JEE MAIN - Chemistry (2025 - 8th April Evening Shift - No. 21)
Consider the following half cell reaction
$$ \text{Cr}_2\text{O}_7^{2-} \, (\text{aq}) + 6\text{e}^- + 14\text{H}^+ \, (\text{aq}) \rightarrow 2\text{Cr}^{3+} \, (\text{aq}) + 7\text{H}_2\text{O} \, (\ell) $$
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 $$E^{\circ}_{\text{Cr}_2\text{O}_7^{2-}, \text{H}^+/\text{Cr}^{3+}} = 1.33\, \text{V}$$, $$\frac{2.303RT}{F} = 0.059\, \text{V}$$.]
Answer
10
Explanation
$$\begin{aligned}
& \mathrm{Cr}_2 \mathrm{O}_{7(\mathrm{aq})}^{-2}+14 \mathrm{H}_{(\mathrm{aq})}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{+3}+7 \mathrm{H}_2 \mathrm{O}_{(\ell)} \\
& \mathrm{E}_{\mathrm{R}}=\mathrm{E}_{\mathrm{R}}^0-\frac{0.059}{6} \log \frac{\left[\mathrm{Cr}^{+3}\right]^2}{\left[\mathrm{Cr}_2 \mathrm{O}_7^{-2}\right]\left[\mathrm{H}^{+}\right]^{14}} \\
& 0=1.33-\frac{0.059}{6} \log \frac{10^{-6}}{\left[\mathrm{H}^{+}\right]^{14}} \\
& \frac{1.33 \times 6}{0.059}=\log \frac{10^{-6}}{[\mathrm{H}]^{14}} \\
& 135.254=-6-14 \log \left[\mathrm{H}^{+}\right] \\
& 141.254=14 \mathrm{pH} \\
& \mathrm{pH}=\frac{141.254}{14}=10.08
\end{aligned}$$
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