−1.63 V + (+2.38 V) = 0.75 V = 0.75
By inspection, 2 moles of electrons are transferred to produce 1 mole of product; therefore n =- Now determine ∆G°:
10 . 2Fe (s) + O 2 (g) + H 2 O (l) → Fe(OH) 2 (s)
To assign oxidation numbers, use the rules given in this chapter. Fe (s) and O 2 (g) have
oxidation numbers of zero, because they are free elements. Hydrogen in H 2 O (l) has an
oxidation number of +1, because oxygen is more electronegative than hydrogen; likewise,
oxygen in H 2 O (l) has an oxidation number of −2. Oxygen and hydrogen in Fe(OH) 2 (s) have the
same oxidation numbers as in H 2 O (l). Each OH group contributes a charge of −1 to Fe(OH) 2
and since there are 2 OH groups, their overall contribution to the compound is −2. Since
Fe(OH) 2 is a neutral compound and thus has no overall charge, the sum of all the oxidation
numbers of the atoms in this compound is zero. Consequently, the Fe in Fe(OH) 2 must possess
a charge of +2 in order to make the overall charge on the compound zero.11 . The balanced equation is:
ClO 3 − + 3H 2 O + 3AsO 2 − → 3AsO 43 − + Cl− + 6H+12 . The rule is that the half-reaction with the greater reduction potential will be the reduction
reaction and the reverse of the other will be the oxidation reaction. Since the first reaction
(reduction of ClO 4 −) has the greater reduction potential, this will proceed as a forward reaction
and ClO 4 − will be reduced. The second reaction (reduction of Ag+), which has the lower
reduction potential, will proceed as a reverse reaction, and Ag (s) will be oxidized to Ag+.
Oxidation occurs at the anode and reduction at the cathode. Thus, the reduction of ClO 4 − to
ClO 3 − will occur at the cathode and the oxidation of Ag (s) to Ag+ will occur at the anode.