Skip to the question

JEE MainChemistry

Reviewed by official_key

+4 marks−1 if incorrectNumericalpyq

Redox Reactions and Electrochemistry: Chemistry | JEE Main

Consider the cell

\mathrm{Pt}(\mathrm{s})\mid\mathrm{H}_2(\mathrm{g})(1\mathrm{~atm})\mid\mathrm{H}^{+}(\mathrm{aq},[\mathrm{H}^{+}]=1)\parallel\mathrm{Fe}^{3+}(\mathrm{aq}), \mathrm{Fe}^{2+}(\mathrm{aq})\mid\mathrm{Pt}(\mathrm{s})

Given

\mathrm{E}_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}}^{\circ}=0.771\mathrm{~V}

and

\mathrm{E}_{\mathrm{H}^{+}/1/2\mathrm{H}_2}^{\circ}=0\mathrm{~V}, \mathrm{T}=298\mathrm{~K}

If the potential of the cell is

0.712\mathrm{~V}

, the ratio of concentration of

\mathrm{Fe}^{2+}

to

\mathrm{Fe}^{3+}

is (Nearest integer)

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

Type the value - units or words beside it are fine.

Step-by-step solutionView

Correct answer

The ratio of concentration of $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}^{3+}$ is 10.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Cell: $\mathrm{Pt}(\mathrm{s})\mid\mathrm{H}_2(1\text{ atm})\mid\mathrm{H}^{+}(1\text{ M})\parallel\mathrm{Fe}^{3+},\mathrm{Fe}^{2+}\mid\mathrm{Pt}(\mathrm{s})$ , $E^\circ_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}} = 0.771\text{ V}$ , $E^\circ_{\mathrm{H}^+/\frac{1}{2}\mathrm{H}_2} = 0\text{ V}$ , $E_{\text{cell}} = 0.712\text{ V}$ at $T = 298\text{ K}$ .

02find

The ratio of concentration of $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}^{3+}$ , i.e., $[\mathrm{Fe}^{2+}]/[\mathrm{Fe}^{3+}]$ .

03strategise

1. Write the half-cell reactions and the net cell reaction: Anode: $\frac{1}{2}\mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{H}^+(\mathrm{aq}) + \mathrm{e}^-$ Cathode: $\mathrm{Fe}^{3+}(\mathrm{aq}) + \mathrm{e}^- \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq})$ Overall: $\frac{1}{2}\mathrm{H}_2(\mathrm{g}) + \mathrm{Fe}^{3+}(\mathrm{aq}) \rightarrow \mathrm{H}^+(\mathrm{aq}) + \mathrm{Fe}^{2+}(\mathrm{aq})$ with $n = 1$ . 2. Calculate $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.771 - 0 = 0.771\text{ V}$ . 3. Apply the Nernst equation at $298\text{ K}$ : $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q$ Here, $Q = \frac{[\mathrm{Fe}^{2+}][\mathrm{H}^+]}{[\mathrm{Fe}^{3+}] P_{\mathrm{H}_2}^{1/2}} = \frac{[\mathrm{Fe}^{2+}]}{[\mathrm{Fe}^{3+}]}$ since $[\mathrm{H}^+] = 1$ and $P_{\mathrm{H}_2} = 1\text{ atm}$ .

04execute

Substitute the numerical values into the Nernst equation: $0.712 = 0.771 - 0.059 \log \frac{[\mathrm{Fe}^{2+}]}{[\mathrm{Fe}^{3+}]}$ $0.059 \log \frac{[\mathrm{Fe}^{2+}]}{[\mathrm{Fe}^{3+}]} = 0.771 - 0.712 = 0.059$ $\log \frac{[\mathrm{Fe}^{2+}]}{[\mathrm{Fe}^{3+}]} = 1$ $\frac{[\mathrm{Fe}^{2+}]}{[\mathrm{Fe}^{3+}]} = 10^1 = 10$

✓verify

Checking the direction of cell potential: $E_{\text{cell}} = 0.712\text{ V} < E^\circ_{\text{cell}} = 0.771\text{ V}$ , which means $Q > 1$ . Hence, the ratio $[\mathrm{Fe}^{2+}]/[\mathrm{Fe}^{3+}] = 10 > 1$ , which is completely consistent.

Hints that build this answer step by step

What is the overall spontaneous cell reaction for this electrochemical cell?
$\frac{1}{2}\mathrm{H}_2(\mathrm{g}) + \mathrm{Fe}^{3+}(\mathrm{aq}) \rightarrow \mathrm{H}^{+}(\mathrm{aq}) + \mathrm{Fe}^{2+}(\mathrm{aq})$
Which form of the Nernst equation applies to this cell at $298\text{ K}$ , where $n = 1$ , $[\mathrm{H}^+] = 1$ , and $P_{\mathrm{H}_2} = 1\text{ atm}$ ?
$E_{\text{cell}} = E^\circ_{\text{cell}} - 0.0591\log_{10}\left(\frac{[\mathrm{Fe}^{2+}]}{[\mathrm{Fe}^{3+}]}\right)$
Substitute $E_{\text{cell}} = 0.712\text{ V}$ and $E^\circ_{\text{cell}} = 0.771\text{ V}$ into the equation to find $\frac{[\mathrm{Fe}^{2+}]}{[\mathrm{Fe}^{3+}]}$ . What is the value?
10

✓ 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: 8 September 2026

Quick checks

Students also ask

Why do we use 0.059 instead of 0.0591 or 2.303 RT/F?

At 298 K, 2.303 RT/F evaluates to ~0.0591 V. Here, 0.771 - 0.712 = 0.059 V, which is designed by the examiner to cancel directly with 0.059.

Answer

The ratio of concentration of $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}^{3+}$ is 10.

Why each option works or fails

Step-by-step solution

given: Cell: $\mathrm{Pt}(\mathrm{s})\mid\mathrm{H}_2(1\text{ atm})\mid\mathrm{H}^{+}(1\text{ M})\parallel\mathrm{Fe}^{3+},\mathrm{Fe}^{2+}\mid\mathrm{Pt}(\mathrm{s})$ , $E^\circ_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}} = 0.771\text{ V}$ , $E^\circ_{\mathrm{H}^+/\frac{1}{2}\mathrm{H}_2} = 0\text{ V}$ , $E_{\text{cell}} = 0.712\text{ V}$ at $T = 298\text{ K}$ .
find: The ratio of concentration of $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}^{3+}$ , i.e., $[\mathrm{Fe}^{2+}]/[\mathrm{Fe}^{3+}]$ .
strategise: 1. Write the half-cell reactions and the net cell reaction: Anode: $\frac{1}{2}\mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{H}^+(\mathrm{aq}) + \mathrm{e}^-$ Cathode: $\mathrm{Fe}^{3+}(\mathrm{aq}) + \mathrm{e}^- \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq})$ Overall: $\frac{1}{2}\mathrm{H}_2(\mathrm{g}) + \mathrm{Fe}^{3+}(\mathrm{aq}) \rightarrow \mathrm{H}^+(\mathrm{aq}) + \mathrm{Fe}^{2+}(\mathrm{aq})$ with $n = 1$ . 2. Calculate $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.771 - 0 = 0.771\text{ V}$ . 3. Apply the Nernst equation at $298\text{ K}$ : $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q$ Here, $Q = \frac{[\mathrm{Fe}^{2+}][\mathrm{H}^+]}{[\mathrm{Fe}^{3+}] P_{\mathrm{H}_2}^{1/2}} = \frac{[\mathrm{Fe}^{2+}]}{[\mathrm{Fe}^{3+}]}$ since $[\mathrm{H}^+] = 1$ and $E^\circ_{\mathrm{H}^+/\frac{1}{2}\mathrm{H}_2} = 0\text{ V}$ .
execute: Substitute the numerical values into the Nernst equation: $0.712 = 0.771 - 0.059 \log \frac{[\mathrm{Fe}^{2+}]}{[\mathrm{Fe}^{3+}]}$ $0.059 \log \frac{[\mathrm{Fe}^{2+}]}{[\mathrm{Fe}^{3+}]} = 0.771 - 0.712 = 0.059$ $\log \frac{[\mathrm{Fe}^{2+}]}{[\mathrm{Fe}^{3+}]} = 1$ $\frac{[\mathrm{Fe}^{2+}]}{[\mathrm{Fe}^{3+}]} = 10^1 = 10$
verify: Checking the direction of cell potential: $E_{\text{cell}} = 0.712\text{ V} < E^\circ_{\text{cell}} = 0.771\text{ V}$ , which means $Q > 1$ . Hence, the ratio $[\mathrm{Fe}^{2+}]/[\mathrm{Fe}^{3+}] = 10 > 1$ , which is completely consistent.