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Balance the following redox reaction if it occurs in acidic solution. What are the coefficients in front of H+ and Fe3+ in the balanced reaction? Fe2+(aq) + MnO4-(aq) → Fe3+(aq) + Mn2+(aq) Balance the following redox reaction if it occurs in acidic solution. What are the coefficients in front of H+ and Fe3+ in the balanced reaction? Fe2+(aq) + MnO4-(aq) → Fe3+(aq) + Mn2+(aq) H+ = 8, Fe3+ = 1 H+ = 8, Fe3+ = 5 H+ = 2, Fe3+ = 3 H+ = 3, Fe3+ = 2 H+ = 5, Fe3+ = 1

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Answer:

H⁺ = 8; Fe³⁺ = 5

Explanation:

In order to balance a redox reaction, we use the ion-electron method.

Step 1: Identify both half-reactions.

Reduction: MnO₄⁻(aq) → Mn²⁺(aq)

Oxidation: Fe²⁺(aq) → Fe³⁺(aq)

Step 2: Balance the masses using H⁺ and H₂O where appropiate.

8 H⁺(aq) + MnO₄⁻(aq) → Mn²⁺(aq) + 4 H₂O(l)

Fe²⁺(aq) → Fe³⁺(aq)

Step 3: Balance the charges adding electrons where appropiate.

8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

Fe²⁺(aq) → Fe³⁺(aq) + 1 e⁻

Step 4: Multiply both half-reactions by numbers that assure that the number of electrons gained is equal to the number of electrons lost.

1 × [8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)]

5 × [Fe²⁺(aq) → Fe³⁺(aq) + 1 e⁻]

Step 5: Add both half-reactions

8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ + 5 Fe²⁺(aq) → Mn²⁺(aq) + 4 H₂O(l) + 5 Fe³⁺(aq) + 5 e⁻

8 H⁺(aq) + MnO₄⁻(aq) + 5 Fe²⁺(aq) → Mn²⁺(aq) + 4 H₂O(l) + 5 Fe³⁺(aq)

Answer:

The coefficient of [tex]\rm H^+[/tex] in the balanced equation is 8 and coefficient of [tex]\rm Fe^3^+[/tex] is 5.

Explanation:

To balance a redox reaction in the acidic medium, the number of atoms of all the elements involved in the reaction must be equal for substrates and product.

The reaction given has:

[tex]\rm Fe^2^+\;+\;MnO_4^-\;\rightarrow\;Fe^3^+\;+\;Mn^2^+[/tex]

The change in the oxidation number of the elements on reaction is seen.

Fe changes its oxidation state from +2 [tex]\rm \rightarrow[/tex] +3, change is +1

The oxidation state of Mn changes from +7 [tex]\rm \rightarrow[/tex] +2, change is -5

There is required 5 atoms of Fe for 1 atom of Mn.

With this, the equation can be written as:

[tex]\rm 5\;Fe^2^+\;+\;1\;MnO_4^-\;\rightarrow\;5\;Fe^3^+\;+\;1\;Mn^2^+[/tex]

The reaction is further balanced with H and O, with formation of [tex]\rm H_2O[/tex].

[tex]\rm 5\;Fe^2^+\;+\;1\;MnO_4^-\;+\;8\;H^+\rightarrow\;5\;Fe^3^+\;+\;1\;Mn^2^+\;+\;4\;H_2O[/tex]

The coefficient in front of [tex]\rm H^+[/tex] is 8 and that of [tex]\rm Fe^3^+[/tex] is 5.

For more information, refer the link:

https://brainly.com/question/2671074?referrer=searchResults

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