Question

Balance the following equation in basic solution, using the lowest possible integer coefficients. What is the coefficient of water? H2O2(aq) + Cr(OH)3(s) ? CrO42-(aq) + H2O(l) a. 1 b. 2 c. 4 d. 6 e. 8

Answer 1

Balanced equation:
`2Cr(OH)_(3)(s)+3H_(2)O_(2)(aq.)+4OH^(-)(aq.)\rightarrow 2CrO_(4)^(2-)(aq.)+8H_(2)O(l)`

The coefficient of water is 8

Step-by-step explanation:

Oxidation:
`Cr(OH)_(3)(s)\rightarrow CrO_(4)^(2-)(aq.)`

Balance O and H in basic medium:
`Cr(OH)_(3)(s)+5OH^(-)(aq.)\rightarrow CrO_(4)^(2-)(aq.)+4H_(2)O(l)`

Balance charge:
`Cr(OH)_(3)(s)+5OH^(-)(aq.)-3e^(-)\rightarrow CrO_(4)^(2-)(aq.)+4H_(2)O(l)` ......(1)

Reduction:
`H_(2)O_(2)(aq.)\rightarrow H_(2)O(l)`

Balance O and H in basic medium:
`H_(2)O_(2)(aq.)\rightarrow 2OH^(-)(aq.)`

Balance charge:
`H_(2)O_(2)(aq.)+2e^(-)\rightarrow 2OH^(-)(aq.)` .....(2)

`[2* equation(1)]+[3* equation(2)]:`

`2Cr(OH)_(3)(s)+3H_(2)O_(2)(aq.)+4OH^(-)(aq.)\rightarrow 2CrO_(4)^(2-)(aq.)+8H_(2)O(l)`

The coefficient of water is 8.