Respuesta :
Answer:
[tex]5{Fe}^{ + 2} +1{MnO_4}^{ - 1} + 8 {H}^{ + } \rightarrow 5{Fe}^{ + 3} + 1{Mn}^{ + 2} + 4H_2O[/tex]
Explanation:
To balance any redox reaction the primary factor is to identify the species undergoing oxidation and reduction.
Below are the steps to identify species undergoing oxidation and reduction:
- Calculate the oxidation number of each atom individually of reactants as well as product.
- find out the change in oxidation number from reactants to products.
- if the oxidation number of products are greater than reactants then the species undergone oxidation.
- if the oxidation number of products are lesser than reactants then the species undergone reduction.
let's calculate oxidation number,
given reaction
[tex]{Fe}^{ + 2} +{MnO_4}^{ - 1} \rightarrow {Fe}^{ + 3} + {Mn}^{ + 2}[/tex]
Of reactant,
[tex]Fe = +2, {MnO_4}^{ - 1} \\ Let \: oxidation \: of \: Mn \: is \: x \: and \: Oxygen -2 \\ {MnO_4}^{ - 1} \rightarrow \: x + 4 \times ( - 2) = - 1 \\ {MnO_4}^{ - 1} \rightarrow \: x + - 8 = - 1 \\ {MnO_4}^{ - 1} \rightarrow \: x = + 8 - 1 \\ {MnO_4}^{ - 1} \rightarrow \: x = + 7 \\ \fbox{ Mn \: +7, O \: -2}[/tex]
Of product,
[tex]\fbox{Fe \: +3, \: Mn \: +2}[/tex]
Now calculate the change in oxidation number
[tex]Fe = +3-(+2) = +1 \\ Mn = +2-(+7) = -5[/tex]
now balance the increase in oxidation number and decrease in oxidation number,
thus we require 5 Fe for 1 MnO4-,
Now putting these numbers as their coefficient,
[tex]5{Fe}^{ + 2} +1{MnO_4}^{ - 1} \rightarrow 5{Fe}^{ + 3} + 1{Mn}^{ + 2}[/tex]
now balance oxygen atom on product side by adding water,
since there are 4 oxygen atoms in reactants 4 water molecules will be there on product side.
[tex]5{Fe}^{ + 2} +1{MnO_4}^{ - 1} \rightarrow 5{Fe}^{ + 3} + 1{Mn}^{ + 2} + 4H_2O[/tex]
now balance Hydrogen atom on reactant side by adding H+ ion, since there are 4 water molecules on product 8 H+ ion will be in reactants!
[tex]5{Fe}^{ + 2} +1{MnO_4}^{ - 1} + 8 {H}^{ + } \rightarrow 5{Fe}^{ + 3} + 1{Mn}^{ + 2} + 4H_2O[/tex]
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Confirming the balanced reaction, by checking number of atoms and charge on reactant vs Product side.
Total number of Fe on reactant and product = 5
Total number of Mn on reactant and product= 1
Total number of Oxygen on reactant and product = 4
Total number of hydrogen on reactant and product= 8
Total charge on reactant side= 5×(+2)-1+8=17
Total charge on product side = 5×(+3)+1×(+2)=17
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