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

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Redox Reactions and Electrochemistry

Consider the following half cell reaction

Cr_{2} O_{7}^{2 -} (aq) + 6 e^{-} + 14 H^{+} (aq) \to 2 Cr^{3 +} (aq) + 7 H_{2} O (l)

The reaction was conducted with the ratio of

\frac{[ Cr ^{3 +} ] ^{2}}{[ Cr _{2} O _{7}^{2 -} ]} = 1 0^{- 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 +}}^{o} = 1.33 V, \frac{2.303 RT}{F} = 0.059 V]

Answer: 10

The Nernst equation for the half-cell is

E = E^{\circ} - \frac{0.059}{n} lo g \frac{[ C r ^{3 +} ] ^{2}}{[ C r _{2} O _{7}^{2 -} ] [ H ^{+} ] ^{14}}

with

n = 6

. Putting

E = 0

gives

0 = 1.33 - \frac{0.059}{6} lo g \frac{1 0 ^{- 6}}{[ H ^{+} ] ^{14}}

. Rearranging,

\frac{1.33 \times 6}{0.059} = lo g (1 0^{- 6}) - 14 lo g [H^{+}]

, that is

135.25 = - 6 + 14 pH

, so

14 pH = 141.25

. The EMF of the half-cell falls to zero at a pH of about

10

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Integer answer