By how much is the oxidizing power of `Cr_2O_7^(2-)"|"Cr^(3+)` couple decreased if the `H^+` concentration is decreased from 1M to `10^(-3`M at `25^(@)` C?

Options

A. 0.001 V
B. 0.207 V
C. 0.441 V
D. 0.414 V

Answer

To determine how much the oxidizing power of the `Cr2O7^(2-) | Cr^(3+)` couple decreases when the `H^+` concentration is decreased from 1 M to `10^(-3)` M at `25°C`, we can use the Nernst equation. Here’s a step-by-step solution:

### Step 1: Write the half-reaction
The half-reaction for the reduction of dichromate ion is:
\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]

### Step 2: Identify the Nernst equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log Q \]
where:
- \( E \) is the cell potential,
- \( E^\circ \) is the standard cell potential,
- \( n \) is the number of moles of electrons transferred,
- \( Q \) is the reaction quotient.

### Step 3: Determine the reaction quotient \( Q \)
For the given half-reaction, the reaction quotient \( Q \) can be expressed as:
\[ Q = \frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}][\text{H}^+]^{14}} \]

### Step 4: Substitute the concentrations into the Nernst equation
Initially, when \( [\text{H}^+] = 1 \, \text{M} \):
\[ Q_1 = \frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}](1)^{14}} = \frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}]} \]

When \( [\text{H}^+] = 10^{-3} \, \text{M} \):
\[ Q_2 = \frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}](10^{-3})^{14}} = \frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}]} \cdot 10^{42} \]

### Step 5: Calculate the change in potential
Using the Nernst equation for both conditions:
1. For \( [\text{H}^+] = 1 \, \text{M} \):
\[ E_1 = E^\circ - \frac{0.0591}{6} \log Q_1 \]

2. For \( [\text{H}^+] = 10^{-3} \, \text{M} \):
\[ E_2 = E^\circ - \frac{0.0591}{6} \log Q_2 \]
\[ E_2 = E^\circ - \frac{0.0591}{6} \log \left( Q_1 \cdot 10^{42} \right) \]
\[ E_2 = E^\circ - \frac{0.0591}{6} \left( \log Q_1 + 42 \right) \]

### Step 6: Find the difference in potential
The difference in potential when changing from \( 1 \, \text{M} \) to \( 10^{-3} \, \text{M} \) is:
\[ \Delta E = E_2 - E_1 \]
\[ \Delta E = - \frac{0.0591}{6} \cdot 42 \]

### Step 7: Calculate the numerical value
Calculating the above expression:
\[ \Delta E = - \frac{0.0591 \times 42}{6} \]
\[ \Delta E = - 0.414 \, \text{V} \]

### Conclusion
The oxidizing power of the `Cr2O7^(2-) | Cr^(3+)` couple decreases by **0.414 V** when the `H^+` concentration is decreased from 1 M to `10^(-3)` M.

---