Redox Reactions — Class 11 MCQs with Answers

Redox Reactions — important questions & MCQs with answers (Class 11 Chemistry)

6 solved questions from this chapter's difficulty levels, each with its answer and explanation. Tap an option to check yourself.

0 of 6 answered

1. Q1 of 6Easy

   Oxidation involves:

   ALoss of electronsBGain of electronsCNo changeDLoss of mass
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   Answer: Loss of electrons

   Explanation: OIL RIG: Oxidation Is Loss of electrons, Reduction Is Gain. Losing electrons raises oxidation number, e.g. Zn → Zn²⁺ + 2e⁻. Confusing this with mass loss is the usual trap in the mass-loss option.

2. Q2 of 6Easy

   Oxidation number of an element in its free (elemental) state is:

   A+1B0C−1DEqual to valency
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   Answer: 0

   Explanation: Any element in its free, uncombined state has oxidation number 0, since no electrons transfer between identical atoms — true for Na, O₂, Fe or P₄ alike. Assigning a nonzero value here, based on the element's usual valency, is the trap.

3. Q3 of 6Medium

   Oxidising agent does:

   AGets reduced (gains electrons)BGets oxidisedCStays the sameDDisappears
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   Answer: Gets reduced (gains electrons)

   Explanation: An oxidising agent causes another species to be oxidised by accepting its electrons, so the agent itself gains electrons and is reduced — its own oxidation number falls. Saying it 'gets oxidised' swaps the agent with the species it acts on.

4. Q4 of 6Medium

   Oxidation number of S in H₂SO₄:

   A0B−2C+4D+6
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   Answer: +6

   Explanation: H is +1 and O is −2, so 2(+1)+S+4(−2)=0 gives S=+6, balancing the neutral formula. This is sulfur's maximum oxidation state, far above the +4 it shows in SO₂.

5. Q5 of 6Hard

   Which couple has the highest standard reduction potential?

   ACl₂/Cl⁻BF₂/F⁻CBr₂/Br⁻DI₂/I⁻
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   Answer: F₂/F⁻

   Explanation: Standard reduction potentials for halogens run F₂/F⁻≈+2.87V > Cl₂/Cl⁻≈+1.36V > Br₂/Br⁻≈+1.09V > I₂/I⁻≈+0.54V, tracking electronegativity down the group. Assuming Cl₂ is strongest, since it's the familiar bleach oxidiser, misses that F₂ tops the series.

6. Q6 of 6Hard

   Disproportionation reaction:

   ANo redoxBTwo different elements oxidisedCSame element is both oxidised and reducedDPure precipitation
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   Answer: Same element is both oxidised and reduced

   Explanation: Disproportionation is when one species of a single element is simultaneously oxidised and reduced: e.g. in 2H₂O₂→2H₂O+O₂, oxygen goes from −1 to −2 (reduced) in H₂O and to 0 (oxidised) in O₂ — one starting state splitting two ways.

Redox Reactions notes: key concepts (Class 11 Chemistry)

Redox chemistry views reactions as electron transfer: oxidation is electron loss, reduction is electron gain, and oxidation numbers track the flow. This chapter fixes rules for assigning oxidation states, spotting oxidising and reducing agents, balancing by two methods, and linking redox to electrode cells.

Oxidation & reduction

Oxidation is loss of electrons and a rise in oxidation number; reduction is gain of electrons and a fall. The two always occur together.

Oxidation number rules

Free elements are 0, a monatomic ion equals its charge, O is usually −2 (−1 in peroxides), H is +1 (−1 in metal hydrides), F always −1.

Sum rule for oxidation numbers

The algebraic sum of oxidation numbers equals the species' net charge — zero for a neutral molecule — which lets you solve for an unknown state.

Oxidising agent

The oxidising agent accepts electrons and is itself reduced, so its oxidation number falls. Strong examples are KMnO₄, K₂Cr₂O₇ and O₃.

Reducing agent

The reducing agent donates electrons and is itself oxidised, so its oxidation number rises. Common ones are active metals, H₂, CO and oxalate.

Redox as electron transfer

Every redox reaction splits into an oxidation half and a reduction half; the electrons lost by one species are exactly gained by the other.

Disproportionation

One element in a single intermediate oxidation state is simultaneously oxidised and reduced, e.g. Cl₂ → Cl⁻ + ClO⁻; it needs an intermediate state.

Stock notation

An element's oxidation number is written as a Roman numeral in parentheses after its name, e.g. iron(III) or manganese(VII) in KMnO₄.

n-factor

For a redox species the n-factor is the number of electrons gained or lost per formula unit; it fixes the equivalent mass and normality.

Oxidation-number method

Balance so the total increase in oxidation number equals the total decrease, then balance O with H₂O and H with H⁺ in acidic medium.

Half-reaction (ion-electron) method

Write separate oxidation and reduction halves, balance atoms then charge with electrons, equalise electrons and add; use OH⁻ in basic medium.

Redox titration

Finds an unknown concentration via a colour change: KMnO₄ is self-indicating (pink endpoint), while K₂Cr₂O₇ needs an external indicator.

Electrochemical cell

A galvanic cell converts redox energy to electricity: oxidation at the anode (−), reduction at the cathode (+), joined by a salt bridge.

Standard electrode potential

Measured against the standard hydrogen electrode (0 V); a more positive E° means a stronger oxidising agent and greater tendency to be reduced.