(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.

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) 0 1+ 1- 2+ 1- 0 reduction (2e-) oxidation (2e-)

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) 0 0 3+ 2- oxidation 2 Al0  Al23+ + 6e- OR Al0  Al3+ + 3e- 2 x 0 = 0 charge TOTAL 2 x 3+ = 6 charge TOTAL

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) 0 3+ 2+ 0 reduction 2 (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