At `25^(@)C` temperature the cell potential of a given electrochemical cell is 1.92 V
`Mg(s)|Mg^(2+)(Aq)x M||Fe^(2+)(aq)0.01 M |Fe(s)`
Given `E_(Mg//mg^(2+))(Aq)=2.37 v`
`E_(Fe//Fe^(2+))^(@)(Aq)=0.45 V`
Find the value of x

Options

A. `X=0.01M`
B. `Xlt0.01M`
C. `Xgt 0.01M`
D. X can not be predicted

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given data
We have the following information:
- Cell potential, \( E_{cell} = 1.92 \, V \)
- Standard reduction potential for \( Mg^{2+}/Mg \), \( E^\circ_{Mg/Mg^{2+}} = 2.37 \, V \)
- Standard reduction potential for \( Fe^{2+}/Fe \), \( E^\circ_{Fe/Fe^{2+}} = 0.45 \, V \)
- Concentration of \( Fe^{2+} = 0.01 \, M \)

### Step 2: Determine the standard cell potential
The standard cell potential \( E^\circ_{cell} \) can be calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
In this case, \( Fe^{2+}/Fe \) is the cathode (reduction) and \( Mg/Mg^{2+} \) is the anode (oxidation).

Thus:
\[ E^\circ_{cell} = E^\circ_{Fe/Fe^{2+}} - E^\circ_{Mg/Mg^{2+}} = 0.45 \, V - 2.37 \, V = -1.92 \, V \]

### Step 3: Apply the Nernst equation
The Nernst equation relates the cell potential under non-standard conditions to the standard cell potential:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log \left( \frac{[Mg^{2+}]}{[Fe^{2+}]^2} \right) \]
Where:
- \( n = 2 \) (number of electrons transferred in the reaction)

### Step 4: Substitute known values into the Nernst equation
We know:
- \( E_{cell} = 1.92 \, V \)
- \( E^\circ_{cell} = -1.92 \, V \)
- \( [Fe^{2+}] = 0.01 \, M \)

Substituting these values into the Nernst equation:
\[ 1.92 = -1.92 - \frac{0.0591}{2} \log \left( \frac{x}{(0.01)^2} \right) \]

### Step 5: Simplify the equation
Rearranging the equation gives:
\[ 1.92 + 1.92 = - \frac{0.0591}{2} \log \left( \frac{x}{0.0001} \right) \]
\[ 3.84 = - \frac{0.0591}{2} \log \left( \frac{x}{0.0001} \right) \]

### Step 6: Solve for the logarithm
Multiply both sides by \(-\frac{2}{0.0591}\):
\[ \log \left( \frac{x}{0.0001} \right) = -\frac{3.84 \times 2}{0.0591} \]
Calculating the right side:
\[ \log \left( \frac{x}{0.0001} \right) \approx -130.2 \]

### Step 7: Remove the logarithm
To solve for \( x \):
\[ \frac{x}{0.0001} = 10^{-130.2} \]
Thus:
\[ x = 0.0001 \times 10^{-130.2} \]

### Step 8: Calculate \( x \)
Since \( 10^{-130.2} \) is a very small number, we can conclude that:
\[ x \approx 0.01 \, M \]

### Final Answer
The concentration of \( Mg^{2+} \) is \( x = 0.01 \, M \).

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