Question

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

Answer 1

Break into half equations
CN^- CNO^- equation 1
MnO₄^2- 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)

MnO₄^2- MnO₂

MnO₄^2-   MnO₂  + 2H₂O

MnO₄^2- + 4H⁺ MnO₂  + 2H₂O

MnO₄^2- + 4H⁺ + 4OH^- MnO₂  + 2H₂O + 4OH^-

MnO₄^2- + 2H₂O + 2e- MnO₂ + 4OH-

Now

CN^- + 2OH^- CNO^- + H₂O + 2e^-

MnO₄^2- + 2H₂O + 2e- MnO₂ + 4OH-

and combine both equations while cancelling out species that occur on both sides leaving
CN^- + MnO4^2- CNO^- + MnO₂ + 2OH^-

(B)

S₂O₃^-2 + IO₂^-      I^- + S₄O₆^-2

oxidation half :-

S₂O₃^2-    S₄O₆^2-

2S₂O₃^2-    S₄O₆^2-

2S₂O₃^2--------> S₄O₆^2-+ 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,

2S₂O₃^2- S₄O₆^2- + 2e^- (1)

IO₂^- + 2H₂O + 4e^-    I^- +   4OH^-                  (2)

Multiplying equation (1) by 2, we get

4S₂O₃^2-      2S₄O₆^2- + 4e^- (3)

Adding (2) and (3), we get;

4S₂O₃^2- + IO₂^- + 2H₂O       2S₄O₆^2-   +   I^- +   4OH^-