In: Chemistry
Balance the following redox reactions. Show the complete balanced reduction half and oxidation half reaction, and label each accordingly. Give the complete balanced reaction.
A.) CN-1 + MnO4-2 ---> CNO-1 + MnO2
B.) S2O3-2 + IO2-1 ---> I- + S4O6-2
Break into half equations
CN-
CNO- equation 1
MnO42-
MnO2 equation 2
balance O using water, then use H+ to balance H, then
cancel H+ using OH- (hence alkali
conditions), then add electrons to balance charges.
CN- + H2O
CNO- (add water)
CN- + H2O CNO- + 2H+ (add H+)
CN- + H2O + 2OH- CNO- + 2H+ + 2OH- (cancel)
CN- + 2OH-
CNO- + H2O + 2e- (add
electrons)
MnO42-
MnO2
MnO42- MnO2 + 2H2O
MnO42- + 4H+ MnO2 + 2H2O
MnO42- + 4H+ + 4OH- MnO2 + 2H2O + 4OH-
MnO42- + 2H2O + 2e-
MnO2 + 4OH-
Now
CN- + 2OH- CNO- + H2O + 2e-
MnO42- + 2H2O + 2e-
MnO2 + 4OH-
and combine both equations while cancelling out species that occur
on both sides leaving
CN- + MnO42-
CNO- + MnO2 + 2OH-
(B)
S2O3-2 + IO2- I- + S4O6-2
oxidation half :-
S2O32- S4O62-
2S2O32- S4O62-
2S2O32--------> S4O62-+ 2e- (1)
reduction half
IO2- I-
IO2- I- + 2H2O
IO2- + 4H+ I- + 2H2O
IO2- + 4H++ 4OH- I- + 2H2O + 4OH-
IO2- + 2H2O + 4e- I- + 4OH- (2)
Now,
2S2O32- S4O62- + 2e- (1)
IO2- + 2H2O + 4e- I- + 4OH- (2)
Multiplying equation (1) by 2, we get
4S2O32- 2S4O62- + 4e- (3)
Adding (2) and (3), we get;
4S2O32- + IO2- + 2H2O 2S4O62- + I- + 4OH-