Question

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10.) Balance the following redox reactions by the ion-electron method: Show your step by step Work, do not just give your answer. a) S2O32- + 12 = 1 +S4062" (in acidic solution) b) Mn2+ + H2O2 → MnO2 + H20 (in basic solution) c) Bi(OH)3 + SnO22- → Sn032 + Bi (in basic solution) d) CIO3 +Cl → Cl2 + ClO2 (in acidic solution)

Answer 1

(10) (a) 5. 02 - + I2 I + S406²- Reduction half; I 2+2e 21 - Oxidation half 25203 5402- + ze Addring these two rxns; I + 25202 - 21 + SAOB Mg2+ + H₂O2 MnO2t H₂O Oxidation half Mn²+ + 404 MnO2 + 2H₂O +2e- Reduction half + 2010 H2O + H₂O +2e → H₂O half reactions; on adding both Mg2+ + 4 + 2040 Mnog + 2H₂O

8:0H) + SnO2- SnO2- + Bi Oxidation hallo SnO2 + 2014- Sno + H₂O +2e Reduction half: BIOH)2 + 3e- Bi 30H Now multiply oxidation half by 3 and reduction half by to balance the number of electrons and then add; we 2 get 3500? + 2 BOH)₂ - 3 5n03 + 3H₂O + 281 C103 + CI- Cl + CO₂ Oxidation half (1) + 20- 2 C1- Reduction half 1,02 H₂O c1o3 + 2H+ + 20 Add gor & clo3 + 24+ Cl + clo2+ H₂O
Common rule in acidic condition :

Firstly balance all the elements other than hydrogen and oxygen . Secondly balance oxygen by adding water molecule on appropriate side . Hydrogen is balanced by H+ . Lastly charge is balanced by adding appropriate number of electrons on appropriate side .

In basic condition :

All the rules remain same except oxygen is balanced by OH- and hydrogen is balanced by adding appropriate number of water molecules. .