Question

Mn(OH)2(s) + MnO4(aq) → MnO42–(aq) (basic solution) When the equation is balanced with smallest whole number coefficients, what is the coefficient for OH–(aq) and on which side of the equation is OH–(aq) present?

Answer 1

Hi, the given equation has some missing parts. Actual equation is- '
`Mn(OH)_(2)(s)+MnO_(4)^(-)(aq.)\rightarrow MnO_(4)^(2-)(aq.)`'

balanced equation:
`Mn(OH)_(2)(s)+4MnO_(4)^(-)(aq.)+6OH^(-)(aq.)\rightarrow 5MnO_(4)^(2-)(aq.)+4H_(2)O(l)`

Step-by-step explanation:

`Mn(OH)_(2)(s)\rightarrow MnO_(4)^(2-)(aq.)`

Balance O and H in basic medium:
`Mn(OH)_(2)(s)+6OH^(-)(aq.)\rightarrow MnO_(4)^(2-)(aq.)+4H_(2)O(l)`

Balance charge:
`Mn(OH)_(2)(s)+6OH^(-)(aq.)-4e^(-)\rightarrow MnO_(4)^(2-)(aq.)+4H_(2)O(l)` ........(1)

`MnO_(4)^(-)(aq.)\rightarrow MnO_(4)^(2-)(aq.)`

Balance charge:
`MnO_(4)^(-)(aq.)+e^(-)\rightarrow MnO_(4)^(2-)(aq.)` .....(2)

`[equation(2)* 4]+[equation (1)]:`

`Mn(OH)_(2)(s)+4MnO_(4)^(-)(aq.)+6OH^(-)(aq.)\rightarrow 5MnO_(4)^(2-)(aq.)+4H_(2)O(l)`

`OH^(-)(aq.)` is present on the left hand side of balanced equation and it's coefficient is 6