Click for the balanced reaction.

• Step 1

  Determine the oxidation number of each species in the unbalanced reaction.

  	Left-hand-side	Right-hand-side
  Cr	+3	+6
  Cl	+1	-1

  Cr in Cr(OH)₃ is oxidized. Cl in ClO^- is reduced.

• Step 2

  Write two separate equations, one using only the substances that are involved in the reduction, and another using only the substances involved in oxidation. (When balanced, these are called half-reaction equations.)

  Oxidation:	Cr(OH)3 -> CrO42-
  Reduction:	ClO- -> Cl-

• Step 3

  Balance each kind of atom other than H and O by using coefficients.

  • They are balanced in these equations.
• Step 4

  Balance oxygen atoms by using H₂O.

  Oxidation:	Cr(OH)3 + H2O -> CrO42-
  Reduction:	ClO- -> Cl- + H2O

• Step 5

  Even though this reaction takes place in basic solution, balance H atoms by using H⁺ ions. We will bring in the OH^- ions in the final step.

  Oxidation:	Cr(OH)3 + H2O -> CrO42- + 5 H+
  Reduction:	ClO- + 2 H+ -> Cl- + H2O

• Step 6

  Balance the charge on each side of the reactions by adding electrons as necessary.
  Electrons appear on the right-hand-side for the oxidation half-reaction.
  Electrons appear on the left-hand-side for the reduction half-reaction.
  

  Oxidation:	Cr(OH)3 + H2O -> CrO42- + 5 H+ + 3 e -
  Reduction:	ClO- + 2 H+ + 2 e- -> Cl- + H2O

• Step 7

  Multiply the half-reactions by the simplest set of whole numbers so that

  electrons gained = electrons lost

  Oxidation:	2 [ Cr(OH)3 + H2O -> CrO42- + 5 H+ + 3 e -]
  Reduction:	3 [ ClO- + 2 H+ + 2 e- -> Cl- + H2O ]

• Step 8

  Combine the oxidation half-reaction with the reduction half-reaction. Cancel electrons and equal amounts of any substance that appears on both sides of the equation.

  2 Cr(OH)3 (s) + 3 ClO- (aq) -> 2 CrO42- (aq) + 3 Cl- (aq) + 4 H+ (aq) + H2O (l)

• Step 9

  Add enough OH^- (aq) ions to both sides of the reaction to neutralize the H⁺ (aq) ions.

  Here we add four OH^- (aq) ions on each side.

  2 Cr(OH)3 (s) + 3 ClO- (aq) + 4 OH- (aq) -> 2 CrO42- (aq) + 3 Cl- (aq) + 4 H+ (aq) + 4 OH- (aq) + H2O (l)

  On the right-hand-side, the four H⁺ combine with the four OH^- ions to form four H₂O.

  2 Cr(OH)3 (s) + 3 ClO- (aq) + 4 OH- (aq) -> 2 CrO42- (aq) + 3 Cl- (aq) + 4 H2O (l) + H2O (l)

  The balanced redox reaction is:

  2 Cr(OH)3 (s) + 3 ClO- (aq) + 4 OH- (aq) -> 2 CrO42- (aq) + 3 Cl- (aq) + 5 H2O (l)

  ClO- is the oxidizing agent.
  Cr(OH)3 is the reducing agent.

• Step 10

  Check to make sure all atoms and charges are balanced, and to see that all substances have whole number coefficients that are in their simplest ratios.

  	Left-hand-side	Right-hand-side
  Cr	2	2
  Cl	3	3
  H	10	10
  O	13	13
  charge	- 7	- 7

  Left-hand-side = Right-hand-side