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Redox Reactions and Electrochemistry: Chemistry | JEE Main

298\text{ K}

, a

1\text{ litre}

solution containing

10\text{mmol}

\text{Cr}_2\text{O}_7^{2-}

and

100\text{mmol}

\text{Cr}^{3+}

shows a

\text{pH}

3.0

. Given:

\text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+}

;

\text{E}^\circ = 1.330\text{ V}

and

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

The potential for the half cell reaction is

x \times 10^{-3}\text{ V}

. The value of

x

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Step-by-step solutionView

Correct answer

The half-cell potential is $0.917\text{ V}$ , which corresponds to $x = 917$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Volume $V = 1\text{ L}$ , $n(\text{Cr}_2\text{O}_7^{2-}) = 10\text{ mmol}$ , $n(\text{Cr}^{3+}) = 100\text{ mmol}$ , $\text{pH} = 3.0$ , $E^\circ = 1.330\text{ V}$ , and $\frac{2.303RT}{F} = 0.059\text{ V}$ .

02find

Find the half-cell potential $E$ in the form $x \times 10^{-3}\text{ V}$ , reporting the integer value $x$ .

03visualise

Write the balanced reduction half-reaction in acidic medium: $14\text{H}^+ + \text{Cr}_2\text{O}_7^{2-} + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$ . Here, the number of electrons transferred is $n = 6$ .

04strategise

Calculate molar concentrations: $[\text{Cr}_2\text{O}_7^{2-}] = \frac{10\text{ mmol}}{1\text{ L}} = 10^{-2}\text{ M}$ , $[\text{Cr}^{3+}] = \frac{100\text{ mmol}}{1\text{ L}} = 10^{-1}\text{ M}$ , $[\text{H}^+] = 10^{-\text{pH}} = 10^{-3}\text{ M}$ . Apply the Nernst equation: $E = E^\circ - \frac{0.059}{6} \log Q$ , where $Q = \frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}][\text{H}^+]^{14}}$ .

05execute

Substitute the values into $Q$ : $Q = \frac{(10^{-1})^2}{(10^{-2})(10^{-3})^{14}} = \frac{10^{-2}}{10^{-2} \times 10^{-42}} = 10^{42}$ . Now compute $E$ : $E = 1.330 - \frac{0.059}{6} \log(10^{42}) = 1.330 - \frac{0.059}{6} \times 42 = 1.330 - 0.059 \times 7 = 1.330 - 0.413 = 0.917\text{ V}$ . Since $E = x \times 10^{-3}\text{ V}$ , $x = 917$ .

✓verify

Notice that $42 / 6 = 7$ is an exact integer, giving $0.059 \times 7 = 0.413\text{ V}$ cleanly. Subtracting from $1.330\text{ V}$ yields $0.917\text{ V} = 917 \times 10^{-3}\text{ V}$ .

Hints that build this answer step by step

What is the balanced reduction half-reaction for dichromate in acidic medium?
$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$
What is the expression for the reaction quotient $Q$ for this half-cell reaction?
$Q = \frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}][\text{H}^+]^{14}}$
Given $[\text{Cr}_2\text{O}_7^{2-}] = 0.01\text{ M}$ , $[\text{Cr}^{3+}] = 0.1\text{ M}$ , and $\text{pH} = 3.0$ ( $[\text{H}^+] = 10^{-3}\text{ M}$ ), what is the value of $\log_{10} Q$ ?
42
Using the Nernst equation $E = E^\circ - \frac{0.059}{n}\log_{10} Q$ , what is the final value of $x$ where $E = x \times 10^{-3}\text{ V}$ ?
917

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why is n = 6 for the reduction of dichromate?

Each Cr atom goes from oxidation state +6 in Cr₂O₇²⁻ to +3 in Cr³⁺ (a decrease of 3). Since there are 2 Cr atoms, the total number of electrons involved is 2 × 3 = 6.

Does water enter the reaction quotient Q?

No, liquid water is the solvent in a dilute aqueous solution, so its activity is taken as 1.

Answer

The half-cell potential is $0.917\text{ V}$ , which corresponds to $x = 917$ .

Why each option works or fails

Step-by-step solution

given: Volume $V = 1\text{ L}$ , $n(\text{Cr}_2\text{O}_7^{2-}) = 10\text{ mmol}$ , $n(\text{Cr}^{3+}) = 100\text{ mmol}$ , $\text{pH} = 3.0$ , $E^\circ = 1.330\text{ V}$ , and $\frac{2.303RT}{F} = 0.059\text{ V}$ .
find: Find the half-cell potential $E$ in the form $x \times 10^{-3}\text{ V}$ , reporting the integer value $x$ .
visualise: Write the balanced reduction half-reaction in acidic medium: $14\text{H}^+ + \text{Cr}_2\text{O}_7^{2-} + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$ . Here, the number of electrons transferred is $n = 6$ .
strategise: Calculate molar concentrations: $[\text{Cr}_2\text{O}_7^{2-}] = \frac{10\text{ mmol}}{1\text{ L}} = 10^{-2}\text{ M}$ , $[\text{Cr}^{3+}] = \frac{100\text{ mmol}}{1\text{ L}} = 10^{-1}\text{ M}$ , $[\text{H}^+] = 10^{-\text{pH}} = 10^{-3}\text{ M}$ . Apply the Nernst equation: $E = E^\circ - \frac{0.059}{6} \log Q$ , where $Q = \frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}][\text{H}^+]^{14}}$ .
execute: Substitute the values into $Q$ : $Q = \frac{(10^{-1})^2}{(10^{-2})(10^{-3})^{14}} = \frac{10^{-2}}{10^{-2} \times 10^{-42}} = 10^{42}$ . Now compute $E$ : $E = 1.330 - \frac{0.059}{6} \log(10^{42}) = 1.330 - \frac{0.059}{6} \times 42 = 1.330 - 0.059 \times 7 = 1.330 - 0.413 = 0.917\text{ V}$ . Since $E = x \times 10^{-3}\text{ V}$ , $[\text{Cr}_2\text{O}_7^{2-}] = \frac{10\text{ mmol}}{1\text{ L}} = 10^{-2}\text{ M}$ .
verify: Notice that $42 / 6 = 7$ is an exact integer, giving $0.059 \times 7 = 0.413\text{ V}$ cleanly. Subtracting from $1.330\text{ V}$ yields $0.917\text{ V} = 917 \times 10^{-3}\text{ V}$ .

Redox Reactions and Electrochemistry: Chemistry | JEE Main

What feels right?

The half-cell potential is $0.917\text{ V}$ , which corresponds to $x = 917$ .

Why each option works or fails

Solution

01given

02find

03visualise

04strategise

05execute

✓verify

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Redox Reactions and Electrochemistry: Chemistry | JEE Main

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