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JEE MAIN 2025
08-04-2025 S2
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$\begin{aligned} & \mathrm{Cr}_2 \mathrm{O}_{7(\mathrm{aq})}^{-2}+14 \mathrm{H}_{(\mathrm{aq})}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{+3}+7 \mathrm{H}_2 \mathrm{O}_{(1)} \\ & \mathrm{E}_{\mathrm{R}}=\mathrm{E}_{\mathrm{R}}^0-\frac{0.059}{6} \log \frac{\left[\mathrm{Cr}^{+3}\right]^2}{\left[\mathrm{Cr}_2 \mathrm{O}_7^{-2}\right]\left[\mathrm{H}^{+}\right]^{14}} \\ & 0=1.33-\frac{0.059}{6} \log \frac{10^{-6}}{\left[\mathrm{H}^{+}\right]^{14}} \\ & \frac{1.33 \times 6}{0.059}=\log \frac{10^{-6}}{[\mathrm{H}]^{14}} \\ & 135.254=-6-14 \log \left[\mathrm{H}^{+}\right] \\ & 141.254=14 \mathrm{pH} \\ & \mathrm{pH}=\frac{141.254}{14}=10.08\end{aligned}$
Question Tags
JEE Main
Chemistry
Medium
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