1. Redox
Reactions
Chemistry / Solutions / PKT – III / Ionic Equilibrium and Redox Reactions ... 12

2. Take-Off
1. b Cl of KClO3 (O, N = +5) is reduced to Cl– and O of KClO3
(O.N. = –2) is oxidised to O2. Hence, redox but not In MnO2 − , the oxidiation state of Mn is + VI
4
disproportionation. Thus, when MnSO4 is converted into MnO2, then the
valency factor is 2, and the equivalent weight of MnSO4
+4 0 will be half of its molecular weight.
2. c S O2 + 2H2S  3S+ 2H2O
→
9. a 2KMnO4 + 3H2SO4 → K2SO4 + 2MnSO4
mol wt. 64 + 3H2O + 5[O]
Eq. wt. of SO2 = = = 16
4 4
SO2 − + [O] → SO2 −  × 5
 3
 4  
+4 +6 2KMnO4 + 3H2SO4 + 5SO3 − → K 2SO4
2
3. b 3SO2 + O3  3S O 3
→
mol wt +2MnSO4 + 3H2O + 5SO2− 4
or
Eq wt. =
2 2MnO4 + 6H+ + 5SO3 − → 2Mn2+ + 5SO2− + 3H2O
− 2
4
5 mole of SO3 − required 2 moles of MnO−
2
4
0 0 +5 –1
4. b I 2 + Cl2  I O3 + Cl
→ – –
2
total no. of moles of e– lost by 1 mole of I2 are 10 1 moles of SO3 − will be required
2 moles of MnO− .
4
5
5. a Equivalent wt = mass of metal which reacts with 8.0 g
oxygen. 10. d W O 0 .9 WO3
0.2 g oxygen react with 1 g metal
1× 8
8.0 g oxygen will react with = 40 gmetal n = 1.8 n = +6
0.2
4 .2
∴ eqivalent wt of metal = 40
1 mol of WO0.9 == 4.2 mol of e–
6. b The total number of moles of e– lost by one mole of 200 × 103 mg == 4.2 mol
K4 [Fe(CN)6 ] are 61
4.2 × 30
30 mg requires == mol
7. c In this reaction, both Fe reduced and S is getting oxidised 200 × 103
1 e – g ain
4.2 × 30
= × 6.023 × 1023 = 3.78 × 1020 e−
Fe S 2 +2 O2 Fe
+3
+ 2 S O2 200 × 103
11. b The metal with higher value of oxidation potential will
1 2e– lo st
replace other metal. Hence, Mg(+2.37) will replace
Al(+1.66).
Mol. wt
Thus, Eq. wt =
11 12. a Metal with minimum standard reduction potential will
be most reactive. Hence, A(–3.05) is most reactive
molecular weight (as a reducing agent).
8. b Equivalent weight =
valency factor
13. a On decreasing the negative value of reducing elec-
If valency factor is 2, then equivalent weight will be trode potential of metals, the reducing character is
equal to its molecular weight. regularly decrease due to decreasing cation forma-
In MnSO4, the oxidation state of Mn is + II tion tendency (i.e., electron donating tendency)
In Mn2O3, the oxidation state of Mn is + III x y z
In MnO2, the oxidation state of Mn is + IV 0.55 V, – 3.03 V – 1.18 V
hence, correct order is
In MnO− , the oxidation state of Mn is + VII
4 y>z>x
Chemistry / Solutions / PKT – III / Ionic Equilibrium and Redox Reactions ... 13

3. 14. a On the basis of reduction-potential (z > y > x) 22. b See the definition of hydrogen electrode.
A spontaneous reaction will have the following
characterstics: 23. d Fe ++ + Sn  Fe + Sn++
→
z reduced and x oxidised
At anode:
y reduced and y oxidised
z reduced and y oxidised Sn  Sn++ + 2e− ; Eo
→ = −0.14 V
Hence, y will oxidise x and not z. Sn++ / Sn
At cathode:
15. b At anode, Fe changes to Fe3+.
At cathode, Fe3+ changes to Fe2+. Fe++ + 2e−  Fe; Eo
→ = −0.44 V
Fe++ / Fe
Eo = −0.30 V
cell
16. d Eo = Eo
cell
o
cathode − Ecathode
= – 0.44 – (–0.76) 24. Sn  Sn4 + + 4 e −
→
= +0.32 V
6e− + Cr2 6  2 Cr 3 +
+
→
17. c Reduction-potentials of the ions are in the order:
∴ meq. of Sn2+ = meq. of K2Cr2O7
Ag+ > Hg2+ > Cu2+ > Mg2+
2 1 1
× 1000 = ×V
Mg2+ (aq) will not be reduced as it reduction potential Esn 10
is much lower than water (– 0.83 V). Hence the se-
1 1
quence of deposition of the metals will be: Ag, Hg, Cu. × 1000 = ×V
118.7 10
18. c Pt / H2(g) / HCl (soln) / AgCl(s) / Ag 4
∴ V = 336.98 ml
19. c From given data,
25. Let the element be A
3 E°Cu2+ / Cu+ = 0.153 V The salt formed is isomorphous with
∴ E°Cu+ / Cu2+ = – 0.153 V
K2SO4 i.e., K2AO4
(i) Cu2+ + 2e–  Cu, E° = +0.337 V,
→
∴ A  A 6 + + 6 e −
→
∴ ∆G°1 = – 2 × 0.337 × F
= 0.674 F ∴ At. wt. of A = Eq. wt × no. of e– lost
= 13.16 × 6
(ii) Cu+  Cu2+ + e– , E° = – 0.153 V,
→ = 78.96
∴ ∆G°2 = – 1 × (– 0.153)F
__________________________________ 26. c For I and IV,
Cu+ + e– → Cu, E° = ? ∴ ∆G°3 = – 1 × E° × F D → D2+ + 2e– ; E° = 1.12 V
ox
___________________________________
or – E°F = ∆G°1 + ∆G°2 A + e – → A – ; E°red = 0.96
= – 0.674 F + 0.153 F so E°cell = E°ox + E°red = 2.08 V
or – E°F = – 0.52 F, It is maximum for all possible combinations
∴ E° = 0.521 V
27. b
20. d Metal with large negative reduction potential will oxidize
readily. 28. d From electrochemical series, the reduction potential
of Fe2+, Br2 and MnO4– are – 0.44 V + 1.056 V and +
21. Fe  Fe++ + 2e−
→ o
∆G1 1.51 V respectively. So order of oxidising tendency is
Fe2+ < Br2 < MnO4–
Fe++  Fe3 + + e− ∆Go
→ 2
__________________ 29. a Fe2+ can reduce Br2 MnO4–, Cu2+ because reduction
Fe  Fe3 + + 3e−
→ ∆G3 = ∆G1 + ∆Go
o
2
potential for these species is greater than reduction
potential of Fe2+
−3EoF = −2E1 F − EoF
3
o
2
30. b
3E o = 2E 1 + E o
3
o
2
We use ∆G0 because it is a state function. 31. a
While Eo is not a state function. 32. d
Chemistry / Solutions / PKT – III / Ionic Equilibrium and Redox Reactions ... 14