🛠️ JEE🧪 Chemistry

Consider the following half cell reaction \({\mathrm{Cr}}_{2}{\mathrm{O}}_{7}^{2-}(\mathrm{aq})+6{\mathrm{e}}^{-}+14{\ma…

Q1

Consider the following half cell reaction
\({\mathrm{Cr}}_{2}{\mathrm{O}}_{7}^{2-}(\mathrm{aq})+6{\mathrm{e}}^{-}+14{\mathrm{H}}^{+}(\mathrm{aq})\to 2{\mathrm{Cr}}^{3+}(\mathrm{aq})+7{\mathrm{H}}_{2}\mathrm{O}(\mathrm{l})\)
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

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

[JEE Main 2025, 8 Apr (Shift 1)]

a

10

b

15

c

14

d

12

🔒
Answer & explanation — PYQ Pass

Options are free to see. Unlock the correct answer and full explanation with Pass.

Unlock · ₹149

Practice more JEE Chemistry PYQs

See every question on Electrochemistry, or browse the full JEE question bank.

See all questions on Electrochemistry →