How to write a redox equation from half equations
There are three crucial steps in balancing redox half equations, which, when mastered, allow any two half equations, no matter how to complex, to be easily balanced. First, we balance the electrons, by determining how many electrons are transferred in each half equation. Take for instance the redox reaction between aluminium and dilute sulphuric acid. Reduction: H2SO4 (aq) + 2e- —> SO4 2- (aq) + H2 (g). Oxidation: Al (s) —> Al3+ (aq) + 3e-. The reduction equation is multiplied by 3 to give 6 electrons transferred. The oxidation equation is multiplied by 2 to give 6 electrons transferred.Next, we add the multiplied equations together and cancel the electrons, so we have:3H2SO4 (aq) + 2Al (s) —> 3SO42- (aq) + 3H2 (g) + 2Al3+ (aq). As SO4 2- and Al3+ are oppositely charged ions, they will attract to form a solution of Al2SO4, so we can see that the species still balance.The third step does not apply to this simple equation, but is important when dealing with more complex redox equations such as those found in transition metal chemistry, and involved cancelling the species that appear on both sides of the equation, so that the equation is left in its empirical form.
