IMO Practice Test — Redox Reactions
12 Questions • 15 min • Olympiad level
15:00
Question 1 of 12
The oxidation number of S in the thiosulphate ion S2O32- (average) is:
+2
+4
+6
-2
Explanation: 2x + 3(-2) = -2 gives average S = +2.
Question 2 of 12
The oxidation number of Fe in K4[Fe(CN)6] is:
+2
+3
+4
+6
Explanation: K = +1 (×4), CN = -1 (×6): 4 + Fe - 6 = 0, so Fe = +2.
Question 3 of 12
In the balanced equation 2MnO4- + 5C2O42- + 16H+ → products, how many CO2 are formed?
5
8
10
16
Explanation: 5 C₂O₄²⁻ each give 2 CO₂ → 10 CO₂.
Question 4 of 12
The oxidation number of Cl in HClO4 is:
+3
+5
+7
-1
Explanation: +1 + x + 4(-2) = 0 gives x = +7.
Question 5 of 12
1 mole of K2Cr2O7 oxidises how many moles of Fe2+ in acidic medium?
2
3
5
6
Explanation: Cr₂O₇²⁻ accepts 6 electrons, so it oxidises 6 Fe²⁺.
Question 6 of 12
The average oxidation number of C in CH3COOH is:
0
+1
-2
+2
Explanation: 2C + 4(+1) + 2(-2) = 0 gives average C = 0.
Question 7 of 12
Which species can act ONLY as an oxidising agent (highest oxidation state)?
SO₂
H₂O₂
KMnO₄
Fe²⁺
Explanation: Mn in KMnO₄ is at its maximum +7, so it can only be reduced.
Question 8 of 12
In the disproportionation 3Br2 + 6OH- → 5Br- + BrO3- + 3H2O, bromine changes from 0 to:
only -1
only +5
both -1 and +5
+1 and +3
Explanation: Br goes to -1 (reduced) and +5 (oxidised) — disproportionation.
Question 9 of 12
The number of electrons transferred when 1 mol H2O2 acts as a reducing agent (to O2) is:
1
2
3
4
Explanation: O goes from -1 to 0 for two O atoms: H₂O₂ → O₂ + 2H⁺ + 2e⁻, so 2 electrons.
Question 10 of 12
Equivalent mass of K2Cr2O7 (M = 294) as an acidic oxidant is:
294
147
98
49
Explanation: 294 / 6 = 49 g equiv⁻¹.
Question 11 of 12
When acidified KMnO4 is titrated, dilute H2SO4 is used rather than HCl because:
HCl is too weak
HCl is oxidised by MnO₄⁻, giving false results
H₂SO₄ is cheaper
HCl forms a precipitate with Mn²⁺
Explanation: MnO₄⁻ oxidises Cl⁻ to Cl₂, consuming extra permanganate and giving erroneous readings.
Question 12 of 12
For the couple with E° = +0.80 V (Ag⁺/Ag) and the SHE, the cell EMF (Ag as cathode) is:
0.00 V
+0.40 V
+0.80 V
+1.60 V
Explanation: EMF = E°(cathode) - E°(anode) = +0.80 - 0.00 = +0.80 V.