Determine the potential of the following cell:
`Pt|H_2(g,0.1 "bar")|H^+(aq,10^(-3)M"||"MnO_4^-)(aq),0.1M)`
`Mn^(2+)(aq,0.01M),H^+(aq,0.01M)|Pt`
Given : `E_(MnO_4^(-)|Mn^(2+))^(@)=1.51V`
(a)1.54V
(b)1.48V
(c)1.84V
(d)none of these

Options

A. 1.54V
B. 1.48V
C. 1.84V
D. none of these

Answer

To determine the potential of the given electrochemical cell, we can follow these steps:

### Step 1: Identify the half-reactions
The cell notation is given as:
\[ \text{Pt} | \text{H}_2(g, 0.1 \text{ bar}) | \text{H}^+(aq, 10^{-3} M) || \text{MnO}_4^-(aq, 0.1 M), \text{Mn}^{2+}(aq, 0.01 M), \text{H}^+(aq, 0.01 M) | \text{Pt} \]

At the anode, hydrogen gas is oxidized:
\[ \text{H}_2(g) \rightarrow 2 \text{H}^+ + 2e^- \]

At the cathode, the reduction of permanganate occurs:
\[ \text{MnO}_4^- + 8 \text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2O \]

### Step 2: Standard Electrode Potentials
The standard electrode potential for the reduction of permanganate is given as:
\[ E^\circ(\text{MnO}_4^-/\text{Mn}^{2+}) = 1.51 \text{ V} \]

The standard electrode potential for the hydrogen half-reaction is:
\[ E^\circ(\text{H}^+/H_2) = 0 \text{ V} \]

### Step 3: Calculate the standard cell potential \(E^\circ_{\text{cell}}\)
The standard cell potential can be calculated using:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
\[ E^\circ_{\text{cell}} = 1.51 \text{ V} - 0 \text{ V} = 1.51 \text{ V} \]

### Step 4: Determine the number of electrons transferred
From the half-reactions:
- The anode reaction involves 2 electrons.
- The cathode reaction involves 5 electrons.

To balance the number of electrons, we multiply the anode reaction by 5 and the cathode reaction by 2:
- Anode: \( 5 \text{H}_2 \rightarrow 10 \text{H}^+ + 10e^- \)
- Cathode: \( 2 \text{MnO}_4^- + 16 \text{H}^+ + 10e^- \rightarrow 2 \text{Mn}^{2+} + 8 \text{H}_2O \)

### Step 5: Write the net reaction
The net reaction combining both half-reactions is:
\[ 5 \text{H}_2 + 2 \text{MnO}_4^- + 6 \text{H}^+ \rightarrow 2 \text{Mn}^{2+} + 8 \text{H}_2O \]

### Step 6: Apply the Nernst Equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]
Where:
- \( n = 10 \) (number of electrons transferred)
- Products: \( [\text{Mn}^{2+}]^2 \) and \( [\text{H}_2O]^8 \) (water is a liquid and not included in the expression)
- Reactants: \( [\text{H}^+]^6 \) and \( [\text{MnO}_4^-]^2 \) and \( P_{\text{H}_2}^5 \)

Substituting the concentrations:
- \( [\text{Mn}^{2+}] = 0.01 \, M \)
- \( [\text{H}^+] = 10^{-3} \, M \)
- \( [\text{MnO}_4^-] = 0.1 \, M \)
- \( P_{\text{H}_2} = 0.1 \, \text{bar} \)

Substituting these values into the Nernst equation:
\[ E_{\text{cell}} = 1.51 - \frac{0.059}{10} \log \left( \frac{(0.01)^2}{(10^{-3})^6 \cdot (0.1)^2 \cdot (0.1)^5} \right) \]

### Step 7: Calculate the logarithm and final potential
Calculating the logarithm:
- The numerator: \( (0.01)^2 = 10^{-4} \)
- The denominator: \( (10^{-3})^6 \cdot (0.1)^2 \cdot (0.1)^5 = 10^{-18} \cdot 10^{-2} \cdot 10^{-5} = 10^{-25} \)

Thus:
\[ \log \left( \frac{10^{-4}}{10^{-25}} \right) = \log(10^{21}) = 21 \]

Now substituting back:
\[ E_{\text{cell}} = 1.51 - \frac{0.059}{10} \cdot 21 \]
\[ E_{\text{cell}} = 1.51 - 0.12459 \]
\[ E_{\text{cell}} \approx 1.48 \text{ V} \]

### Final Answer
The potential of the cell is **1.48 V**.