Question

(30 pts) What are the oxidation numbers of chromium in each of the following compounds: K-CrF [Cr(NHs)6]Cl CrCl2 NH)2Cr20 Cr 40 pts) For each of the following reactions, circle the chemical species that is oxidized nd draw a box around the chemical species that is reduced: 3 Cu (s) + 2 NOS (aq) + 8 H+ (aq) → 3 Cu2+ (aq) + 2 NO (g) + 41400) Pb(OH)42. (aq) + 2H' (aq) + CIO' (aq) → PbOz(s) + cr (aq) + 3H00 2 HNO3 + 2 KMnO4 + 3 H2O2 → 3 O2 + 2 MnO2 + 4 H,0 + 2 Kno, ) In the second reaction above, what is the reducing agent ?

Answer 1

Assume oxidation station of Cr = x , thrn

K2CrF6

2(+1) + x + 6(-1) = 0

x = +4

(NH4)2Cr2O7

2(+1) + 2x + 7(-2) = 0

x = +3

[Cr(NH3)6]Cl3

x + 6*(0) + 3(-1) = 0

x = +3

Cr

x = 0

CrCl2

x + 2(-1) = 0

x = +2

Cs3CrO4

3(+1) + x + 4(-2) = 0

x = +5

3 Cu(s) + 2 NO3- (aq) + 8 H+ (aq) -------> 3 Cu+2 (aq) + 2 NO (g) + 4 H2O(l)

In above reaction Cu has 0 oxidation state by loosing 2 electrons it is converted to Cu+2 so Cu undergone oxidation.

Pb(OH)4 2- (aq) + 2 H+ (aq) + ClO- (aq) ------> PbO2 (s) + Cl- (aq) + 3 H2O (l)

In the above reaction Pb in Pb(OH)4 2- has +2 oxidation state which looses 2 electrons and converted to Pb +4 in PbO2. Pb undergone oxidation.

2 HNO3 + 2 KMNO4 + 3 H2O2 ------> 3 O2 + 2 MnO2 + 4 H2O + 2 KNO3

In the above reaction O has -1 oxidation state in H2O2 is changed to 0 in O2 molecule so Oxygen undergone oxidation.

IN the second reaction Pb+2 undergone oxidation so it reduces the other species hence it is the reducing agent.

Pb(OH)4 2- (aq) + 2 H+ (aq) + ClO- (aq) ------> PbO2 (s) + Cl- (aq) + 3 H2O (l)