To determine the potential of the given electrochemical cell, we will follow these steps:
### Step 1: Identify the Anode and Cathode Reactions
The cell notation indicates that the left side is the anode and the right side is the cathode.
**Anode Reaction:**
At the anode, hydrogen gas is oxidized:
\[
\text{H}_2(g) \rightarrow 2\text{H}^+(aq) + 2e^-
\]
The standard electrode potential for this reaction is \( E^\circ_{\text{anode}} = 0 \, \text{V} \).
**Cathode Reaction:**
At the cathode, the reduction of permanganate ion occurs:
\[
\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2O(l)
\]
The standard electrode potential for this reaction is given as \( E^\circ_{\text{cathode}} = 1.51 \, \text{V} \).
### Step 2: Write the Overall Cell Reaction
To balance the number of electrons transferred, we multiply the anode reaction by 5:
\[
5\text{H}_2(g) \rightarrow 10\text{H}^+(aq) + 10e^-
\]
Now, we can combine the anode and cathode reactions:
\[
2\text{MnO}_4^-(aq) + 10\text{H}^+(aq) + 5\text{H}_2(g) \rightarrow 2\text{Mn}^{2+}(aq) + 8\text{H}_2O(l) + 10\text{H}^+(aq)
\]
After canceling out the \(10\text{H}^+\) ions, the overall reaction simplifies to:
\[
2\text{MnO}_4^-(aq) + 5\text{H}_2(g) \rightarrow 2\text{Mn}^{2+}(aq) + 8\text{H}_2O(l) + 6\text{H}^+(aq)
\]
### Step 3: Calculate the Standard Cell Potential
The standard cell potential (\(E^\circ_{\text{cell}}\)) is calculated using:
\[
E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}
\]
Substituting the values:
\[
E^\circ_{\text{cell}} = 1.51 \, \text{V} - 0 \, \text{V} = 1.51 \, \text{V}
\]
### Step 4: Apply the Nernst Equation
The Nernst equation is given by:
\[
E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q
\]
Where \(n\) is the number of moles of electrons transferred (which is 10) and \(Q\) is the reaction quotient.
**Calculate Q:**
The reaction quotient \(Q\) is given by:
\[
Q = \frac{[\text{Mn}^{2+}]^2 \cdot [\text{H}^+]^6}{[\text{MnO}_4^-]^2 \cdot P_{\text{H}_2}^5}
\]
Substituting the concentrations:
- \([\text{Mn}^{2+}] = 0.01 \, \text{M}\)
- \([\text{H}^+] = 10^{-3} \, \text{M}\)
- \([\text{MnO}_4^-] = 0.1 \, \text{M}\)
- \(P_{\text{H}_2} = 0.1 \, \text{bar}\)
Thus,
\[
Q = \frac{(0.01)^2 \cdot (10^{-3})^6}{(0.1)^2 \cdot (0.1)^5}
\]
Calculating \(Q\):
\[
Q = \frac{(10^{-4}) \cdot (10^{-18})}{(10^{-2}) \cdot (10^{-5})} = \frac{10^{-22}}{10^{-7}} = 10^{-15}
\]
### Step 5: Substitute into the Nernst Equation
Now substituting into the Nernst equation:
\[
E_{\text{cell}} = 1.51 \, \text{V} - \frac{0.0591}{10} \log(10^{-15})
\]
Calculating the log term:
\[
\log(10^{-15}) = -15
\]
Thus,
\[
E_{\text{cell}} = 1.51 \, \text{V} - \frac{0.0591}{10} \cdot (-15)
\]
\[
E_{\text{cell}} = 1.51 \, \text{V} + 0.885 \, \text{V}
\]
\[
E_{\text{cell}} \approx 1.48 \, \text{V}
\]
### Final Answer
The potential of the cell is approximately:
\[
\boxed{1.48 \, \text{V}}
\]
