Question

calculate the oxidation number method for kmno4+na2so3 gives Mno2+na2so4+koh​

Answer 1

We have to balance given redox reaction with help of oxidation number method.

O.N of Mn in KMnO4 = +7

O.N of Mn in MnO2 = +4

hence, MnO4^- is reduced into MnO2 .

O.N of S in Na2SO3 = +4

O.4 of S in Na2SO4 = +6

hence, SO3²- is oxidised into SO4²- .

step 1 :- let's balance reduction reaction,

$KMn^{+7}O_4+3e^-\rightarrow Mn^{+4}O_2+KOH$

step 2 : let's balance the oxidation reaction,

$Na_2S^{+4}O_3\rightarrow Na_2S^{+6}O_4+2e^-$

for balancing, multiply (2) with equation (1) and 3 with equation (2) and then add both equation.

$2KMnO_4+3Na_2SO_3\rightarrow2MnO_2+2KOH+3Na_2SO_4$.

now to Balance 2 hydrogen and one oxygen add one H2O in left side of chemical equation.

i.e.,$2KMnO_4+3Na_2SO_3+H_2O\rightarrow2MnO_2+2KOH+3Na_2SO_4$.

This is balanced chemical equation.

Answer 2

Answer:

steps and how to solve with explaination