Balance the following redox reaction in an acidic medium.
Mn 2+ + BiO3- --> MnO4- + Bi 3+
First, we need to write the half-reactions for each species involved in the reaction.
1. Oxidation half-reaction:
Mn 2+ --> MnO4- (oxidation state change of Mn from +2 to +7)
2. Reduction half-reaction:
BiO3- --> Bi 3+ (oxidation state change of Bi from +5 to +3)
Next, we balance each half-reaction separately:
Oxidation half-reaction:
Mn 2+ --> MnO4-
Mn 2+ --> MnO4-
First, balance the number of Mn atoms by adding 5 H2O molecules to the product side:
Mn 2+ + 5 H2O --> MnO4-
Then, balance the charge by adding 8 H+ ions to the product side:
Mn 2+ + 5 H2O + 8 H+ --> MnO4- + 8 H+
Reduction half-reaction:
BiO3- --> Bi 3+
BiO3- --> Bi 3+
First, balance the number of Bi atoms by adding 2 H2O molecules to the product side:
BiO3- + 2 H2O --> Bi 3+
Then, balance the charge by adding 6 H+ ions to the reactant side:
BiO3- + 2 H2O + 6 H+ --> Bi 3+
Now we combine the two half-reactions, ensuring that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction. In this case, 5 electrons are lost in the oxidation half-reaction and 6 electrons are gained in the reduction half-reaction. To balance the electron transfer, we multiply the oxidation half-reaction by 6 and the reduction half-reaction by 5 before adding them together:
6(Mn 2+ + 5 H2O + 8 H+ --> MnO4- + 8 H+) + 5(BiO3- + 2 H2O + 6 H+ --> Bi 3+)
gives:
6Mn2+ + 30 H2O + 48 H+ --> 6MnO4- + 48 H+ + 5BiO3- + 10 H2O + 30 H+ --> 5Bi3+
Finally, simplifying the equation by removing the spectator ions from both sides gives the balanced redox reaction in an acidic medium as:
6Mn 2+ + 8 H2O + 5 BiO3- + 48 H+ --> 6 MnO4- + 5 Bi 3+ + 30 H2O