At 298 K, a 1 litre solution containing 10 mmol of $\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}$ and 100 mmol of $\mathrm{Cr}^{3+}$ shows a pH of 3.0. Given : $\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-} \rightarrow \mathrm{Cr}^{3+}$ ; $\mathrm{E}^{0}=1.330 \mathrm{~V}$ and $\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}$ The potential for the half cell reaction is $\mathrm{x} \times 10^{-3} V$. The value of $x$ is ______.

JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Redox Reactions and Electrochemistry

At 298 K, a 1 litre solution containing 10 mmol of

Cr_{2} O_{7}^{2 -}

and 100 mmol of

Cr^{3 +}

shows a pH of 3.0.

Given :

Cr_{2} O_{7}^{2 -} \to Cr^{3 +}

;

E^{0} = 1.330 V

and

\frac{2.303 RT}{F} = 0.059 V

The potential for the half cell reaction is

x \times 1 0^{- 3} V

. The value of

x

is ______.

Answer: 917

The reduction half-cell is

C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2C r^{3 +} + 7 H_{2} O

, with

n = 6

. Taking

[C r_{2} O_{7}^{2 -}] = 1 0^{- 2} M

[C r^{3 +}] = 1 0^{- 1} M

and

[H^{+}] = 1 0^{- 3} M

at pH

3

, the Nernst equation reads

E = E^{0} - \frac{0.059}{6} lo g \frac{[ C r ^{3 +} ] ^{2}}{[ C r _{2} O _{7}^{2 -} ] [ H ^{+} ] ^{14}} .

The reaction quotient is

\frac{( 1 0 ^{- 1} ) ^{2}}{( 1 0 ^{- 2} ) ( 1 0 ^{- 3} ) ^{14}} = \frac{1 0 ^{- 2}}{1 0 ^{- 44}} = 1 0^{42}

, so

E = 1.330 - \frac{0.059}{6} \times 42 = 1.330 - 0.413 = 0.917 V .

Expressed as

x \times 1 0^{- 3} V

, the value of

x

917

JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Integer answer