Question

Balance the following basic solution

H20+CrO4+Cn-->OCN+Cr(OH)3+h20

Answer 1

Solution :-

Half reactions

CrO4^2-   ----- > Cr(OH)3

CN- ------ > OCN-

Balancing the oxygen by adding H2O

CrO4^2-   ----- > Cr(OH)3 + H2O

H2O + CN- ------ > OCN-

Balancing hydrogens by adding H+

5H^+ + CrO4^2-   ----- > Cr(OH)3 + H2O

H2O + CN- ------ > OCN- + 2H^+

Balancing electronic charge by adding electrons

3e- +5H^+ + CrO4^2-   ----- > Cr(OH)3 + H2O

H2O + CN- ------ > OCN- + 2H^+ +2e-

Equalizing electrons by multiplying equation 2 by 3 and equation 1 by 2

6e- +10H^+ + 2CrO4^2-   ----- > 2Cr(OH)3 + 2H2O

3H2O + 3CN- ------ > 3OCN- + 6H^+ +6e-

Adding equation by canceling the similar species on opposite sides

4H^+ + 2CrO4^2- + H2O + 3CN^- ----- > 2Cr(OH)3 + 3OCN^-

Now adding OH- to both sides

4OH^- + 4H^+ + 2CrO4^2- + H2O + 3CN^- ----- > 2Cr(OH)3 + 3OCN^- + 4OH-

Final equation

5H2O + 2CrO4^2- + 3CN^- ----- > 2Cr(OH)3 + 3OCN^- + 4OH^-