Calculate the electrode potential at `25^(@)C` of `Cr^(3+),Cr_(2)O_(7)^(2-)` electrode at `pOH=11` in solution of 0.01 M both in `Cr^(3+)` and `Cr_(2)O_(7)^(2-)`
`Cr_(2)O_(7)^(2-)+14H^(+)+6eto2Cr^(3+)+7H_(2)O`
`E^(0)=1.33V`.

Explanation

To calculate the electrode potential of the `Cr^(3+), Cr_(2)O_(7)^(2-)` electrode at `25°C`, we will use the Nernst equation. The reaction given is:

\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]

The standard electrode potential \(E^{\circ} = 1.33 \, \text{V}\).

### Step 1: Determine the pH and H\(^+\) concentration
Given that \(pOH = 11\), we can find the pH using the relationship:

\[ pH + pOH = 14 \]

So,

\[ pH = 14 - pOH = 14 - 11 = 3 \]

Now, we can calculate the concentration of H\(^+\):

\[ [\text{H}^+] = 10^{-pH} = 10^{-3} \, \text{M} \]

### Step 2: Write the Nernst equation
The Nernst equation is given by:

\[ E = E^{\circ} - \frac{0.0591}{n} \log Q \]

where:
- \(E\) is the electrode potential,
- \(E^{\circ}\) is the standard electrode potential,
- \(n\) is the number of electrons transferred in the reaction,
- \(Q\) is the reaction quotient.

### Step 3: Identify \(n\) and calculate \(Q\)
From the balanced reaction, we see that \(n = 6\) (6 electrons are transferred).

Now, we can calculate \(Q\):

\[ Q = \frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}][\text{H}^+]^{14}} \]

Substituting the concentrations:
- \([\text{Cr}^{3+}] = 0.01 \, \text{M}\)
- \([\text{Cr}_2\text{O}_7^{2-}] = 0.01 \, \text{M}\)
- \([\text{H}^+] = 10^{-3} \, \text{M}\)

So,

\[ Q = \frac{(0.01)^2}{(0.01)(10^{-3})^{14}} \]

Calculating \(Q\):

\[ Q = \frac{0.0001}{0.01 \times 10^{-42}} = \frac{0.0001}{10^{-40}} = 10^{40} \]

### Step 4: Substitute values into the Nernst equation
Now substituting \(E^{\circ}\), \(n\), and \(Q\) into the Nernst equation:

\[ E = 1.33 - \frac{0.0591}{6} \log(10^{40}) \]

Calculating the logarithm:

\[ \log(10^{40}) = 40 \]

Now substituting this back into the equation:

\[ E = 1.33 - \frac{0.0591}{6} \times 40 \]

Calculating:

\[ E = 1.33 - \frac{0.0591 \times 40}{6} \]

\[ E = 1.33 - \frac{2.364}{6} \]

\[ E = 1.33 - 0.394 \]

\[ E \approx 0.936 \, \text{V} \]

### Final Answer
The electrode potential at \(25°C\) of the `Cr^(3+), Cr_(2)O_(7)^(2-)` electrode is approximately **0.936 V**.

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