Calculate the potential of an indicator electrode versus the standard hydrogen electrode, which originally contained `0.1M MnO_(4)^(-)` and `0.8M H^(+)` and which was treated with `Fe^(2+)` necessary to reduce `90%` of the `MnO_(4)` to `Mn^(2+)`
`MnO_(4)^(-) +8H^(+) +5e rarr Mn^(2+) +H_(2)O, E^(@) = 1.51V`

Explanation

To calculate the potential of the indicator electrode versus the standard hydrogen electrode, we will use the Nernst equation. Here’s a step-by-step solution:

### Step 1: Write the Nernst Equation
The Nernst equation is given by:
\[ E = E^0 - \frac{0.059}{n} \log \left( \frac{[\text{Products}]}{[\text{Reactants}]} \right) \]
where:
- \( E \) = cell potential
- \( E^0 \) = standard electrode potential (given as 1.51 V)
- \( n \) = number of electrons transferred in the reaction (5 for this reaction)

### Step 2: Identify Initial Concentrations
We start with:
- Initial concentration of \( \text{MnO}_4^- \) = 0.1 M
- Initial concentration of \( \text{H}^+ \) = 0.8 M

### Step 3: Calculate the Change in Concentrations
Since 90% of \( \text{MnO}_4^- \) is reduced to \( \text{Mn}^{2+} \):
- Amount of \( \text{MnO}_4^- \) reduced = \( 0.1 \times 0.9 = 0.09 \) M
- Remaining \( \text{MnO}_4^- \) = \( 0.1 - 0.09 = 0.01 \) M
- Concentration of \( \text{Mn}^{2+} \) produced = \( 0.09 \) M

### Step 4: Substitute Values into the Nernst Equation
Now we can substitute the values into the Nernst equation:
\[ E = 1.51 - \frac{0.059}{5} \log \left( \frac{[\text{Mn}^{2+}]}{[\text{MnO}_4^-][\text{H}^+]^8} \right) \]
Substituting the concentrations:
\[ E = 1.51 - \frac{0.059}{5} \log \left( \frac{0.09}{0.01 \times (0.8)^8} \right) \]

### Step 5: Calculate the Logarithmic Term
Calculate \( (0.8)^8 \):
\[ (0.8)^8 = 0.16777216 \]
Thus,
\[ E = 1.51 - \frac{0.059}{5} \log \left( \frac{0.09}{0.01 \times 0.16777216} \right) \]
\[ = 1.51 - \frac{0.059}{5} \log \left( \frac{0.09}{0.0016777216} \right) \]
\[ = 1.51 - \frac{0.059}{5} \log (53.6) \]

### Step 6: Calculate the Logarithm
Using a calculator:
\[ \log (53.6) \approx 1.729 \]
Now substitute back:
\[ E = 1.51 - \frac{0.059}{5} \times 1.729 \]
\[ = 1.51 - 0.00706 \times 1.729 \]
\[ = 1.51 - 0.0122 \]
\[ = 1.395 \text{ V} \]

### Final Answer
The potential of the indicator electrode versus the standard hydrogen electrode is approximately:
\[ E \approx 1.395 \text{ V} \]

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