Balance the following oxidationreduction reaction using the

Balance the following oxidation-reduction reaction using the half-reaction method. The reaction is carried out in sodium hydroxide.

-

C14H12 + MnO4 2- ---> 2C7H5O2 - + MnO2

Solution

C14H12 -------> 2C7H5O2-

adding H2O to balance O.

C14 H12 + 4H2O ------> 2C7H5O2- , H+ added on to balance H

C14H12 + 4H2O -----> 2C7H5O2- + 10H+ , now same amount OH- added both sides

C14H12 + 10OH- ----> 2C7H5O2- + 6H2O , charge also balanced,

second half

MnO4 2- ---> MnO2

MnO42- ------> MnO2 + 2H2O ,

MnO4 2- + 4H+ ----> MnO2 + 2H2O ,

MnO4 2- + 4H2O --------> MnO2 + 2H2O + 4OH- ,

MnO4 2- + 2H2O + 2e- --------> MnO2 + 4OH- now adding both half reactions

C14H12 + 2e-(aq)+ 6OH-(aq)+ MnO42-(aq) ----> 2C7H5O2-(aq)+ MnO2 +4 H2O