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

Answer:

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

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)

$[2\times equation(1)]+[3\times equation(2)]:$

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

The coefficient of water is 8.