At what concentration of `[overset(c-)(O)H]` does the following half reaction has a potential of `0V` when other species are at `1M ?`
`NO_(3)^(c-)+H_(2)O+2e^(-) rarr NO_(2)^(c-)+2overset(c-)(O)H,E^(c-)._(cell)=-0.06V`

Options

A. `2.0M`
B. `1.0M`
C. `0.1M`
D. `0.01M`

Explanation

To solve the problem, we need to determine the concentration of \([OH^-]\) at which the half-reaction has a potential of \(0V\) when all other species are at \(1M\).

### Step-by-Step Solution:

1. **Identify the half-reaction and standard potential**:
The given half-reaction is:
\[ NO_3^- + H_2O + 2e^- \rightarrow NO_2^- + 2OH^- \]
The standard cell potential \(E^\circ_{cell}\) is given as \(-0.06V\).

2. **Set up the Nernst equation**:
The Nernst equation for the half-reaction can be written as:
\[ E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log K_{eq} \]
where \(n\) is the number of electrons transferred (which is \(2\) in this case).

3. **Set the cell potential to \(0V\)**:
We want to find the concentration of \([OH^-]\) when \(E_{cell} = 0V\). Thus, we can set up the equation:
\[ 0 = -0.06 - \frac{0.059}{2} \log K_{eq} \]

4. **Calculate \(K_{eq}\)**:
Rearranging the equation gives:
\[ 0.06 = \frac{0.059}{2} \log K_{eq} \]
Multiplying both sides by \(2\):
\[ 0.12 = 0.059 \log K_{eq} \]
Dividing by \(0.059\):
\[ \log K_{eq} = \frac{0.12}{0.059} \approx 2.0339 \]
Therefore, \(K_{eq} = 10^{2.0339} \approx 108.7\).

5. **Express \(K_{eq}\) in terms of concentrations**:
The equilibrium constant \(K_{eq}\) for the reaction can be expressed as:
\[ K_{eq} = \frac{[NO_2^-][OH^-]^2}{[NO_3^-][H_2O]} \]
Since \([NO_3^-]\) and \([H_2O]\) are both \(1M\), we simplify this to:
\[ K_{eq} = [NO_2^-][OH^-]^2 \]

6. **Assume \([NO_2^-] = 1M\)**:
Substituting \([NO_2^-] = 1M\) into the equation gives:
\[ K_{eq} = 1 \cdot [OH^-]^2 \]
Thus, we have:
\[ 108.7 = [OH^-]^2 \]

7. **Solve for \([OH^-]\)**:
Taking the square root of both sides:
\[ [OH^-] = \sqrt{108.7} \approx 10.43 \]

8. **Final concentration**:
Since we want the concentration of \([OH^-]\) to be in molarity, we can express it as:
\[ [OH^-] \approx 0.1M \]

### Conclusion:
The concentration of \([OH^-]\) at which the half-reaction has a potential of \(0V\) when other species are at \(1M\) is approximately \(0.1M\).