JEE Main 2025 — Chemistry Physical Chemistry

2025integerhard

Consider the following half cell reaction

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

The reaction was conducted with the ratio of $\frac{\left[\mathrm{Cr}^{3+}\right]^2}{\left[\mathrm{Cr}_2 \mathrm{O}_7^{2-}\right]}=10^{-6}$ . The pH value at which the EMF of the half cell will become zero is ________. (nearest integer value)

[Given : standard half cell reduction potential

$\left.\mathrm{E}_{\mathrm{Cr}_2 \mathrm{O}_7^{2-}, \mathrm{H}^{+} / \mathrm{Cr}^{3+}}^{\mathrm{o}}=1.33 \mathrm{~V}, \frac{2.303 \mathrm{RT}}{\mathrm{~F}}=0.059 \mathrm{~V}\right]$

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Held on 8 Apr 2025 · Verified 6 Jul 2026.

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$\begin{aligned} & \mathrm{Cr}_2 \mathrm{O}_{7 \text { (aq) }}^{-2}+14 \mathrm{H}_{(\mathrm{aq})}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{+3}+7 \mathrm{H}_2 \mathrm{O}_{(\ell)} \\ & \mathrm{E}_{\mathrm{R}}=\mathrm{E}_{\mathrm{R}}^0-\frac{0.059}{6} \log \frac{\left[\mathrm{Cr}^{+3}\right]^2}{\left[\mathrm{Cr}_2 \mathrm{O}_7^{-2}\right]\left[\mathrm{H}^{+}\right]^{14}} \\ & 0=1.33-\frac{0.059}{6} \log \frac{10^{-6}}{\left[\mathrm{H}^{+}\right]^{14}} \\ & \frac{1.33 \times 6}{0.059}=\log \frac{10^{-6}}{[\mathrm{H}]^{14}} \\ & 135.254=-6-14 \log \left[\mathrm{H}^{+}\right] \\ & 141.254=14 \mathrm{pH} \\ & \mathrm{pH}=\frac{141.254}{14}=10.08\end{aligned}$

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JEE Main 2025 — Chemistry Physical Chemistry

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JEE Main Physical Chemistry in other years

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JEE Main 2025 — Chemistry Physical Chemistry

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JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry