The standard electrode potential for the following reaction is `+1.33V`. What is the potential at `pH =2.0 ?`
`Cr_(2)O_(7)^(2-)(aq,1M)+14H^(+)(aq)+6e^(-) rarr 2Cr^(3+)(aq. 1M)+7H_(2)O(l)`

Options

A. `+1.820V`
B. `+1.990V`
C. `+1.608V`
D. `+1.0542V`

Explanation

To find the potential of the cell at pH = 2.0 for the given reaction, we can use the Nernst equation. The reaction is:

\[ \text{Cr}_2\text{O}_7^{2-} (aq, 1M) + 14 \text{H}^+ (aq) + 6 e^- \rightarrow 2 \text{Cr}^{3+} (aq, 1M) + 7 \text{H}_2\text{O} (l) \]

### Step-by-Step Solution:

1. **Identify the Standard Electrode Potential (E°)**:
The standard electrode potential (E°) for the reaction is given as +1.33 V.

2. **Determine the Number of Electrons Transferred (n)**:
From the balanced equation, we see that 6 electrons are transferred in the reaction. Thus, \( n = 6 \).

3. **Calculate the Concentration of H⁺ Ions**:
Given that the pH is 2.0, we can calculate the concentration of H⁺ ions using the formula:
\[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-2} \text{ M} \]

4. **Set Up the Nernst Equation**:
The Nernst equation is given by:
\[ E = E° - \frac{0.059}{n} \log \left( \frac{[\text{Products}]}{[\text{Reactants}]} \right) \]
For our reaction, the products are \( [\text{Cr}^{3+}]^2 \) and the reactants are \( [\text{H}^+]^{14} \) and \( [\text{Cr}_2\text{O}_7^{2-}] \).

Since both \( [\text{Cr}^{3+}] \) and \( [\text{Cr}_2\text{O}_7^{2-}] \) are 1 M, we can substitute these values into the equation:
\[ E = 1.33 - \frac{0.059}{6} \log \left( \frac{(1)^2}{(10^{-2})^{14} \cdot (1)} \right) \]

5. **Calculate the Logarithmic Term**:
\[ \log \left( \frac{1}{(10^{-2})^{14}} \right) = \log(10^{28}) = 28 \]

6. **Substitute Back into the Nernst Equation**:
\[ E = 1.33 - \frac{0.059}{6} \cdot 28 \]

7. **Calculate the Final Potential**:
\[ E = 1.33 - \frac{0.059 \cdot 28}{6} \]
\[ E = 1.33 - \frac{1.652}{6} \approx 1.33 - 0.2753 \approx 1.0547 \text{ V} \]

### Final Answer:
The potential at pH = 2.0 is approximately **1.054 V**.

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