In acid medium, `MnO_(4)^(c-)` is an oxidizing agent.
`MnO_(4)^(c-)+8H^(o+)+5e^(-) rarr Mn^(2+)+4H_(2)O`
If `H^(o+)` ion concentration is doubled, electrode potential of the half cell `MnO_(4)^(c-), Mn^(2+)|Pt` will

Options

A. Increase by `28.36mV`
B. Decrease by `28.36mV`
C. Increase by `14.23mV`
D. Decrease by `142.30mV`

Explanation

To solve the problem, we need to analyze how the electrode potential of the half-cell changes when the concentration of \( H^+ \) ions is doubled. The half-reaction given is:

\[ \text{MnO}_4^- + 8H^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4H_2O \]

### Step 1: Write the Nernst Equation
The Nernst equation for the half-cell reaction can be expressed as:

\[ E = E^\circ - \frac{0.0591}{n} \log Q \]

where:
- \( E \) is the electrode potential,
- \( E^\circ \) is the standard electrode potential,
- \( n \) is the number of electrons transferred (which is 5 in this case),
- \( Q \) is the reaction quotient.

### Step 2: Determine the Reaction Quotient \( Q \)
For the given half-reaction, the reaction quotient \( Q \) can be expressed as:

\[ Q = \frac{[\text{Mn}^{2+}][H_2O]^4}{[\text{MnO}_4^-][H^+]^8} \]

### Step 3: Analyze the Effect of Doubling \( H^+ \) Concentration
If the concentration of \( H^+ \) ions is doubled, the new concentration becomes \( 2[H^+] \). Therefore, the new reaction quotient \( Q' \) will be:

\[ Q' = \frac{[\text{Mn}^{2+}][H_2O]^4}{[\text{MnO}_4^-][(2[H^+])]^8} \]

This can be simplified to:

\[ Q' = \frac{[\text{Mn}^{2+}][H_2O]^4}{[\text{MnO}_4^-][2^8 \cdot [H^+]^8}} = \frac{1}{256} \cdot Q \]

### Step 4: Substitute \( Q' \) into the Nernst Equation
Now we substitute \( Q' \) into the Nernst equation:

\[ E' = E^\circ - \frac{0.0591}{5} \log Q' \]

Substituting \( Q' \):

\[ E' = E^\circ - \frac{0.0591}{5} \log \left(\frac{1}{256} Q\right) \]

### Step 5: Simplify the Equation
Using the properties of logarithms, we can rewrite this as:

\[ E' = E^\circ - \frac{0.0591}{5} \left( \log \left(\frac{1}{256}\right) + \log Q \right) \]

Since \( \log \left(\frac{1}{256}\right) = -\log(256) = -8 \log(2) \):

\[ E' = E^\circ - \frac{0.0591}{5} \left(-8 \log(2) + \log Q\right) \]

### Step 6: Calculate the Change in Electrode Potential
The change in electrode potential \( \Delta E \) can be calculated as:

\[ \Delta E = E' - E = \frac{0.0591}{5} \cdot 8 \log(2) \]

Calculating \( 8 \log(2) \):

Using \( \log(2) \approx 0.301 \):

\[ \Delta E \approx \frac{0.0591}{5} \cdot 8 \cdot 0.301 \approx 0.0288 \text{ V} = 28.8 \text{ mV} \]

### Final Answer
Thus, the electrode potential of the half-cell \( \text{MnO}_4^-, \text{Mn}^{2+} | \text{Pt} \) will increase by approximately **28.8 mV**.

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