Please do ALL the parts (a) S203(ag) Cu(OH)2(s) S032(a)CuOH(s) (b) NOa)H2g)- NO(
ID: 574950 • Letter: P
Question
Please do ALL the parts
(a) S203(ag) Cu(OH)2(s) S032(a)CuOH(s) (b) NOa)H2g)- NO(G) Oxidation reaction Oxidation reaction chemPad Help chemPad Help Greek Greek Your answer contains an ambiguous or incomplete reaction equation. Check all the components on the reactant-side of the equation. Reduction reaction chemPad Help Greek Reduction reaction chemPad Help Net reaction chemPad Help Net reaction Greek chemPad 9 Help Greek- (c) Sc(s) Cr0-(aa) Sc(OH)3(s)Cr(OH)4 (d) Sn(OH)2(s) + O2(g) SnO2(s) Oxidation reaction Oxidation reaction chemPad Help chemPad Help Greek Greek Reduction reaction Reduction reaction chemPad Help chemPad Help Greek Greek Net reaction Net reaction chemPad Help chemPad Help Greek GreekExplanation / Answer
a)
Oxidation reaction: S2O32- (aq) + 3 H2O (l) --------> 2 SO32- (aq) + 6 H+ (aq) + 4 e-
Reduction reaction: Cu(OH)2 (aq) + e- -------> Cu(OH) (s) + OH- (aq)
Net Reaction:
Multiply the reduction reaction by 4 and add to the oxidation half
S2O32- (aq) + 3 H2O (l) + 4 Cu(OH)2 (aq) --------> 2 SO32- (aq) + 6 H+ (aq) + 4 Cu(OH) (s) + 4 OH- (aq)
Combine 4 H+ and 4 OH- to form 4 H2O and write
S2O32- (aq) + 4 Cu(OH)2 (aq) + 3 H2O (l) -------> 2 SO32- (aq) + 4 Cu(OH) + 4 H2O (l)
Cancel the common ions/molecules from both sides and get
S2O32- (aq) + 4 Cu(OH)2 (aq) --------> 2 SO32- (aq) + 4 Cu(OH) (s) + H2O (l)
b) Oxidation reaction: H2 (g) ------> 2 H+ (aq) + 2 e-
Reduction reaction: NO2- (aq) + 2 H+ (aq) + e- --------> NO (g) + H2O (l)
Net Reaction:
Multiply the reduction reaction by 2 and add to the oxidation half.
H2 (g) + 2 NO2- (aq) + 4 H+ (aq) --------> 2 H+ (aq) + 2 NO (g) + 2 H2O (l)
Cancel the common ions/molecules from both sides and obtain
H2 (g) + 2 NO2- (aq) + 2 H+ (aq) -------> 2 NO (g) + 2 H2O (l)
c) Oxidation reaction: Sc (s) + 3 OH- (aq) --------> Sc(OH)3 (s) + 3 e-
Reduction reaction: Cr2O72- (aq) + 7 H2O (l) + 6 e- ---------> 2 Cr(OH)4- (aq) + 6 OH- (aq)
Net Reaction:
Multiply the oxidation reaction by 2 and add to the reduction reaction and obtain
2 Sc (s) + 6 OH- (aq) + Cr2O72- (aq) + 7 H2O (l) -------> 2 Sc(OH)3 (s) + 2 Cr(OH)4- (aq) + 6 OH- (aq)
Cancel out the common ions/molecules from both sides and obtain
2 Sc (s) + Cr2O72- (aq) + 7 H2O (l) -------> 2 Sc(OH)3 (s) + 2 Cr(OH)4- (aq)
d) Oxidation Reaction: Sn(OH)2 (s) -------> SnO2 (s) + 2 H+ (aq) + 2 e-
Reduction Reaction: O2 (g) + 4 H+ (aq) + 4 e- -------> 2 H2O (l)
Net Reaction:
Multiply the oxidation reaction by 2 and add to the reduction reaction and obtain
2 Sn(OH)2 (s) + O2 (g) + 4 H+ (aq) -------> 2 SnO2 (s) + 4 H+ (aq) + 2 H2O (l)
Cancel out the common ion/molecules from both sides and obtain
2 Sn(OH)2 (s) + O2 (g) --------> 2 SnO2 (s) + 2 H2O (l)