Question

1) production of lead from the ore galena,pbs, involoves several steps. In the first step's overall reaction, Pbs is oxidized to PbO and SO2 by (O2)i.e.,

2PbS + 3O2 --> 2PbO + 2SO2

a)Which element is oxidized?

b)Which element is reduced?

(c) Write the balanced equations for the oxidation and reduction half reactions?

Must show all the steps used

Use H+ to balance the charge

Oxidation:

Reduction:

(d) Use these two half reactions to generate a balanced equation for the oxidation of Pbs:

2) Evaluate each of the pairs shown in the table for solubility over the entire composition range. Place an X under the expected solubility for each of the pairs. Indicate which of the Hume-Rothery rules are not met in each case where complete solubility is not supported.

pair	Complete solubility	Limited solubility	No Solubility	H-R Rules Not Met
Cr-Fe
Ti-Fe
Fe-Cu
Ni-Cu
Cu-Co
Ag-Pt

Construct a table using the data

element Radius crystal structure EN

the data

element Radius Crystal structure Electronegativity

Ni 115 FCC 1.8

C 77   

H 32

O 73

Ti 132 HCP 1.54

Ag 134 FCC 1.93

Al 118 FCC 1.61

Cu 117 FCC 1.90

Co 116 HCP 1.88

Cr 118 BCC 1.66

Fe 117 BCC 1.83

Pt 130 FCC 2.28

Au 134 FCC 2.40

Zn 125 HCP 1.65

Ir 127 FCC 2.2

3) a) A face centered cubic lattice contains an interstitial space in the body center. For a host element of 200pm radius, calculate the radius of the octahedral interstitial space.

b) which elements fit in this space?

Answer 1

1a) Sulfur is oxidised

1b) Oxygen

1c) Oxidation: Pb²⁺ + S^2- ------> Pb²⁺ + S⁴⁺ +6e^-

Reduction: Pb²⁺ + O₂ + 4e^- -------> Pb²⁺ + 2O^2-

1d) Adding 2*oxidation reaction to 3*reduction reaction (excluding the lead ions) balances the no. of electrons on both the sides, hence yields the balanced equation

2Pb²⁺ + 2S^2- +3O₂ ------> 2Pb²⁺ + 2S⁴⁺ + 6O^2-

Or 4PbS + 3/2 O2 ------> 2PbO + 2SO₂