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^(-30`M at `25^(@)` C?

Options

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

Answer

To solve the problem of how much the oxidizing power of the `Cr2O7^(2-) | Cr^(3+)` couple decreases when the concentration of `H^+` is decreased from 1M to `10^(-30)` M at `25°C`, we can follow these steps:

### Step 1: Write the half-reaction
The half-reaction for the reduction of dichromate ion to chromium ion is:
\[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]

### Step 2: Identify the number of electrons transferred
In the above half-reaction, we see that 6 electrons are transferred (N = 6).

### Step 3: Write the Nernst equation
The Nernst equation for this half-reaction can be written as:
\[ E = E^0 - \frac{0.059}{n} \log \left( \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}] \cdot [H^+]^{14}} \right) \]
Where:
- \( E \) is the cell potential,
- \( E^0 \) is the standard cell potential,
- \( n \) is the number of electrons transferred (6 in this case).

### Step 4: Calculate the cell potential at 1M `H^+`
When the concentration of `H^+` is 1M, the equation simplifies to:
\[ E_1 = E^0 - \frac{0.059}{6} \log \left( \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}] \cdot (1)^{14}} \right) \]
This reduces to:
\[ E_1 = E^0 - \frac{0.059}{6} \log \left( \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}]} \right) \]

### Step 5: Calculate the cell potential at `10^(-30)` M `H^+`
Now, when the concentration of `H^+` is decreased to `10^{-30}` M, the equation becomes:
\[ E_2 = E^0 - \frac{0.059}{6} \log \left( \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}] \cdot (10^{-30})^{14}} \right) \]
This simplifies to:
\[ E_2 = E^0 - \frac{0.059}{6} \log \left( \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}]} \cdot 10^{-420} \right) \]

### Step 6: Subtract the two potentials
Now, we can find the difference in potential:
\[ E_2 - E_1 = \left( E^0 - \frac{0.059}{6} \log \left( \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}]} \cdot 10^{-420} \right) \right) - \left( E^0 - \frac{0.059}{6} \log \left( \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}]} \right) \right) \]
The \( E^0 \) terms cancel out:
\[ E_2 - E_1 = - \frac{0.059}{6} \left( \log \left( \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}]} \cdot 10^{-420} \right) - \log \left( \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}]} \right) \right) \]
This simplifies to:
\[ E_2 - E_1 = - \frac{0.059}{6} \cdot (-420) \]
\[ E_2 - E_1 = \frac{0.059 \times 420}{6} \]

### Step 7: Calculate the final value
Calculating this gives:
\[ E_2 - E_1 = \frac{24.78}{6} = 4.13 \text{ volts} \]

### Step 8: Conclusion
Thus, the oxidizing power of the `Cr2O7^(2-) | Cr^(3+)` couple decreases by approximately **0.414 volts** when the `H^+` concentration is decreased from 1M to `10^{-30}` M.