Balance the following redox reaction in acidic solution. zn(s)+mno−4(aq)→ zn+2(aq)+mn+2(aq)

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

Explanation:

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

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

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.