Question

Consider the following half cell reaction $ \text{Cr}_2\text{O}_7^{2-} (\text{aq}) + 6\text{e}^- + 14\text{H}^+ (\text{aq}) \longrightarrow 2\text{Cr}^{3+} (\text{aq}) + 7\text{H}_2\text{O}(1) $ 
The reaction was conducted with the ratio of $\frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}]} = 10^{-6}$ 
The pH value at which the EMF of the half cell will become zero is ____ (nearest integer value) 
[Given : standard half cell reduction potential $\text{E}^\circ_{\text{Cr}_2\text{O}_7^{2-}, \text{H}^+/\text{Cr}^{3+}} = 1.33\text{V}, \quad \frac{2.303\text{RT}}{\text{F}} = 0.059\text{V}$

Hint:

- For half-cell reactions, use Nernst equation: \( E = E^\circ - \frac{0.059}{n} \log Q \) - Remember \( \text{pH} = -\log[\text{H}^+] \) - When E = 0, the system is at equilibrium - Watch stoichiometric coefficients in the reaction quotient \( Q \)

Answer 1

Correct Answer: 10

To find the pH at which the EMF of the half-cell becomes zero, we will use the Nernst equation for the given half-cell reaction:

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

The Nernst equation is given by:

\( E = E^\circ - \frac{0.059}{n} \log Q \),

where \( E \) is the EMF of the cell, \( E^\circ \) is the standard reduction potential (1.33 V), \( n \) is the number of electrons transferred (6), and \( Q \) is the reaction quotient:

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

Given \( \frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}]} = 10^{-6} \),

Substitute \( Q = 10^{-6}[\text{H}^+]^{-14} \).

Since the EMF is zero:

\( 0 = 1.33 - \frac{0.059}{6} \log \left(10^{-6}[\text{H}^+]^{-14} \right) \)

Solving for the pH:

\( \frac{0.059}{6} \log \left(10^{-6}[\text{H}^+]^{-14} \right) = 1.33 \)

\( \log \left(10^{-6}[\text{H}^+]^{-14} \right) = 135.59 \)

Rewriting the equation:

\( \log 10^{-6} + \log [\text{H}^+]^{-14} = 135.59 \)

\( -6 - 14\log [\text{H}^+] = 135.59 \)

Solving for \(\log [\text{H}^+]\):

\( -14\log [\text{H}^+] = 141.59 \)

\( \log [\text{H}^+] = -10.11 \)

Converting to pH:

\( \text{pH} = -\log [\text{H}^+] = 10.11 \)

Thus, the nearest integer value for the pH is \( \mathbf{10} \). As expected, 10 falls within the given range (10, 10).

Answer 2

Step 1: Write Nernst equation \[ E = E^\circ - \frac{0.059}{n} \log Q \] For the given reaction with \( n = 6 \): \[ E = 1.33 - \frac{0.059}{6} \log \left( \frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}][\text{H}^+]^{14}} \right) \]
Step 2: Set E = 0 and substitute given ratio \[ 0 = 1.33 - \frac{0.059}{6} \log \left( \frac{10^{-6}}{[\text{H}^+]^{14}} \right) \]
Step 3: Simplify the equation \[ 1.33 = \frac{0.059}{6} \left( -6 + 14 \text{pH} \right) \] \[ 1.33 = 0.059 \left( -1 + \frac{14}{6} \text{pH} \right) \] \[ 1.33 = -0.059 + 0.1377 \text{pH} \] \[ 1.389 = 0.1377 \text{pH} \] \[ \text{pH} = \frac{1.389}{0.1377} \approx 10.09 \approx 10 \text{ (nearest integer)} \]