1. (1+) (X) 4(2-) = 0 : X = Cl = 7+ 2(1+) 2(X) 7(2-) = 0 : X = Cr = 6+
RUBRIC FOR ELECTROCHEM-REDOX EXAM OF 4/2/07
1) For NaClO4
The Na is 1+, group one rule
The O is 2-, oxygen one rule
(1+) (X) 4(2-) = 0 : X = Cl = 7+
ALGEBRA RULE: the oxidation states of all atoms in a compound MUST equal the total charge of the compound.
RULE, halogens are 1- in binary compounds, however they show multiple states in polyatomic ions from table E. Must solve with algebra rule.
2) K2Cr2O7
The O is 2-, oxygen one rule
The K is 1+, group one rule
RULE, transition metals (B groups) show multiple states and you Must solve with the algebra rule if the periodic table gives more than one oxidation state..
2(1+) 2(X) 7(2-) = 0 : X = Cr = 6+
ALGEBRA RULE: the oxidation states of all atoms in a compound MUST equal the total charge of the compound.

2. Mg(s) + 2H+ (aq) + 2 Cl- (aq) Mg2+ (aq) + 2Cl- (aq) + H2 (g)
6) Mgo is oxidized into Mg2+
Mg(s) + 2H+ (aq) + 2 Cl- (aq) Mg2+ (aq) + 2Cl- (aq) + H2 (g)
reduction (2e-)
oxidation (2e-)

3. 2 Al(s) + 3 O2 (g)  2 Al2O3(s) OR Al0  Al3+ + 3e-
13) Alo is oxidized into Al3+, O is 2- in Al2O3
2 Al(s) + 3 O2 (g)  2 Al2O3(s)
oxidation
2 Al0  Al e- OR
Al0  Al e-
2 x 0 = 0 charge TOTAL
2 x 3+ = 6 charge TOTAL

4. Zn(s) + Cr 3+ (aq)  Zn2+ (aq) + Cr(s) 0 3+ 2+ 0
RUBRIC FOR 15 to 20
Zn(s) + Cr 3+ (aq)  Zn2+ (aq) + Cr(s)
reduction (3e-) = 6e-
oxidation 3 (2e-) = 6e-
3 Zn(s) + 2 Cr3+(aq)  3 Zn2+(aq) + 2 Cr(s)
Zn0  Zn2+ + 2e- (oxidation)
Cr3+ + 3e-  Cr0 (reduction)
Releases e- at ANODE
Gains e- at CATHODE