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 electrode potential at pH = 12.0 for the given reaction, we will use the Nernst equation. Here’s a step-by-step solution:

### Step 1: Identify the standard electrode potential (E°)
The standard electrode potential (E°) for the reaction is given as:
\[ E° = -0.57 \, \text{V} \]

### Step 2: Calculate the concentration of hydroxide ions (OH⁻)
Since we are given the pH, we can find the concentration of hydrogen ions (H⁺) and then use it to find the concentration of hydroxide ions (OH⁻).

The relationship between pH and H⁺ concentration is:
\[ \text{pH} = -\log[H^+] \]
Given that pH = 12, we can calculate [H⁺]:
\[ [H^+] = 10^{-\text{pH}} = 10^{-12} \, \text{M} \]

Using the water dissociation constant:
\[ K_w = [H^+][OH^-] = 10^{-14} \]
We can find [OH⁻]:
\[ [OH^-] = \frac{K_w}{[H^+]} = \frac{10^{-14}}{10^{-12}} = 10^{-2} \, \text{M} \]

### Step 3: Write the Nernst equation
The Nernst equation is given by:
\[ E_{cell} = E°_{cell} - \frac{0.0591}{n} \log Q \]
Where:
- \( n \) = number of moles of electrons transferred in the reaction (which is 4 in this case)
- \( Q \) = reaction quotient

### Step 4: Calculate the reaction quotient (Q)
From the reaction:
\[ \text{TeO}_3^{2-} + 3H_2O + 4e^- \rightarrow \text{Te} + 6OH^- \]
The reaction quotient \( Q \) can be expressed as:
\[ Q = \frac{[OH^-]^6}{[\text{TeO}_3^{2-}][H_2O]^3} \]
Assuming the concentration of water is constant and equal to 1 M, we have:
\[ Q = \frac{(10^{-2})^6}{1} = 10^{-12} \]

### Step 5: Substitute values into the Nernst equation
Now we can substitute the values into the Nernst equation:
\[ E_{cell} = -0.57 - \frac{0.0591}{4} \log(10^{-12}) \]

Calculating the logarithm:
\[ \log(10^{-12}) = -12 \]

Thus:
\[ E_{cell} = -0.57 - \frac{0.0591}{4} \times (-12) \]
\[ E_{cell} = -0.57 + \frac{0.0591 \times 12}{4} \]
\[ E_{cell} = -0.57 + 0.1773 \]
\[ E_{cell} = -0.3927 \, \text{V} \]

### Step 6: Round to appropriate significant figures
Rounding to two decimal places, we get:
\[ E_{cell} \approx -0.39 \, \text{V} \]

### Final Answer:
The potential at pH = 12.0 is approximately:
\[ E_{cell} \approx -0.39 \, \text{V} \]

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