How do you balance this?

Question

3. [1pt]


Balance the following equation by the half-reaction method in an acidic or basic solution as indicated.


OCl - + Cr(OH)4- --> CrO42- + Cl - ( basic )


You must enter the correct coefficients of the above reactants and products as a single group of integers.


Example:

C2H5OH(aq) + Cr2O72-(aq) --> CH3CO2H(aq) + Cr3+(aq)


the balanced equation is

3C2H5OH(aq) + 2Cr2O72-(aq) + 16H3O+(aq) --> 3CH3CO2H(aq) + 4Cr3+(aq) + 27H2O(l)

Answer 1

Put spaces between your species and their charges.
Chromium +3 (in Cr(OH)4 -1 ) is being oxidized to +6 in the chromate ion
The oxidizing agent, Cl+1 in the OCl -1 is reduced to the -1 level as the chloride.
So,
2 Cr+3 > 2 CrO4 -2 + 6e
3 OCl- + 6e > 3 Cl-
To balance out the oxygen with water, hydroxyl ion is used, we already have 8 from the Cr(OH)4 - ion

10 (OH) + 3O > 8O + 5 H2O
So 2[Cr(OH)4]-1 + 2 OH- + 3 OCl - >
........................ 2 CrO4 -2 + 3 Cl- + 5 H2O

Answer 2

Put spaces between your species and their charges.
Chromium +3 (in Cr(OH)4 -1 ) is being oxidized to +6 in the chromate ion
The oxidizing agent, Cl+1 in the OCl -1 is reduced to the -1 level as the chloride.
So,
2 Cr+3 > 2 CrO4 -2 + 6e
3 OCl- + 6e > 3 Cl-
To balance out the oxygen with water, hydroxyl ion is used, we already have 8 from the Cr(OH)4 - ion

10 (OH) + 3O > 8O + 5 H2O
So 2[Cr(OH)4]-1 + 2 OH- + 3 OCl - >
........................ 2 CrO4 -2 + 3 Cl- + 5 H2O

Answer 3

Here's a REALLY good site where you can plug in an equation (but only if you get stuck! :-)... make sure to read the tips on how to type them into the box.