For the following electrochemical cell at `298K`
`Pt(s)+H_(2)(g,1bar) |H^(+) (aq,1M)||M^(4+)(aq),M^(2+)(aq)|Pt(s)`
`E_(cell) = 0.092 V` when `([M^(2+)(aq)])/([M^(4+)(aq)])=10^(x)`
Guven, `E_(M^(4+)//M^(2+))^(@) = 0.151V, 2.303 (RT)/(F) = 0.059`
The value of x is-

Options

A. `-2`
B. `-1`
C. 1
D. 2

Answer

To solve the given electrochemical cell problem, we will follow these steps:

### Step 1: Write the half-cell reactions
The half-cell reactions for the electrochemical cell are:
- **Anode (oxidation)**:
\[ H_2(g) \rightarrow 2H^+(aq) + 2e^- \]
- **Cathode (reduction)**:
\[ M^{4+}(aq) + 2e^- \rightarrow M^{2+}(aq) \]

### Step 2: Write the net cell reaction
By combining the half-cell reactions, we cancel the electrons:
\[ H_2(g) + M^{4+}(aq) \rightarrow 2H^+(aq) + M^{2+}(aq) \]

### Step 3: Apply the Nernst equation
The Nernst equation is given by:
\[ E_{cell} = E^\circ_{cell} - \frac{RT}{nF} \ln Q \]
Where:
- \( E_{cell} = 0.092 \, V \)
- \( E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \)
- \( n = 2 \) (number of electrons transferred)
- \( Q = \frac{[H^+]^2}{P_{H_2} \cdot [M^{4+}]} \)

### Step 4: Calculate \( E^\circ_{cell} \)
From the problem, we know:
- \( E^\circ_{M^{4+}/M^{2+}} = 0.151 \, V \)
- \( E^\circ_{H^+/H_2} = 0 \, V \)

Thus:
\[ E^\circ_{cell} = E^\circ_{M^{4+}/M^{2+}} - E^\circ_{H^+/H_2} = 0.151 - 0 = 0.151 \, V \]

### Step 5: Substitute values into the Nernst equation
Substituting the known values into the Nernst equation:
\[ 0.092 = 0.151 - \frac{0.059}{2} \log \left( \frac{[M^{2+}]}{[M^{4+}]} \right) \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ \frac{0.059}{2} \log \left( \frac{[M^{2+}]}{[M^{4+}]} \right) = 0.151 - 0.092 \]
\[ \frac{0.059}{2} \log \left( \frac{[M^{2+}]}{[M^{4+}]} \right) = 0.059 \]
\[ \log \left( \frac{[M^{2+}]}{[M^{4+}]} \right) = 2 \]

### Step 7: Solve for the ratio
From the logarithmic equation:
\[ \frac{[M^{2+}]}{[M^{4+}]} = 10^2 \]
Thus:
\[ \frac{[M^{2+}]}{[M^{4+}]} = 100 \]

### Step 8: Relate to x
Given that:
\[ \frac{[M^{2+}]}{[M^{4+}]} = 10^x \]
We can conclude:
\[ x = 2 \]

### Final Answer
The value of \( x \) is \( 2 \).

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