The voltage of a cell whose half-cells are given below is
`Mg^(2+)+2e^(-)rarrMg(s),E^(@)=-2.37V`
`Cu^(2+)+2e^(-)rarrCu(s),E^(@)=+0.34V`
standard EMF of the cell is

Options

A. `-2.03V`
B. 1.36V
C. 2.71V
D. 2.03V

Answer

To find the standard EMF (electromotive force) of the cell, we will follow these steps:

### Step 1: Identify the half-reactions and their standard reduction potentials.
The half-reactions provided are:
1. \( \text{Mg}^{2+} + 2e^- \rightarrow \text{Mg}(s) \), \( E^\circ = -2.37 \, \text{V} \) (Anode)
2. \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}(s) \), \( E^\circ = +0.34 \, \text{V} \) (Cathode)

### Step 2: Determine which half-reaction occurs at the cathode and which at the anode.
- The half-reaction with a higher reduction potential occurs at the cathode, where reduction takes place.
- The half-reaction with a lower reduction potential occurs at the anode, where oxidation takes place.

In this case:
- Copper (Cu) has a higher reduction potential (+0.34 V) and will be reduced at the cathode.
- Magnesium (Mg) has a lower reduction potential (-2.37 V) and will be oxidized at the anode.

### Step 3: Use the formula for standard EMF of the cell.
The standard EMF of the cell can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

### Step 4: Substitute the values into the formula.
Substituting the values we have:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{Cu}} - E^\circ_{\text{Mg}} = (+0.34 \, \text{V}) - (-2.37 \, \text{V}) \]

### Step 5: Simplify the equation.
Calculating this gives:
\[ E^\circ_{\text{cell}} = 0.34 \, \text{V} + 2.37 \, \text{V} = 2.71 \, \text{V} \]

### Conclusion:
The standard EMF of the cell is \( 2.71 \, \text{V} \).

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