The standard electrode potential for the following reaction is `+1.33` V. What is the potential at pH=2.0?
`Cr_(2)O_(7)^(2-)(aq,1M)+14H^(+)(aq)+6e^(-)to2Cr^(3+)(aq,1M)+7H_(2)O(l)`

Options

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

Answer

To find the potential at pH = 2.0 for the given reaction, we will 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 1: Identify the standard electrode potential (E°)
The standard electrode potential (E°) for the reaction is given as +1.33 V.

### Step 2: Determine the number of electrons transferred (n)
From the balanced reaction, we see that 6 electrons (e^-) are involved in the reduction of dichromate ions. Thus, \( n = 6 \).

### Step 3: Calculate the concentration of \(\text{H}^+\) ions at pH = 2
We know that:
\[ \text{pH} = -\log[\text{H}^+] \]
Thus,
\[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-2} \, \text{M} \]

### Step 4: Write the Nernst equation
The Nernst equation is given by:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]

### Step 5: Substitute the values into the Nernst equation
The products of the reaction are \( \text{Cr}^{3+} \) and \( \text{H}_2\text{O} \), and the reactants are \( \text{Cr}_2\text{O}_7^{2-} \) and \( \text{H}^+ \). The concentrations are:
- \([\text{Cr}^{3+}] = 1 \, \text{M}\)
- \([\text{Cr}_2\text{O}_7^{2-}] = 1 \, \text{M}\)
- \([\text{H}^+] = 10^{-2} \, \text{M}\)

Now substituting into the Nernst equation:
\[ E_{cell} = 1.33 \, \text{V} - \frac{0.0591}{6} \log \left( \frac{(1)^2}{(1)(10^{-2})^{14}} \right) \]

### Step 6: Simplify the logarithmic term
The logarithmic term simplifies as follows:
\[ \log \left( \frac{1}{10^{-28}} \right) = \log(10^{28}) = 28 \]

### Step 7: Substitute back into the Nernst equation
Now substituting this back into the equation:
\[ E_{cell} = 1.33 \, \text{V} - \frac{0.0591}{6} \times 28 \]

### Step 8: Calculate the potential
Calculating the second term:
\[ \frac{0.0591 \times 28}{6} = 0.2758 \]

Now substituting this value:
\[ E_{cell} = 1.33 \, \text{V} - 0.2758 \]
\[ E_{cell} = 1.0542 \, \text{V} \]

### Final Answer
The potential at pH = 2.0 is approximately **1.0542 V**.

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