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 solve the problem, we will follow these steps:

### Step 1: Determine the initial concentrations
Initially, we have:
- Concentration of \( \text{MnO}_4^- \) = 0.1 M
- Concentration of \( \text{H}^+ \) = 0.8 M

### Step 2: Calculate the remaining concentration of \( \text{MnO}_4^- \)
Since 90% of \( \text{MnO}_4^- \) is reduced, only 10% remains:
\[ \text{Remaining } \text{MnO}_4^- = 0.1 \, \text{M} \times 0.10 = 0.01 \, \text{M} \]

### Step 3: Calculate the concentration of \( \text{Mn}^{2+} \) produced
The amount of \( \text{Mn}^{2+} \) produced is 90% of the initial \( \text{MnO}_4^- \):
\[ \text{Mn}^{2+} = 0.1 \, \text{M} \times 0.90 = 0.09 \, \text{M} \]

### Step 4: Use the Nernst equation
The Nernst equation is given by:
\[ E = E^0 - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]
where:
- \( E^0 \) = 1.51 V (given)
- \( n \) = 5 (number of electrons transferred)

The products are \( \text{Mn}^{2+} \) and the reactants are \( \text{MnO}_4^- \) and \( \text{H}^+ \):
\[ E = 1.51 - \frac{0.0591}{5} \log \left( \frac{[\text{Mn}^{2+}] }{[\text{MnO}_4^-] \cdot [\text{H}^+]^8} \right) \]

### Step 5: Substitute the concentrations into the Nernst equation
Substituting the values we have:
\[ E = 1.51 - \frac{0.0591}{5} \log \left( \frac{0.09}{0.01 \cdot (0.8)^8} \right) \]

### Step 6: Calculate \( (0.8)^8 \)
Calculating \( (0.8)^8 \):
\[ (0.8)^8 \approx 0.1678 \]

### Step 7: Substitute this value into the equation
Now we substitute this back into the Nernst equation:
\[ E = 1.51 - \frac{0.0591}{5} \log \left( \frac{0.09}{0.01 \cdot 0.1678} \right) \]

### Step 8: Calculate the logarithm
Calculating the ratio:
\[ \frac{0.09}{0.01 \cdot 0.1678} = \frac{0.09}{0.001678} \approx 53.6 \]
Now calculate the logarithm:
\[ \log(53.6) \approx 1.729 \]

### Step 9: Substitute the logarithm back into the equation
Substituting this value back:
\[ E = 1.51 - \frac{0.0591}{5} \times 1.729 \]

### Step 10: Calculate \( E \)
Calculating:
\[ E = 1.51 - 0.01182 \times 1.729 \approx 1.51 - 0.0204 \approx 1.4896 \, \text{V} \]

### Final Answer
The potential of the indicator electrode versus the standard hydrogen electrode is approximately:
\[ \boxed{1.39 \, \text{V}} \]