Question

Balance the following redox equations by the ion-electron method:? A) H2O2 + Fe 2+ ---> Fe 3+ + H2O (in the acidic solution) C) CN- + MnO4- ---> CNO- +MnO2 (in basic solution) E) S2O2/3- + I2 ---> I- + S4O2/6- (in acidic solution)

Answer 1

We balance the given reactions above by following the rules in balancing redox reactions in acidic or basic solutions. Balance the atoms aside from the O and H atoms. Then we balance the Os and Hs by adding H2O or H+. Finally, we balance the total charge of the reactant and product by adding e-. We do as follows:

A) H2O2 + Fe 2+ ---> Fe 3+ + H2O (in the acidic solution)
2H+ + H2O2 + Fe 2+ ---> Fe 3+ + 2H2O
e- + 2H+ + H2O2 + Fe 2+ ---> Fe 3+ + 2H2O

C) CN- + MnO4- ---> CNO- +MnO2 (in basic solution)
CN- + MnO4- ---> CNO- +MnO2 + H2O
2H+ + CN- + MnO4- ---> CNO- +MnO2 + H2O
2OH- + 2H+ + CN- + MnO4- ---> CNO- +MnO2 + H2O + 2OH-
2H2O + CN- + MnO4- ---> CNO- +MnO2 + H2O + 2OH-
e- + H2O + CN- + MnO4- ---> CNO- +MnO2 + 2OH-

E) S2O2/3- + I2 ---> I- + S4O2/6- (in acidic solution)
2S2O2/3- + I2 ---> 2I- + S4O2/6-
4H+ + 2S2O2/3- + I2 ---> 2I- + S4O2/6- + 2H2O
6e- + 4H+ + 2S2O2/3- + I2 ---> 2I- + S4O2/6- + 2H2O