Question

Write balanced net ionic equations for the following reactions. Supply H^+ and/ or H2O as needed to obtain balance.

2+ 3+ 2+ 2+ b Cro)+CuCr+ +Cu(s) Cr(s) Cu+ Cu(s) -+ NO3+ Zn(s) + NO2(g) + Mn2+ + Se042 T3Co(CNT(CN)64 2+

Answer 1

(a) The half reactions are Oxidation Sn+2 Reduction: Co+3+e- _-> Co+2 -> Sn+4 2 e Balance electrons: Oxidation Sn+2 Reduction: 2 Co+3+ 2 e Sn+42 e -> 2 Co+2 Combine: Sn+2 +2 Co+3 2 e---..>Snt4 2e- 2Co+2 Sn+22 Co*3 >Sn+4 + 2Co12 (b) The half reactions are Oxidation : Cr (s) -> Cr+3 +3 e Reduction: Cu*eCu (s) Balance electrons: Oxidation :Cr (s) -> Cr+3 3 e Reduction: 3 Cut +3 e> 3 Cu (s) Combine: Cr (s) + 3 Cu* +3 e > Cr+3 3 e3 Cu (s) Cr (s) +3 Cu*Cr3 Cu (s)

(#C) The half reactions are Reduction Oxidation Zn Zn+2 No3 NO NO Balance atoms other than H and O Zn -> Zn+2 3 -> NO Balance Oxygen: Zn >Zn+2 NO3 > NO2 + H2O Add H,O on less side) Balance Hydrogen : Zn> Zn+2 (Add H+ on less side) Balance charge Zn>Zn+2+ 2 e Combine: -> NO2+ H2O > 2 NO2 H2O Zn Zn+2 +2 e 2NO2-+4H+ + 2 e 2 NO4H+ 2 e--> Combine: 2 NO3+ Zn+ 4H* 2 NO2 + 2 H20 > 2 NO2 +Znt2+ 2 H20

The half reactions are Oxidation Reduction Mno-> Mn*2. Balance atoms other than H and O Balance Oxygen: H2SeO3 H2O -> SeO4 Mno-> Mn (Add H2O on less side) Balance Hydrogen: H2SeO H2OSe04 H+ 8HMn++ 4 H2O (Add Ht on less side) Balance charge:H2Se03 H2O> SeO4+4 H Combine: 5 H2SeO3 5 H2O3 Mn04 24 H*+ 15 e The balanced equation is 36e Mn+2+ 4 H2O Mno4 + 8 H+5 e-> 3 MnO4+ 24 H*+ 15 e 5 H2SeO 5 H2O ----> 5 Seo+ 20 H+ + 15e e>3 Mn+2+12 H2O -> 5 SeO20 H 15e 3 Mn+2+ 12 H2O 5 H2SeO3 3 MnO4+ 4 H+> 5 Se04 3 Mn2+ 7 H20

(HE) The half reactions are Oxidatiorn Ti+3 > TiO+2 Reduction Co(CNCo(CN64 Balance atoms otherthan H and O Co(CNCo(CN) Balance Oxygen: Ti3H2> Tio 2 (Add H2O on less side) Co(CNCo(CNS4 Balance Hydrogen : Ti+3+ H2O--> Ti0%+2H+ (Add H+ on less side) Co(CN)6Co(CN)6 Balance charge Ti3 H2Tio2+2HCo(CN)6 1e>Co(CN)6 4 Combine: Ti+3 + H2O > Ti0% + 2H+ + e- Co(N31 e>Co(CN)4 Combine: Ti+3 + Co(CN)6-3 + H2O > Ti0% + 2H++ Co(CN)6-4