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CBSE · Class 11 · Chemistry

Redox Reactions

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
In terms of electrons, oxidation is the:
  1. gain of protons
  2. loss of neutrons
  3. gain of electrons
  4. loss of electrons
2.
The oxidation number of iron in Fe3O4 is:
  1. +8/3
  2. −8/3
  3. +11/3
  4. +5/3
3.
In the reaction 2CuO + C → 2Cu + CO2, the reducing agent is:
  1. CuO
  2. Cu
  3. C
  4. CO2
4.
The redox reaction 4KClO3 → 3KClO4 + KCl is an example of:
  1. non-metal displacement
  2. decomposition
  3. combination
  4. disproportionation
5.
When Cl2 changes to Cl−, each Cl2:
  1. gains 1 electron
  2. loses 2 electrons
  3. gains 2 electrons
  4. gains 3 electrons
6.
In Stock notation, Hg2Cl2 is named:
  1. mercury(I) chloride
  2. mercury(IV) chloride
  3. mercury(II) chloride
  4. mercury(III) chloride
7.
Using standard reduction potentials (Zn2+/Zn −0.76 V, K+/K −2.93 V, Au3+/Au +1.4 V, H+/H2 0 V), the strongest reducing agent is:
  1. H2
  2. Zn
  3. Au
  4. K
8.
Assertion (A): Tin cannot displace chromium from a solution of its salt.
Reason (R): Oxidation is the loss of electrons.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
Define oxidation and reduction in terms of oxidation number.
10.
Find the oxidation number of each element in NO3−.
11.
Balance by the ion-electron method (acidic medium): Zn + NO3− → Zn2+ + NH4+
12.
How many moles of electrons are gained when 3 mol of MnO4− changes to Mn2+? How many coulombs is this? (1 mol of electrons = 96,500 C)
13.
Write the formula of chromium(III) chloride, and the oxidation number of chromium in it.
14.
A cell is made from the couples Cr3+/Cr (E° = −0.74 V) and Fe2+/Fe (E° = −0.44 V). Which electrode is the anode, which the cathode, and what is E°cell?
15.
In Fe2O3 + 3CO → 2Fe + 3CO2, which element is oxidised and which is reduced? Show the change in oxidation number.
16.
Will iron displace aluminium from a solution of its salt? Give a reason. (E°: Fe2+/Fe −0.44 V, Al3+/Al −1.66 V)
Section C — Word problems4 × 4 = 16 marks
17.
Write the rules for finding the oxidation number of an element in a compound or ion.
18.
Balance the reaction between permanganate and manganese(II) in acidic medium by the ion-electron method, showing both half reactions: MnO4− + Mn2+ → MnO2
19.
Find the oxidation number of nitrogen in N2H4, NO2, NH2OH, NO, and arrange them in increasing order of it. Nitrogen ranges from −3 to +5: which of these can act only as an oxidising agent, and which only as a reducing agent?
20.
A galvanic cell is made from Zn in a 1 M solution of Zn2+ and Fe in a 1 M solution of Fe2+. (E°: Zn2+/Zn −0.76 V, Fe2+/Fe −0.44 V) (a) Write the cell notation. (b) Write the half reactions and the cell reaction. (c) Find E°cell. (d) In which direction do electrons flow in the wire?
AmbifulWorksheets
Practice that never repeats
CBSE · Class 11 · Chemistry

Answer key

Redox Reactions · AW-11C-RDX-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) loss of electrons—
2(a) +8/3the oxidation numbers add up to the charge on the species
3(c) CC is oxidised: C 0 → +4
4(d) disproportionationone element is both oxidised and reduced
5(c) gains 2 electronsCl: 0 → −1, for 2 atoms
6(a) mercury(I) chlorideHg is +1
7(d) Kthe lowest E°: its reduced form is most easily oxidised
8(b) Both A and R are true, but R is not the correct explanation of A.A is true: E°: Sn −0.14 V, Cr −0.74 V. R is true but does not explain A.
9Oxidation is an increase in the oxidation number of an element; reduction is a decrease.—
10N +5, O −2the sum is the charge, −1
114Zn + NO3− + 10H+ → 4Zn2+ + NH4+ + 3H2Ooxidation: Zn → Zn2+ + 2e−; reduction: NO3− + 10H+ + 8e− → NH4+ + 3H2O
1215 mol; 1,447,500 Ceach MnO4− gains 5 electrons
13CrCl3; +3the Roman numeral is the oxidation number
14Anode Cr, cathode Fe; E°cell = +0.3 Vthe lower E° is oxidised (anode); E°cell = E°cathode − E°anode
15Oxidised: C +2 → +4; reduced: Fe +3 → 0an increase in oxidation number is oxidation
16No: Fe has the higher E°, so it cannot reduce Al3+.the metal with the lower E° is the stronger reducing agent
17An element in the free state has 0. A monatomic ion has its charge. Oxygen is −2 (−1 in peroxides, +2 in OF2). Hydrogen is +1 (−1 in metal hydrides). Fluorine is always −1; alkali metals +1; alkaline earth metals +2. The sum of oxidation numbers is zero for a neutral compound and equals the charge for an ion.—
18Oxidation: Mn2+ + 2H2O → MnO2 + 4H+ + 2e−. Reduction: MnO4− + 4H+ + 3e− → MnO2 + 2H2O. Overall: 2MnO4− + 3Mn2+ + 2H2O → 5MnO2 + 4H+.balance O with H2O and H with H+ (in basic medium then add OH−), then electrons; multiply so the electrons cancel
19N2H4 −2 < NH2OH −1 < NO +2 < NO2 +4. Only an oxidising agent (highest state, can only be reduced): none of them. Only a reducing agent (lowest state, can only be oxidised): none of them.charge balance with O −2 and H +1
20(a) Zn | Zn2+ || Fe2+ | Fe. (b) Anode: Zn → Zn2+ + 2e−; cathode: Fe2+ + 2e− → Fe; cell: Zn + Fe2+ → Zn2+ + Fe. (c) +0.32 V. (d) From Zn to Fe.lower E° is the anode; E°cell = E°cathode − E°anode