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

Answer:

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

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)

$[equation(2)\times 4]+[equation (1)]:$

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

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