Balance each of the following in BASE:
a) CN-+ MnO4-®CNO-+ MnO2(hint: treat CN as a UNIT with a single oxidation number…this shortcut will not show you which of C or N is being oxidized or reduced, but it will show you whether oneof them is being oxidized or reduced, which will give you the info you need. This is harder than something I’d put on the exam).
b) Mn2++ H2O2®MnO2+ H2O (hint: H2O2has oxygen in the RARE -1 oxidation state. One more of these outliers, just for practice.)
c) Bi(OH)3+ SnO22-®SnO32-+ Bi
d) CrO42-(aq)+ Cu(s)®Cr(OH)3(s)+ Cu(OH)2(s)
You have the balances:
a) 3CN- + 2MnO4- + 2H + → 3CNO- + 2MnO2 + H2O
b) Mn2 + + H2O2 → MnO2 + 2H +
c) 2Bi (OH) 3 + 3SnO22- → 3SnO32- + 2Bi + 3H2O
d) 2CrO42- + 3Cu + 4H2O + 4H + → 2Cr (OH) 3 + 3Cu (OH) 2
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Balance each of the following in BASE: a) CN-+ MnO4-®CNO-+ MnO2(hint: treat CN as a UNIT with...
Balance the following redox equations by the half-reaction method: (a) Mn2+ + H2O2 → MnO2 + H2O (in basic solution) (b) Bi(OH)3 + SnO22− → SnO32− + Bi (in basic solution) (c) Cr2O72− + C2O42− → Cr3+ + CO2 (in acidic solution) (d) ClO3− + Cl− → Cl2 + ClO2 (in acidic solution) (e) Mn2+ + BiO3− → Bi3+ + MnO4− (in acidic solution)