Question

Question 8.18 Balance the following redox reactions by ion-electron method:
(a) MnO4(-) (aq) + I– (aq)→MnO2 (s) + I2(s) (in basic medium)
(b) MnO4(-) (aq) + SO2 (g)→Mn2+ (aq) + HSO4(-) (aq) (in acidic solution)
(c) H2O2 (aq) + Fe2+ (aq)→Fe3+ (aq) + H2O (l) (in acidic solution)
(d)Cr2O7(2-) + SO2(g)→Cr3+ (aq) + SO4(2-) (aq) (in acidic solution)
Class XI Redox Reactions Page 274

Answer 1

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Adding the two half reactions, we have the net balanced redox reaction as:

6l-(aq) + 2MnO-4(aq) + 4H2O(l) → 3l2(s) + 2MnO2(s) + 8OH-(aq)

(b) Following the steps as in part (a), we have the oxidation half reaction as:

SO2(g) + 2H2O(l) → HSO-4(aq) + 3H+(aq) + 2e-(aq)

And the reduction half reaction as:

MnO-4(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)

Multiplying the oxidation half reaction by 5 and the reduction half reaction by 2, and then by adding them, we have the net balanced redox reaction as:

2MnO-4(aq) + 5SO2(g) + 2H2O(l) + H+(aq) → Mn2+(aq) + HSO-4(aq)

(c) Following the steps as in part (a), we have the oxidation half reaction as:

Fe2+(aq) → Fe3+(aq) + e-

And the reduction half reaction as:

H2O2(aq) + 2H+(aq) + 2e- → 2H2O(l)

Multiplying the oxidation half reaction by 2 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:

H2O2(aq) + 2Fe2+(aq) + 2H+(aq) → 2Fe3+(aq) + 2H2O(l)

(d) Following the steps as in part (a), we have the oxidation half reaction as:

SO2(g) + 2H2O(l) → SO2-4(aq) + 4H+ (aq) + 2e-

And the reduction half reaction as:

Cr2O2-7(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 3SO2-4(aq) + H2O(l)

Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:

Cr2O2-7(aq) + 3SO2(g) + 2H+(aq) → 2Cr3+(aq) + 3SO2-4(aq) + H2O(l)

Answer 2

Answer:

Adding the two half reactions, we have the net balanced redox reaction as:

6l-(aq) + 2MnO-4(aq) + 4H2O(l) → 3l2(s) + 2MnO2(s) + 8OH-(aq)

(b) Following the steps as in part (a), we have the oxidation half reaction as:

SO2(g) + 2H2O(l) → HSO-4(aq) + 3H+(aq) + 2e-(aq)

And the reduction half reaction as:

MnO-4(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)

Multiplying the oxidation half reaction by 5 and the reduction half reaction by 2, and then by adding them, we have the net balanced redox reaction as:

2MnO-4(aq) + 5SO2(g) + 2H2O(l) + H+(aq) → Mn2+(aq) + HSO-4(aq)

(c) Following the steps as in part (a), we have the oxidation half reaction as:

Fe2+(aq) → Fe3+(aq) + e-

And the reduction half reaction as:

H2O2(aq) + 2H+(aq) + 2e- → 2H2O(l)

Multiplying the oxidation half reaction by 2 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:

H2O2(aq) + 2Fe2+(aq) + 2H+(aq) → 2Fe3+(aq) + 2H2O(l)

(d) Following the steps as in part (a), we have the oxidation half reaction as:

SO2(g) + 2H2O(l) → SO2-4(aq) + 4H+ (aq) + 2e-

And the reduction half reaction as:

Cr2O2-7(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 3SO2-4(aq) + H2O(l)

Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:

Cr2O2-7(aq) + 3SO2(g) + 2H+(aq) → 2Cr3+(aq) + 3SO2-4(aq) + H2O(l)