Online Test — Redox Reactions
18 Questions • 15 min • Chapter MCQ
15:00
Question 1 of 18
Oxidation is best defined as:
gain of electrons
loss of electrons
gain of protons
loss of protons
Explanation: Oxidation Is Loss (OIL) of electrons.
Question 2 of 18
The oxidation number of Mn in MnO4- is:
+4
+5
+6
+7
Explanation: x + 4(-2) = -1 gives x = +7.
Question 3 of 18
In Zn + CuSO4 → ZnSO4 + Cu, the oxidising agent is:
Zn
CuSO₄
ZnSO₄
SO₄²⁻
Explanation: Cu²⁺ is reduced to Cu, so CuSO₄ is the oxidising agent.
Question 4 of 18
The oxidation number of S in H2SO4 is:
+2
+4
+6
+8
Explanation: 2(+1) + x + 4(-2) = 0 gives x = +6.
Question 5 of 18
Which is an example of a disproportionation reaction?
Zn + 2HCl → ZnCl₂ + H₂
2H₂O₂ → 2H₂O + O₂
C + O₂ → CO₂
AgNO₃ + NaCl → AgCl + NaNO₃
Explanation: Oxygen at -1 is both reduced to -2 and oxidised to 0 — disproportionation.
Question 6 of 18
The oxidation number of H in NaH is:
+1
-1
0
+2
Explanation: In metal hydrides hydrogen is -1.
Question 7 of 18
How many electrons are gained per MnO4- in acidic medium?
3
4
5
7
Explanation: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.
Question 8 of 18
While balancing a half-reaction in acidic medium, oxygen is balanced by adding:
O₂
OH⁻
H₂O
H⁺
Explanation: O atoms are balanced with H₂O, then H with H⁺.
Question 9 of 18
In 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O, the number of HNO3 molecules is:
6
8
10
4
Explanation: The balanced equation uses 8 HNO₃.
Question 10 of 18
The n-factor of K2Cr2O7 as an oxidant in acidic medium is:
2
3
6
7
Explanation: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.
Question 11 of 18
Equivalent mass of KMnO4 (M = 158) in acid is:
158
79
52.7
31.6
Explanation: 158 / 5 = 31.6 g equiv⁻¹.
Question 12 of 18
In iodometry, liberated I2 is titrated with:
KMnO₄
K₂Cr₂O₇
Na₂S₂O₃
NaOH
Explanation: Thiosulphate: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻.
Question 13 of 18
The standard reduction potential of the standard hydrogen electrode is:
0.00 V
+0.34 V
-0.76 V
+1.10 V
Explanation: SHE is the reference, defined as 0.00 V.
Question 14 of 18
Which reaction is a comproportionation?
2H₂O → 2H₂ + O₂
5Cl⁻ + ClO₃⁻ + 6H⁺ → 3Cl₂ + 3H₂O
Cl₂ + 2NaOH → NaCl + NaClO + H₂O
Zn + Cu²⁺ → Zn²⁺ + Cu
Explanation: Two chlorine species (-1 and +5) form one intermediate state (0) — comproportionation.
Question 15 of 18
The reducing agent in MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O is:
MnO₂
HCl
MnCl₂
H₂O
Explanation: Cl goes -1 to 0 (oxidised), so HCl is the reducing agent.
Question 16 of 18
The dichromate ion Cr2O72- has chromium in the oxidation state:
+3
+4
+6
+7
Explanation: 2x + 7(-2) = -2 gives x = +6.
Question 17 of 18
For a basic-medium half-reaction balanced first as acidic, you then add:
H⁺ to one side
OH⁻ equal to H⁺ on both sides
extra O₂
starch
Explanation: Add OH⁻ equal to the H⁺ on both sides; H⁺ + OH⁻ form H₂O.
Question 18 of 18
The EMF of the Daniell cell (Zn²⁺/Zn = -0.76 V, Cu²⁺/Cu = +0.34 V) is:
+0.42 V
+1.10 V
-1.10 V
+0.34 V
Explanation: EMF = +0.34 - (-0.76) = +1.10 V.