Unit 4: Topic 9 - Oxidation-Reduction (Redox) Reactions

Question 1

Re-define oxidation and reduction, and give examples of elements undergoing
oxidation/reduction via half-equations.

Answer 1

Oxidation is the process of an element losing electrons and as a result gaining charge. Reduction is
the opposite: an element gains electrons and loses charge.
Let us consider the reduction of permanganate ion to manganese (ii) ion. In MnO4-, since oxygen has
an oxidation number of 2-, we can solve for the oxidation number in manganese, which is 7+. Clearly
the oxidation number of Mn in Mn2+ is 2+. Therefore, in order to reduce the charge from 7+ to 2+, we
must reduce charge by 5, or add 5 electrons: Mn7(+) + 5e- —> Mn 2+. Oxygen is not included here
because the oxidation number stays at -2.
Oxidation of sulfite to sulfate: in SO3 2-, sulfur has oxidation number 4+, and 6+ in SO4 2-. Therefore,
two electrons are lost, shown by S(4+) —> S(6+) + 2e-.

Question 2

In addition to the conservation of mass while balancing redox equations, what
else should be kept in mind?

Answer 2

When balancing redox reactions, keep in mind that there are also electrons being transferred between
elements. To account for this, we want to make sure the net amount of electrons being transferred is
0: i.e. if there are x electrons being gained due to reduction, there should be exactly x electrons lost to
oxidation.

Question 3

Balance the equation for when permanganate ion reacts with sulfite ion in acidic
solution to produce manganese (ii) ion and sulfate ion

Answer 3

In order to balance a full oxidation-reduction equation, we need to combine the half equations in such a way such that the net electron transfer is 0, i.e. the number of electrons being oxidized equals the number of electrons being reduced. Since there are 5 electrons gained per mole of Manganese and 2 electrons lost per mole of sulfur, we should have 2 moles of manganese and 5 moles of sulfur (or anything of this ratio) in our balanced equation. Thus, we should have 2MnO4- + 10e- + 16H+ —> 8H2O + 2Mn 2+ and 5SO3(2-) + 5H2O —> 5SO4 (2-) + 10H+ + 10e-. Combine these equations to arrive at a balanced reaction: 2MnO4- + 5SO3(2-) + 6H+ —> 2Mn(2+) + 5SO4(2-) + 3H2O.