Question

FeSO_4 (aq) + Mg (s) rightarrow Fe (s) + MgSO_4 (aq) Then, write balanced half-reactions describing the oxidation and reduction that happen in this reaction.

Answer 1

Ans. Step 1: Oxidation half reaction: An increase in oxidation number of the chemical species is called its oxidation. Note that Mg (Solid state, oxidation number = 0) is being oxidized to Mg²⁺ (Oxn. No. of Mg in MgSO₄ = +2).

            Mg(s) ----->Mg²⁺ + 2e^-                             - reaction 1

Method: Balance the number of atoms on both side of equation. Add electrons to RHS of equation to balance charge.

Step 2: Reduction half reaction: A decrease in oxidation number of the chemical species is called its reduction. Note that Fe in FeSO₄ (oxidation number = +2) is being reduced to metallic Fe (Oxn. No. of Fe metal = 0).

            Fe²⁺ + 2e^- ---> Fe                                        - reaction 2

Method: Balance the number of atoms on both side of equation. Add electrons to LHS of equation to balance charge.

Step 3: Check if the number of electrons in reaction 1 is equal to that of reaction 2 on the opposite side of the reaction.

Yes, the number of electrons is equal. Now, it is said to be ‘balanced reaction’. We don’t need go forwards anymore. So, the balanced half reactions are as follow-

            Mg(s) ----->Mg²⁺ + 2e^-                             - Oxidation half reaction

            Fe²⁺ + 2e^- ---> Fe                                        - Reduction half reaction

# Note: In number of electrons in the two half-reactions were NOT equal- say reaction 1 had 2 electrons and reaction 2 had 5 electrons.

Then, multiply the reaction 1 with number of electrons in reaction 2; and vice versa. The new multiplied reactions would be the balanced-half reactions.