Can somebody help me balance this equation

Question

Cr(OH)3 (s) ClO- (aq) OH -(aq) -----> CrO42- (aq) Cl- (aq) H2O(l)

Answer 1

This is a redox reaction occurring in a basic solution.   The trick is to balance it as if it were acidic and then add enough hydroxide ions to each side to neutralize all of the hydrogen ions.

I prefer the half-reaction method where water is used to balance oxygen and then use H+ to balance the hydrogen atoms.  In this case we use an extra step to use OH- to neutralize H+ and make the solution basic.

Multiply each half-reaction by a constant so that the number of electrons gained and lost is the same.

Cr(OH)3(s) + ClO- + OH -   ---> CrO4^2- +  Cl- + H2O(l)

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

3( 2H+ + ClO- + 2e- --> Cl- + H2O )

--------------------------------------...

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

add 10 OH- to each side

4OH- + 2Cr(OH)3 + 3ClO- + 8H2O --> 2CrO4^2- + 3Cl- + 13H2O

Simplify the water molecules

4OH- + 2Cr(OH)3 + 3ClO-  --> 2CrO4^2- + 3Cl- + 5H2O

Answer 2

2Cr(OH)3 + 3ClO- + 4OH- --> 2CrO4^2- + 3Cl- + 5H2O