Consider the following half cell reaction Cr 2 O 7 2- ( aq )+6 e - +14 H + ( aq ) rightarrow 2 Cr 3+ ( aq )+7 H 2 O ( I ) The reaction was conducted with the ratio of frac [ Cr 3+ ] 2 [ Cr 2 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 Cr 2 O 7 2- , H + / Cr 3+ 0 =1.33 V , frac 2.303 RT F =0.059 V ]

Chemistry · Electrochemistry

JEE Main 2025 — 8 April, Evening Shift — Question 17

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{I})

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 $\left.\mathrm{E}_{\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}, \mathrm{H}^{+} / \mathrm{Cr}^{3+}}^{0}=1.33 \mathrm{~V}, \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}\right]$

Answer: 10

Numerical answer — enter this value.

Step-by-step solution

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

Using Nernst equation

\begin{aligned} & \mathrm{E}_{\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} / \mathrm{H}^{+} / \mathrm{Cr}^{3+}}=\mathrm{E}_{\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} / \mathrm{H}^{+} / \mathrm{Cr}^{3+}}^{0}-\frac{2.303 \mathrm{RT}}{6 \mathrm{~F}} \log \frac{\left[\mathrm{Cr}^{3+}\right]^{2}}{\left[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right]\left[\mathrm{H}^{+}\right]} 14 \\& 0=\mathrm{E}_{\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}, \mathrm{H}^{+} / \mathrm{Cr}^{3+}}^{0}-\frac{0.059}{6} \log \left(10^{-16}\left[\mathrm{H}^{+}\right]^{-14}\right) \\& 0=1.33+0.059-\frac{0.059 \times 14}{6} \mathrm{pH} \\& \mathrm{pH}=\frac{1.389 \times 6}{0.059 \times 14}=10.10 \simeq 10 \end{aligned}

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 8 April, Evening Shift
Subject: Chemistry
Chapter: Electrochemistry
Topic: Nernst Equation and Electrochemical Series