Question

In: Chemistry

Balance the following redox reactions. Show the complete balanced reduction half and oxidation half reaction, and...

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

Solutions

Expert Solution

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-  


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