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Answer:

2MnO₄⁻ + 5Zn + 16H⁺ → 2Mn²⁺ + 8H₂O + 5Zn²⁺

Explanation:

To balance a redox reaction in an acidic medium, we simply follow some rules:

  1. Split the reaction into an oxidation and reduction half.
  2. By inspecting, balance the half equations with respect to the charges and atoms.
  3. In acidic medium, one atom of H₂O is used to balance up each oxygen atom and one H⁺ balances up each hydrogen atom on the deficient side of the equation.
  4. Use electrons to balance the charges. Add the appropriate numbers of electrons the side with more charge and obtain a uniform charge on both sides.
  5. Multiply both equations with appropriate factors to balance the electrons in the two half equations.
  6. Add up the balanced half equations and cancel out any specie that occur on both sides.
  7. Check to see if the charge and atoms are balanced.

Solution

                            Zn + MnO₄⁻ → Zn²⁺ + Mn²⁺

The half equations:

                      Zn → Zn²⁺                          Oxidation half

                      MnO₄⁻ → Mn²⁺                  Reduction half

Balancing of atoms(in acidic medium)

                     Zn → Zn²⁺

                    MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O

Balancing of charge

                   Zn → Zn²⁺ + 2e⁻

                    MnO₄⁻ + 8H⁺ + 5e⁻→ Mn²⁺ + 4H₂O

Balancing of electrons

         Multiply the oxidation half by 5 and reduction half by 2:

                          5Zn → 5Zn²⁺ + 10e⁻

                        2MnO₄⁻ + 16H⁺ + 10e⁻→ 2Mn²⁺ + 8H₂O

Adding up the two equations gives:

              5Zn + 2MnO₄⁻ + 16H⁺ + 10e⁻ → 5Zn²⁺ + 10e⁻ + 2Mn²⁺ + 8H₂O

The net equation gives:

         5Zn + 2MnO₄⁻ + 16H⁺ → 5Zn²⁺ + 2Mn²⁺ + 8H₂O

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The balanced redox reaction equation is 5Zn(s) + 2MnO4^−(aq) +  16H^+(aq) ----> 5Zn^+2(aq) + 2Mn^2+(aq) + 8H2O(l)

What is redox reaction?

The term redox reaction refers to a reaction where there is loss or gain  of electrons. The first step is to break up the reaction into half equation.

The oxidation half equation is;

Zn(s) -----> Zn^+2(aq) + 2e   ---- 1

The reduction half equation is

MnO4^−(aq) + 5e  + 8H^+(aq) ----> Mn^2+(aq) + 4H2O(l) ------- 2

Next we multiply equation (1) by 5 and equation (2) by 2

5Zn(s) -----> 5Zn^+2(aq) + 10e --- 3

2MnO4^−(aq) + 10e  + 16H^+(aq) ----> 2Mn^2+(aq) + 8H2O(l)

Next we write the equations together omitting the electrons

5Zn(s) + 2MnO4^−(aq) +  16H^+(aq) ----> 5Zn^+2(aq) + 2Mn^2+(aq) + 8H2O(l)

Learn more about reaction equations: https://brainly.com/question/3045247

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