Respuesta :
Answer:
Balanced equation: [tex]2Cr(OH)_{3}(s)+3H_{2}O_{2}(aq.)+4OH^{-}(aq.)\rightarrow 2CrO_{4}^{2-}(aq.)+8H_{2}O(l)[/tex]
The coefficient of water is 8
Explanation:
Oxidation: [tex]Cr(OH)_{3}(s)\rightarrow CrO_{4}^{2-}(aq.)[/tex]
Balance O and H in basic medium: [tex]Cr(OH)_{3}(s)+5OH^{-}(aq.)\rightarrow CrO_{4}^{2-}(aq.)+4H_{2}O(l)[/tex]
Balance charge: [tex]Cr(OH)_{3}(s)+5OH^{-}(aq.)-3e^{-}\rightarrow CrO_{4}^{2-}(aq.)+4H_{2}O(l)[/tex] ......(1)
Reduction: [tex]H_{2}O_{2}(aq.)\rightarrow H_{2}O(l)[/tex]
Balance O and H in basic medium: [tex]H_{2}O_{2}(aq.)\rightarrow 2OH^{-}(aq.)[/tex]
Balance charge: [tex]H_{2}O_{2}(aq.)+2e^{-}\rightarrow 2OH^{-}(aq.)[/tex] .....(2)
[tex][2\times equation(1)]+[3\times equation(2)]:[/tex]
[tex]2Cr(OH)_{3}(s)+3H_{2}O_{2}(aq.)+4OH^{-}(aq.)\rightarrow 2CrO_{4}^{2-}(aq.)+8H_{2}O(l)[/tex]
The coefficient of water is 8.
