1) Determine the oxidation number of each element in the following compounds: a) Scandium oxide (ScaO,) Sc C b) Gallium chloride (GaCl) Ga CI c) Hydrogen phosphate ion H 10 d) lodine trifluoride 2. Use the half-reaction method to balance the following redox reactions: (6 points each) a) Mn (aq) + NaBiO,(s) > Bis*(aq) + MnO, (aq) + Na*(a) (in acidic solution) O b) NO, (aq) + Al(s) > NH,(aq) + AI(OH)4 (in basic solution) 3. Use the oxidation number method to balance the following redox equations. Identify the oxidizing agent and reducing agent. E a) PbS(s) + Oz(g) - PbO(s) + SOz(g) b) KCIO, (aq) + HBr(aq) > Brz (1) + HO(1) + KCI(aq)

1. (a) Sc = +2 ; O = -2

   (b) Ga = -1 ; Cl = +3

   (c) H = +1 ; P = +5

   (d) I = +3 ; F = -1

2. (a) Mn²⁺_(aq) + 4NaBiO_(s) + H⁺_(aq) ------> 4Bi³⁺_(aq) + MnO₄^-_(aq) +4Na⁺_(aq) + H⁺_(aq) 

    (b) NO₃^-_(aq) + Al_(s) + OH^-_(aq) ------> NH_3(aq) + Al(OH)₄^-_(aq) 

3. (a) Reducing agent - PbS ; Oxidizing agent - O₂ 

          Balanced equation: 2PbS_(s) + 3O_2(g) ------> 2PbO_(s) + 2SO_2(g) 

    (b) Oxidizing agent - KClO₃ ; Reducing agent - HBr 

          Balanced equation: KClO_3(aq) + 6HBr_(aq) -------> 3Br_2(l) + 3H₂O_(l) + KCl_(aq) 

Step-by-step explanation

1. (a) Scandium oxide (Sc₂O₂)

Sc₂O₂ is a neutral compound, hence, it has a net charge of zero (0). Therefore, the sum of the oxidation numbers of its constituent ions (i.e. Sc and O ions) is equal to zero (0). Oxygen has 2 valence electrons and thus, the oxide ion (O^2-) has an oxidation state of -2. Its oxidation number is -2 because O is more electronegative than Sc.

Therefore, 

Oxidation number of O = -2

Total oxidation number of O^2- in Sc₂O₂ = 2*-2 = -4

Total oxidation number of Sc = 0 - (-4) = +4

Oxidation number of Sc = +4/2 = +2

 

   (b) Gallium chloride (GaCl₃)

Cl is a halogen and therefore has 1 valence electron. Since Cl is more electronegative than Ga, it has an oxidation state of -1. The net charge of GaCl₃ is zero because it is a neutral compound. 

Therefore,

Oxidation number of Cl = -1

Total oxidation number of Cl = -1*3 = -3

Oxidation number of Ga = 0 - (-3) = +3

 

   (c) Hydrogen phosphate ion

Hydrogen phosphate ion (HPO₄^2-) has a net of -2. Hydrogen has 1 valence electron and tends to be more electropositive than the phosphate ion (PO₄^3-). 

Hence,

Oxidation number of H = -2 - (-3) = +1

In the PO₄^3- ion, O is more electronegative than P and therefore it has an oxidation state of -2. 

Hence,

Total oxidation number of O = -2*4 = -8

Oxidation number of P = -3 - (-8) = +5

 

   (d) Iodine trifluoride

Iodine trifluoride (IF₃) is a neutral compound with a net charge of zero. Both I and F are halogens, with valence electron of 1 each. F is more electronegative than I and thus, have a and oxidation state of -1. 

Hence,

Oxidation number of F = -1

Total oxidation number of F = -1*3 = -3

Oxidation number of I = 0 - (-3) = +3

 

2. (a) Mn²⁺_(aq) + NaBiO_(s) ------> Bi³⁺_(aq) + MnO₄^-_(aq) +Na⁺_(aq) 

Half reactions for the above reaction:

Mn²⁺_(aq) -----> Mn⁷⁺_(aq) + 5e^- 

NaBiO_(s) ------> Na⁺_(aq) + Bi³⁺_(aq) + 4O^2-_(aq) + H⁺_(aq)  + 4e^- 

Balancing both sides of the half-reactions,

4NaBiO_(s) ------> 4Na⁺_(aq) + 4Bi³⁺_(aq) + 4O^2-_(aq) + H⁺_(aq)  + 16e^- 

Mn²⁺_(aq) -----> Mn⁷⁺_(aq) + 5e^- 

From the above 2 balanced half-reactions, the balanced overall reaction will be as follows:

Mn²⁺_(aq) + 4NaBiO_(s) + H⁺_(aq) ------> 4Bi³⁺_(aq) + MnO₄^-_(aq) +4Na⁺_(aq) + H⁺_(aq) 

 

    (b) NO₃^-_(aq) + Al_(s) ------> NH_3(aq) + Al(OH)₄^-_(aq) 

Half-reactions for the above reaction:

NO₃^-_(aq) + OH^-_(aq) ------> N³⁺_(aq) + 4OH^-_(aq) + 2e^-

Al_(s) -------> Al³⁺_(aq) + 3e^- 

Balancing the half-reactions:

NO₃^-_(aq) + OH^-_(aq) ------> N³⁺_(aq) + 4OH^-_(aq) + 2e^-

Al_(s) -------> Al³⁺_(aq) + 3e^- 

Hence, the overall balanced equation is:

NO₃^-_(aq) + Al_(s) + OH^-_(aq) ------> NH_3(aq) + Al(OH)₄^-_(aq) 

 

3. (a) PbS_(s) + O_2(g) ------> PbO_(s) + SO_2(g) 

From the above reaction, the oxidation number of S changes from -2 in PbS to +4 in SO₂ while that of O changes from zero (0) in O₂ to -2 in SO₂. Since the oxidation number of S (from the PbS) increases -2 to +4, it is oxidized. That means PbS is the reducing agent. On the other hand, the oxidation number of O decreases from zero (0) to -2. That means that O is reduced, hence, O₂ is the oxidizing agent. 

 

Hence, the balanced equation is:

 2PbS_(s) + 3O_2(g) ------> 2PbO_(s) + 2SO_2(g) 

 

    (b) KClO_3(aq) + HBr_(aq) -------> Br_2(l) + H₂O_(l) + KCl_(aq) 

From the above reaction, the oxidation number of Cl changes from +5 in KClO₃ to -1 in KCl while that of Br changes from -1 in HBr to zero (0) in Br₂. Since the oxidation number of Cl (from the KClO₃) decreases +5 to -1, it is reduced. That means KClO₃ is the oxidizing agent. On the other hand, the oxidation number of Br increases from -1 to zero (0). That means that Br is oxidized, hence, HBr is the reducing agent. 

 

Hence, the balanced equation is:

KClO_3(aq) + 6HBr_(aq) -------> 3Br_2(l) + 3H₂O_(l) + KCl_(aq)