Question

At $298 \,K$, a 1 litre solution containing $10\, m\,mol$ of $Cr _2 O _7{ }^{2-}$ and $100\, m\,mol$ of $Cr ^{3+}$ shows a pH of 30.

Given: $Cr _2 O _7^{2-} \rightarrow Cr ^{3+} ; E ^{\circ}=1330 V$ and $\frac{2303 RT }{ F }=0059 V$.

The potential for the half cell reaction is $x \times 10^{-3} V$. The value of $x$ is ___ .

Answer 1

Correct Answer: 917

The Nernst equation for the reaction \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \) is given by: \[ E = E^\circ - \frac{0.059}{n} \log \left( \frac{[\text{Products}]}{[\text{Reactants}]} \right) \] Given data: - \( E^\circ = 1.330 \, \text{V} \), \( n = 6 \), \( \text{pH} = 3 \), - \( [\text{H}^+] = 10^{-\text{pH}} = 10^{-3} \), - \( [\text{Cr}_2\text{O}_7^{2-}] = \frac{10}{1000} = 0.01 \, \text{M} \), - \( [\text{Cr}^{3+}] = \frac{100}{1000} = 0.1 \, \text{M} \). Substitute into the Nernst equation: \[ E = 1.330 - \frac{0.059}{6} \log \left( \frac{(0.1)^2}{(0.01)(10^{-3})^{14}} \right) \] \[ E = 1.330 - \frac{0.059}{6} \log \left( \frac{0.01}{0.01 \times 10^{-42}} \right) \] \[ E = 1.330 - \frac{0.059}{6} \log (10^{42}) \] \[ E = 1.330 - \frac{0.059}{6} \times 42 \] \[ E = 1.330 - 0.413 = 0.917 \, \text{V} \] Since \( E = x \times 10^{-3} \), we have \( x = 917 \).