Mc Ques Eng Redox Rxns Ans

Public Exam Multiple Choice Questions - Redox Reactions Suggested Answers :

1.

A

2.

B

3.

B

4.

C

5.

C

6.

B

7.

C

8.

D

9.

A

10 .

B

11.

D

12 .

B

13 .

D

14 .

C

15 .

C

16 .

D

17 .

B

18 .

C

19 .

A

20 .

D

21 .

A

22 .

C

23 .

C

24 .

B

25 .

B

26 .

C

27 .

D

28 .

A

29 .

B

30 .

B

31 .

B

32 .

A

33 .

C

34 .

C

Explanation of M.C. Questions :

1. A : O.N. of Cu changes from +2 (in CuSO4) to 0 (in Cu) B : O.N. of Fe changes from +2 (in FeCl2) to +3 (in FeCl3) C : no change in O.N. of Pb (+2) D : no. change in O.N. of Mg (+2) 2. O.N. change of Fe : +3 (in Fe2O3) to 0 (in Fe) ⇒ undergoes reduction (oxidizing agent) O.N. change in C : +2 (in CO) to +4 (in CO2) ⇒ undergoes oxidation (reducing agent) (2) is correct statement. Ans. : B 3. O.N. of N in N2 = 0 O.N. of N in NH3 = -3 O.N. of N in NO = +2 O.N. of N in NO2 = +4 O.N. of N in HNO3 = +5 O.N. change in step 1 = 3 O.N. change in step 2 = 5 O.N. change in step 3 = 2 O.N. change in step 4 = 1 Step 2 involves the largest change in O.N. Ans. : B 4. The electronic arrangement of X = 2,8,8,2 X is calcium, which is reducing agent. It belongs to Period 4. When it burns in oxygen, a brick red flame is observed. (2) and (3) are correct statements. Ans. : C

5. A : Cl2 is an oxidizing agents which can be reduced by Na2SO3. Cl2(g) + SO32-(aq) + H2O(l) → 2Cl-(aq) + SO42-(aq) + 2H+(aq) S is a reducing agent which can’t react with Na2SO3. B : Na is a stronger reducing agent than Mg because Na is in a higher position in E.C.S. C : P and Cl are non-metals, the compound formed from them is covalent. The formula can be PCl3 or PCl5. D : 2Mg(s) + O2(g) → 2MgO(s) MgO(s) + H2O(l) → Mg(OH)2 Mg(OH)2 is alkaline / basic. This means that MgO is a basic oxide. 6. O.N. of C changes from 0 to +4 O.N. change = 4 O.N. of N changes from +5 to +4 O.N. change = 1 O.N. of S changes from +4 to +6 O.N. change = 2 O.N. of Mn changes from +7 to +2 O.N. change = 5 Ans. : B 7. O.N. of Cl changes from 0 (in Cl2) to –1 (in NaCl) ⇒ undergoes reduction (oxidizing agent) O.N. of Cl changes from 0 (in Cl2) to +1 (in NaOCl) ⇒ undergoes oxidation (reducing agent) Cl2 undergoes reduction and oxidation at the same time. Ans. : C 8. A reducing agent undergoes oxidation, which involves in an increase in O.N. A : O.N. of S changes from +4 (in SO2) to 0 (in S) B : no change in O.N. of Pb (+2) C : no change in O.N. of H and Cl D : O.N. of Br changes from –1 (in KBr) to 0 (in Br2)

9.

The equation is constructed in the following way : at zinc : Zn → Zn2+ + 2e- ------------------------------------ (i) at graphite : 2NH4+ + 2e- → 2NH3 + H2 ------------------ (ii) 2MnO2 + H2 → Mn2O3 + H2O -------------- (iii) (i) + (ii) + (iii), Zn + 2NH4+ + 2MnO2 → Zn2+ + 2NH3 + Mn2O3 + H2O x = 1, y = 2, z = 3 Ans. : A

10. A : no change in O.N. of Cu (+2) B : O.N. of S changes from +4 (in SO2) to 0 (in S) C : O.N. of N changes from –3 (in NH3) to 0 (in N2) D : O.N. of Zn changes from 0 (in Zn) to +2 (in Zn(NO3)2) 11. A : correct. Formulae of bromine and chlorine are Br2 and Cl2 respectively. B : correct. They belong to Group VII which means that they have 7 e-s in their outermost shell. C : correct. They form Br- and Cl- respectively. D : wrong. Cl2 is a stronger oxidizing agent than Br2. Cl2 is in a lower position in E.C.S. than Br2. 12. dilute nitric acid acts as an oxidizing agent which can compete with MnO4-/H+ to react with reducing agent. Ans. : B 14. (1) : it reacts KMnO4/H+(aq) to carry out addition reaction and form ethan-1,2-diol CH2=CH2 → HO-CH2-CH2-OH (2) : CuSO4 can’t react with KMnO4/H+(aq). (3) : FeSO4 is reducing agent with is oxidized by KMnO4/H+(aq). 5Fe2+(aq) + MnO4-(aq) + 8H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) (1) and (3) react with KMnO4/H+(aq) Ans. : C

15. The chemical equation involved : Al(s) + 3Ag+(aq) → Al3+(aq) + 3Ag(s) statement (1) : correct. Al is a reactive metal and Ag+ is a less reactive metal ion. Reactive metal can displace less reactive metal ion. It is a displacement reaction. statement (2) : it’s a process which coat a layer of Al2O3 on the surface of aluminium. statement (3) : correct. O.N. of Al changes from 0 to +3 (oxidation) O.N. of Ag changes from +1 to 0 (reduction) (1) and (3) are correct statements. Ans. : C 16. O.N. of Cl changes from 0 (in Cl2) to –1 (in KCl) ------------------- (i) O.N. of Cl changes from 0 (in Cl2) to +5 (in KClO3) ---------------- (ii) There is no change in O.N. of K (+1) statement (1) : wrong. statement (2) : correct. From (i). statement (3) : correct. From (ii). (2) and (3) are correct statements. Ans. : D 18. A : O.N. of Fe changes from +3 (in Fe2(SO4)3) to +2 (in FeSO4) O.N. of S changes from -2 (in H2S) to 0 (in S) B : O.N. of Al changes from 0 (in Al) to +3 (in AlCl3) O.N. of H changes from +1 (in HCl) to 0 (in H2) C : there is no change in O.N. of each of the species D : O.N. of Cl changes from +5 (in KClO3) to -1 (in KCl) O.N. of O changes from -2 (in KClO3) to 0 (in O2) 19. A : O.N. of N = -3 B : O.N. of N = +2 C : O.N. of N = +1 D : O.N. of N = 0 20. statement (1) : correct. O.N. of X changes from 0 (in X) to -2 (in X2-) statement (2) : correct. statement (3) : correct. no. of protons does not change when X changes to X2-. (1), (2) and (3) are correct statements. Ans. : D

21. the overall ionic equation : 16H+(aq) + 2MnO4-(aq) + 5C2O42-(aq) → 2Mn2+(aq) + 8H2O(l) + 10CO2(g) no. of moles of MnO -4 2 = no. of moles of C 2 O 24- 5 2 no. of moles of MnO4- = 5 = 0.4 mol Ans. : A 22. A : there is no change in O.N. of every species B : there is no change in O.N. of every species C : O.N. of Fe changes from +2 (in FeSO4) to +3 (in Fe2O3) – oxidation O.N. of S changes from +6 (in FeSO4) to +4 (in SO2) – reduction D : there is no change in O.N. of every species 24. let x be O.N. of Pb then x + 4 × (-1) = -2 ∴ x = +2 Ans. : B 25. A : there is no change in O.N. of every species B : O.N. of P changes from +3 (in PCl3) to +5 (in PCl5) – oxidation O.N. of Cl changes from 0 (in Cl2) to -1 (in PCl5) – reduction C : there is no change in O.N. of every species D : there is no change in O.N. of every species 26. let x be O.N. of V then x + 3 × (-2) + 1 = 0 ∴ x = +5 Ans. : C

27. statement (1) : correct. FeSO3 is pale green and Fe2(SO3)3 is yellowish brown. statement (2) : correct. KMnO4/H+ can only react with FeSO4. 5Fe2+(aq) + MnO4-(aq) + 8H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) statement (3) : correct. Both can react with ammonia to give insoluble hydroxides but they are of different colour. Fe2+(aq) + 2OH-(aq) → Fe(OH)2(s) (dirty green)

Fe (aq) + 3OH (aq) → Fe(OH)3(s) 3+

-

(yellowish brown)

(1), (2) and (3) are correct statements. Ans. : D 28. (1) : correct. The solution contains mobile ions but water does not. It thus conducts electricity better. SO2(g) + H2O(l) → H2SO3(aq) H2SO3(aq) → 2H+(aq) + SO32-(aq) (2) : correct. Sulphite solution is reducing agent and Fe2(SO4)3 is oxidizing agent. They can carry out redox reaction. SO32-(aq) + 2Fe3+(aq) + H2O(l) → SO42-(aq) + 2Fe2+(aq) + 2H+(aq) (3) : wrong. KBr(aq) is reducing agent. It can’t react with sulphite solution. (1) and (2) are correct statements. Ans. : A 29. A : wrong. It’s an exothermic reaction. B : correct. Ca(s) + 2H2O(l) → Ca(OH)2(aq) + H2(g) O.N. of Ca changes from 0 (in Ca) to +2 (in Ca(OH)2) O.N. of H changes from +1 (in H2O) to 0 (in H2) C : wrong. No explosion observed. D : wrong. It doesn’t burn in water. 30. A : wrong. Na is reducing B : correct. Hg is toxic. C : wrong. D : wrong. K2Cr2O7 is oxidizing 31. A : O.N. of Fe changes from +3 to 0 B : there is no change in O.N. of Cu C : there is a loss of O in the process, thus it’s reduction. D : O.N. of S changes from +6 to +4

32. X (magnesium) is metal but Y (sulphur) and Z (chlorine) are non-metals. A : correct. They give an ionic compound with the formula of XZ2. B : wrong. Z is a stronger oxidizing agent than Y. In E.C.S., Cl2 is in lower position on the L.H.S. than S. C : wrong. X has a giant metallic structure. D : wrong. Y can’t conduct electricity even in molten state because it doesn’t contain mobile ions. 33. A : SO32-(aq) + I2(aq) + H2O(l) → SO42-(aq) + 2I-(aq) + 2H+(aq) the colour of solution changes from reddish brown (I2) to colourless (I-). B : 5SO32-(aq) + 2MnO4-(aq) + 6H+(aq) → 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l) the colour of solution changes from purple (MnO4-) to colourless (Mn2+). C : there is no reaction between Na2SO3 and Cr2(SO4)3. D : SO32-(aq) + 2Fe3+(aq) + H2O(l) → SO42-(aq) + 2Fe2+(aq) + 2H+(aq) the colour of solution changes from yellowish brown (Fe3+) to pale green (Fe2+). 34. let x be O.N. of S then 2x + 2 × 1 + 7 × (-2) = 0 ∴ x = +6 Ans. : C

---------- End of Analysis ----------