Lecture 27
Balancing Redox Reactions - Continued
•Example —> balance the following equation:
•The first step would be to assign the oxidation states and determine the elements
that are oxidized and reduce (write half reactions)
•We know that on the reactant side Ce has a 4+ charge, H has a 1+, As has a 3+
and O has a 2- charge
•On the product side Ce has a 3+, H has a 1+, As has a 5+, and O has a 2- charge
•The half reactions of the equation are: (1) e— + Ce4+ (aq) —> Ce3+ (aq) and!
(2) H3AsO3 (aq) —> H3AsO4 (aq) + e—
•Next you are required to balance the half-reactions
•To do this, first balance all the elements besides H and O then add the number of
electrons transferred to the product side for oxidation and to the reactant side for
reduction
•Remember that in acid solutions, add water to the side that is O deficient and add
an H+ to the side that is H deficient (balance O first then H)
•The third step is to multiply the half reactions by factors so the electrons lost in
oxidation equals the electrons gained in reduction
•The final step is to add the half reactions and cancel identical species that appear
on both sides