The standard electrode potential for the following reaction is `-0.57` V. What is the potential at pH=12.0 ?
`TeO_(3)^(2-)(aq,1M)+3H_(2)O(l)+4e^(-)toTe(s)+6OH^(-)(aq)`

Options

A. `-017 V`
B. `+0.21 V`
C. `-0.39 V`
D. ``+1.95 V`

Answer

To find the potential at pH = 12.0 for the given reaction, we will use the Nernst equation. Here are the steps to solve the problem:

### Step 1: Identify the given values
- Standard electrode potential (E°) = -0.57 V
- pH = 12.0
- The reaction is:
\[ \text{TeO}_3^{2-} (aq, 1M) + 3 \text{H}_2\text{O} (l) + 4 e^- \rightarrow \text{Te} (s) + 6 \text{OH}^- (aq) \]

### Step 2: Calculate the concentration of OH⁻ ions
- First, we need to find the concentration of H⁺ ions from the pH:
\[ \text{pH} = -\log[\text{H}^+] \]
Therefore,
\[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-12} \]

- Using the relation \( K_w = [\text{H}^+][\text{OH}^-] = 10^{-14} \):
\[ [\text{OH}^-] = \frac{K_w}{[\text{H}^+]} = \frac{10^{-14}}{10^{-12}} = 10^{-2} \text{ M} \]

### Step 3: Apply the Nernst equation
The Nernst equation is given by:
\[ E = E° - \frac{0.0591}{n} \log \frac{[\text{products}]}{[\text{reactants}]} \]
Here, \( n = 4 \) (the number of electrons transferred).

- The products are Te (solid) and OH⁻ ions, and the reactants are TeO₃²⁻ (1 M) and water (1 M, but its concentration is not included in the Nernst equation).
- The concentration of the products (OH⁻) raised to the power of its stoichiometric coefficient:
\[ [\text{products}] = [\text{OH}^-]^6 = (10^{-2})^6 = 10^{-12} \]

- The concentration of the reactants (TeO₃²⁻):
\[ [\text{reactants}] = [\text{TeO}_3^{2-}] = 1 \text{ M} \]

### Step 4: Substitute values into the Nernst equation
\[ E = -0.57 - \frac{0.0591}{4} \log \frac{10^{-12}}{1} \]
\[ E = -0.57 - \frac{0.0591}{4} \log(10^{-12}) \]
\[ E = -0.57 - \frac{0.0591}{4} \times (-12) \]
\[ E = -0.57 + 0.0591 \times 3 \]
\[ E = -0.57 + 0.1773 \]
\[ E = -0.3927 \text{ V} \approx -0.39 \text{ V} \]

### Final Answer
The potential at pH = 12.0 is approximately **-0.39 V**.

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