JEE Main 20258 AprilEvening Shift
- Exam
- JEE Main
- Year
- 2025
- Date
- 8 April
- Shift
- Evening Shift
- Subject
- Chemistry
- Chapter
- Electrochemistry
JEE Main 2025 · 8 April · Evening Shift · Chemistry · Electrochemistry
Consider the following half cell reaction Cr_2 O_7^ 2- (aq)+6 e^ -...
Consider the following half cell reaction
The reaction was conducted with the ratio of . The pH value at which the EMF of the half cell will become zero is ________. (nearest integer value)
[Given : standard half cell reduction potential
The reaction was conducted with the ratio of . The pH value at which the EMF of the half cell will become zero is ________. (nearest integer value)
[Given : standard half cell reduction potential
Correct answer
0
Step-by-step solution
\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}