Question

balance equation by ion electron method MnO4+SnO2 2+H2O -MnO2+ SnO3 2 +OH (alkaline medium)​

Answer 1

Answer:

balance equation by ion electron method MnO4+SnO2 2+H2O -MnO2+ SnO3 2 +OH (alkaline medium)​

Refer (v)

Answer 2

Answer:

The balanced equation is $3SnO_{2}^{2-}+2MnO_{4}^{-}+H_2O\rightarrow 3SnO_{3}^{2-} +2MnO_2+2OH^-$

Explanation:

Given equation is

$MnO_4+SnO_2+H_2O\rightarrow MnO_2+SnO_3+OH^-$

We have to balance the given equation.

We know that Balanced chemical equation

• A balanced chemical equation is an equation where the number of atoms of each type in the reaction is the same on both reactants and product sides.
• The mass, as well as the change, are equal in a balanced chemical equation.

Now we find the Half  reaction Oxidation is

$SnO_{2}^{2-}\rightarrow SnO_{3}^{2-}\\\\SnO_{2}^{2-}+OH^-\rightarrow SnO_{3}^{2-}+H_2O\\\\SnO_{2}^{2-}+2OH^-\rightarrow SnO_{3}^{2-}+H_2O+2e^-$

Now we find the Half reaction Reduction is

$MnO_{4}^{-}\rightarrow MnO_2\\\\MnO_{4}^{-}\rightarrow MnO_2+2OH^-\\\\MnO_{4}^{-}+2H_2O\rightarrow MnO_2+2OH^-\\\\MnO_{4}^{-}+2H_2O+3e^-\rightarrow MnO_2+4 OH^-$

Now add both the reaction,

$[SnO_{2}^{2-}+2OH^- \rightarrow SnO_{3}^{2-}+H_2O+2e^-]\times3\\\underline{[MnO_{4}^{-}+2H_2O+3e^-\rightarrow MnO_2+4OH^-]\times2}\\3SnO_{2}^{2-}+6OH^-+2MnO_{4}^{-}+4H_2O\rightarrow 3SnO_{3}^{2-}+3H_2O+2MnO_2+8OH^-$

Further simplifying the above equation, we get

$3SnO_{2}^{2-}+2MnO_{4}^{-}+H_2O\rightarrow 3SnO_{3}^{2-} +2MnO_2+2OH^-$

Hence, the balanced equation is $\bold{3SnO_{2}^{2-}+2MnO_{4}^{-}+H_2O\rightarrow 3SnO_{3}^{2-} +2MnO_2+2OH^-}$

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