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

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 \[ 1.33 = \frac{0.059}{6} \left( -6 + 14 \text{pH} \right) \] \[ 1.33 = 0.059 \left( -1 + \frac{14}{6} \text{pH} \right) \] \[ 1.33 = -0.059 + 0.1377 \text{pH} \] \[ 1.389 = 0.1377 \text{pH} \] \[ \text{pH} = \frac{1.389}{0.1377} \approx 10.09 \approx 10 \text{ (nearest integer)} \]