Organic Chemistry Multiple Choice Questions on “Redox Reactions”.
1. KMnO4 reacts with oxalic acid according to the equation given below. Here 20 mL of 0.1 M KMnO4 is equivalent to how many mL and molarity of H2C2O4?
2MnO4– + 5C2O4– + 16H+ → 2Mn+1 + 10CO2 + 8H2O
a) 20 mL of 0.5 M H2C2O4
b) 50 mL of 0.5 M H2C2O4
c) 50 mL of 0.1 M H2C2O4
d) 20 mL of 0.1 M H2C2O4
Answer: c
Clarification: In any reaction number of equivalent reacted are equal
∴ neq (KMnO4) = neq (C2 O42-)
Also, neq = n × nf
In given problem:
∴ neq(KMnO4) = neq (C2 O42-)
n × nf = (n × nf) C2 O4–
Also, (M × V × nf) KMnO4 = 0.1 × 20 × 5 = 10
And, (n × nf) C2 O4– = 50 × 0.1 × 2 = 10
2. Excess of KI reacts with CuSO4 solution and then Na2S2O3 solution is added to it. Which of the statement is incorrect for this reaction?
a) Na2S2O3 is oxidized
b) CuI2 is formed
c) Cu2I2 is formed
d) Evolved I2 is reduced
Answer: b
Clarification: (1) KI + CuSO4 → I2
(2) I2 + Na2S2O3 → NaI + Na2S2O6
(a) Na2S2O3 is oxidized to Na2S2O6
(b) Cu2I2 is formed as CuI2 disproportionate.
3. How many number of moles of KMnO4 will react with 180 gm H2C2O4 according to given reaction?
KMnO4 + H2C2O4 → 2C02 + Mn2+
a) 4/5
b) 2/5
c) 1/5
d) 4/3
Answer: a
Clarification: neqKMnO4 = neqH2C2O4
n × nf = n × nf
n × 5 = w⁄M × 2
n = 180⁄90 × 2⁄5 = 4⁄5.
4. What is the mass of K2Cr2O7 required to produce 254 gm I2 from KI solution? Given: K2Cr2O7 + 2KI → 2Cr3+ + I2
a) 49 g
b) 98 g
c) 9.8 g
d) 4.9 g
Clarification:
neqK2Cr2O7 = neqI2
n × nf = n × nf
w⁄M × 6 = w⁄M × 2
w/294 × 6= 254/254 × 2
w = 98g.
5. What is the volume of 0.05 M KMnO4 which will react with 50 ml of 0.1 M H2S in acidic medium (H2S → S02)?
a) 60 ml
b) 6 ml
c) 12 ml
d) 120 ml
Answer: d
Clarification:
neqKMnO4 = neq H2S
M × V × nf = M × V × nf
0.05 × V × 5 = 0.1 × 50 × 6
V = 120 mL.
