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Redox Reactions and Electrochemistry: Chemistry | JEE Main

Q: Why does only Fe2+ contribute to the n-factor?

In , iron is in the oxidation state. The ions already present in do not undergo oxidation; only the moles of are oxidized to , each transferring electron.

Q: How do we know the lattice arrangement of O2- ions in non-stoichiometric FeO?

Wüstite ( ) inherits the rock-salt lattice of stoichiometric , where ions form the ccp (cubic closed packed) array while cations occupy octahedral holes.

When

\mathrm{Fe}_{0.93}\mathrm{O}

is heated in presence of oxygen, it converts to

\mathrm{Fe}_2\mathrm{O}_3

. The number of correct statement/s from the following is A. The equivalent weight of

\mathrm{Fe}_{0.93}\mathrm{O}

\frac{\text{Molecular weight}}{0.79}

B. The number of moles of

\mathrm{Fe}^{2+}

and

\mathrm{Fe}^{3+}

in 1 mole of

\mathrm{Fe}_{0.93}\mathrm{O}

0.79

and

0.14

respectively C.

\mathrm{Fe}_{0.93}\mathrm{O}

is metal deficient with lattice comprising of cubic closed packed arrangement of

\mathrm{O}^{2-}

ions D. The \% composition of

\mathrm{Fe}^{2+}

and

\mathrm{Fe}^{3+}

\mathrm{Fe}_{0.93}\mathrm{O}

85\%

and

15\%

respectively

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Step-by-step solutionView

Correct answer

Statements B, C, and D are correct, giving a total of 3 correct statements.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Non-stoichiometric compound $\mathrm{Fe}_{0.93}\mathrm{O}$ converts to $\mathrm{Fe}_2\mathrm{O}_3$ upon heating in presence of oxygen.

02strategise

1. Determine moles of $\mathrm{Fe}^{2+}$ and $\mathrm{Fe}^{3+}$ per mole of $\mathrm{Fe}_{0.93}\mathrm{O}$ by electrical neutrality: $2x + 3(0.93 - x) = 2$ . 2. Calculate n-factor for $\mathrm{Fe}_{0.93}\mathrm{O} \rightarrow \mathrm{Fe}_2\mathrm{O}_3$ (only $\mathrm{Fe}^{2+}$ is oxidized to $\mathrm{Fe}^{3+}$ ). 3. Check crystal structure and metal deficiency defect of wüstite. 4. Calculate molar % composition of iron species.

03execute

Solving for $x$ (moles of $\mathrm{Fe}^{2+}$ ): $2x + 3(0.93 - x) = 2 \implies -x + 2.79 = 2 \implies x = 0.79$ . Thus, $\mathrm{Fe}^{3+} = 0.93 - 0.79 = 0.14$ . Statement B is correct.

04execute

During oxidation to $\mathrm{Fe}_2\mathrm{O}_3$ , iron is in $+3$ state. Each $\mathrm{Fe}^{2+}$ loses $1\,e^-$ , while $\mathrm{Fe}^{3+}$ does not change oxidation state. Total electrons lost per formula unit of $\mathrm{Fe}_{0.93}\mathrm{O}$ is $0.79 \times 1 = 0.79$ . Thus, $n\text{-factor} = 0.79$ and Equivalent weight $= \frac{\text{Molar mass}}{0.79}$ . Statement A is correct.

05execute

FeO has a rock-salt (NaCl) structure where oxide ions ( $\mathrm{O}^{2-}$ ) form an FCC/ccp lattice, and iron cations occupy octahedral voids. Metal deficiency arises because three $\mathrm{Fe}^{2+}$ ions are replaced by two $\mathrm{Fe}^{3+}$ ions to maintain charge neutrality, leaving cation vacancies. Statement C is correct.

06execute

Percentage of $\mathrm{Fe}^{2+} = \frac{0.79}{0.93} \times 100 \approx 84.95\% \approx 85\%$ . Percentage of $\mathrm{Fe}^{3+} = \frac{0.14}{0.93} \times 100 \approx 15.05\% \approx 15\%$ . Statement D is correct.

✓verify

All four statements A, B, C, and D are valid and correct. Hence, the number of correct statements is 4.

Hints that build this answer step by step

Using electrical neutrality for $\mathrm{Fe}_{0.93}\mathrm{O}$ , what are the molar amounts of $\mathrm{Fe}^{2+}$ and $\mathrm{Fe}^{3+}$ per formula unit?
$0.79\text{ mol }\mathrm{Fe}^{2+}\text{ and }0.14\text{ mol }\mathrm{Fe}^{3+}$
What are the percentages of $\mathrm{Fe}^{2+}$ and $\mathrm{Fe}^{3+}$ out of total iron content?
$\approx 85\%\text{ and }\approx 15\%$
When $\mathrm{Fe}_{0.93}\mathrm{O}$ is converted to $\mathrm{Fe}_2\mathrm{O}_3$ , what is the n-factor and corresponding equivalent weight?
$n = 0.79$ , so equivalent weight is $\frac{\text{Molecular weight}}{0.79}$
How many of the statements A, B, C, and D are correct?
4

✓ 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 does only Fe2+ contribute to the n-factor?

In $\mathrm{Fe}_2\mathrm{O}_3$ , iron is in the $+3$ oxidation state. The $\mathrm{Fe}^{3+}$ ions already present in $\mathrm{Fe}_{0.93}\mathrm{O}$ do not undergo oxidation; only the $0.79$ moles of $\mathrm{Fe}^{2+}$ are oxidized to $\mathrm{Fe}^{3+}$ , each transferring $1$ electron.

How do we know the lattice arrangement of O2- ions in non-stoichiometric FeO?

Wüstite ( $\mathrm{Fe}_{1-x}\mathrm{O}$ ) inherits the rock-salt lattice of stoichiometric $\mathrm{FeO}$ , where $\mathrm{O}^{2-}$ ions form the ccp (cubic closed packed) array while cations occupy octahedral holes.

Answer

Statements B, C, and D are correct, giving a total of 3 correct statements.

Why each option works or fails

Step-by-step solution

given: Non-stoichiometric compound $\mathrm{Fe}_{0.93}\mathrm{O}$ converts to $\mathrm{Fe}_2\mathrm{O}_3$ upon heating in presence of oxygen.
strategise: 1. Determine moles of $\mathrm{Fe}^{2+}$ and $\mathrm{Fe}^{3+}$ per mole of $\mathrm{Fe}_{0.93}\mathrm{O}$ by electrical neutrality: $2x + 3(0.93 - x) = 2$ . 2. Calculate n-factor for $\mathrm{Fe}_{0.93}\mathrm{O} \rightarrow \mathrm{Fe}_2\mathrm{O}_3$ (only $\mathrm{Fe}^{2+}$ is oxidized to $\mathrm{Fe}^{3+}$ ). 3. Check crystal structure and metal deficiency defect of wüstite. 4. Calculate molar % composition of iron species.
execute: Solving for $x$ (moles of $\mathrm{Fe}^{2+}$ ): $2x + 3(0.93 - x) = 2 \implies -x + 2.79 = 2 \implies x = 0.79$ . Thus, $\mathrm{Fe}^{3+} = 0.93 - 0.79 = 0.14$ . Statement B is correct.
execute: During oxidation to $\mathrm{Fe}_2\mathrm{O}_3$ , iron is in $+3$ state. Each $\mathrm{Fe}^{2+}$ loses $1\,e^-$ , while $\mathrm{Fe}^{3+}$ does not change oxidation state. Total electrons lost per formula unit of $\mathrm{Fe}_{0.93}\mathrm{O}$ is $0.79 \times 1 = 0.79$ . Thus, $n\text{-factor} = 0.79$ and Equivalent weight $= \frac{\text{Molar mass}}{0.79}$ . Statement A is correct.
execute: FeO has a rock-salt (NaCl) structure where oxide ions ( $\mathrm{O}^{2-}$ ) form an FCC/ccp lattice, and iron cations occupy octahedral voids. Metal deficiency arises because three $\mathrm{Fe}^{2+}$ ions are replaced by two $\mathrm{Fe}^{3+}$ ions to maintain charge neutrality, leaving cation vacancies. Statement C is correct.
execute: Percentage of $\mathrm{Fe}^{2+} = \frac{0.79}{0.93} \times 100 \approx 84.95\% \approx 85\%$ . Percentage of $\mathrm{Fe}^{3+} = \frac{0.14}{0.93} \times 100 \approx 15.05\% \approx 15\%$ . Statement D is correct.
verify: All four statements A, B, C, and D are valid and correct. Hence, the number of correct statements is 4.

Redox Reactions and Electrochemistry: Chemistry | JEE Main

What feels right?

Statements B, C, and D are correct, giving a total of 3 correct statements.

Why each option works or fails

Solution

01given

02strategise

03execute

04execute

05execute

06execute

✓verify

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