag^{+}]^2} \]
Substituting the concentrations:
- \( [Mg^{2+}] = 0.1 \, M \)
- \( [Ag^{+}] = 1.0 \times 10^{-3} \, M \)

Thus:
\[ Q = \frac{0.1}{(1.0 \times 10^{-3})^2} = \frac{0.1}{1.0 \times 10^{-6}} = 1.0 \times 10^{5} \]

### Step 4: Calculate the cell potential under non-standard conditions
Using the Nernst equation:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log(Q) \]
where \( n \) is the number of moles of electrons transferred in the balanced reaction. Here, \( n = 2 \).

Now substituting the values:
\[ E_{cell} = 3.17 \, V - \frac{0.0591}{2} \log(1.0 \times 10^{5}) \]
Calculating \( \log(1.0 \times 10^{5}) = 5 \):
\[ E_{cell} = 3.17 \, V - \frac{0.0591}{2} \times 5 \]
\[ E_{cell} = 3.17 \, V - 0.0591 \times 2.5 \]
\[ E_{cell} = 3.17 \, V - 0.14775 \, V \]
\[ E_{cell} = 3.02225 \, V \approx 3.02 \, V \]

### Final Answer
The e.m.f. of the cell is approximately **3.02 V**.

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