Balancing Redox Reaction in Acidic Solution: Step-by-Step Guide with Example

Balance the redox reaction below if it takes place in an acidic solution.Fe2+ (aq) + MnO4- (aq) → Fe3+ (aq) + Mn2+ (aq)

5Fe^2+(aq)+8H^+(aq)+MnO4^-(aq)=Mn^2+(aq)+4H2O(l)+5Fe^3+(aq)

The first step in balancing a redox reaction is to separate it into two half-reactions: the oxidation half-reaction and the reduction half-reaction.

Oxidation half-reaction:
Fe2+ (aq) → Fe3+ (aq)

Reduction half-reaction:
MnO4- (aq) → Mn2+ (aq)

Next, balance each half-reaction separately:

Oxidation half-reaction:
Fe2+ (aq) → Fe3+ (aq) + e-

Reduction half-reaction:
MnO4- (aq) + 8 H+ (aq) + 5 e- → Mn2+ (aq) + 4 H2O (l)

In this step, we balanced the oxidation half-reaction by adding an electron to the product side, and balanced the reduction half-reaction by adding H+ ions and water molecules to the reactant side, and balancing the charge on both sides.

Now, we need to balance the number of electrons transferred in each half-reaction. To do this, we multiply the oxidation half-reaction by 5, so that the number of electrons on both sides are equal:

5 Fe2+ (aq) → 5 Fe3+ (aq) + 5 e-

Now that the number of electrons is balanced, we can combine the two half-reactions to get the overall balanced equation:

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

So, the balanced redox reaction in an acidic solution is:

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

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