For a particular redox reaction Cr is oxidized to CrO42Solut

For a particular redox reaction Cr is oxidized to CrO42

oxidation half reaction

Cr(s) --------> CrO42- which can be balanced as

4H2O +Cr ---------------> CrO42- +8H+ +6e- ..............(1)

Reduction Half reaction

Cu2+ ----------> Cu+ which can be balanced as

Cu2+ +e- --------> Cu+ ...................(2)

multiplying equaion (2) by 6 and then adding equation (1) and (2) , we get

4H2O +Cr +6 Cu2+ ----------------> 6Cu+ + CrO42- + 8H+

but the reaction condition should be basic so adding 8OH- to both sides . we get..

8OH- + 4H2O +Cr +6 Cu2+ ----------------> 6Cu+ + CrO42- + 8H+ +8OH-

======> that is

OH- + 4H2O +Cr +6 Cu2+ ----------------> 6Cu+ + CrO42- + 8H2O

========> hence

8OH- +Cr +6 Cu2+ ----------------> 6Cu+ + CrO42- + 4H2O

it is the final equation.