Question

1. A voltaic cell is constructed that is based on the following reaction:
Sn2+(aq)+Pb(s)→Sn(s)+Pb2+(aq).

a. If the concentration of Sn2+ in the cathode compartment is 1.50 M and the cell generates an emf of 0.22 V , what is the concentration of Pb2+ in the anode compartment?

b. If the anode compartment contains [SO2−4]= 1.50 M in equilibrium with PbSO4(s), what is the Kspof PbSO4?

2. A voltaic cell is constructed with two silver-silver chloride electrodes, each of which is based on the following half-reaction:
AgCl(s)+e−→Ag(s)+Cl−(aq).
The two cell compartments have [Cl−]=1.49×10⁻² M and [Cl−]= 3.00 M , respectively.

a. What is the cell emf for the concentrations given?

Answer 1

1.

Cell Reaction -

Sn²⁺(aq) + Pb ----> Sn + Pb²⁺(aq)

Reduction half :-

Sn²⁺(aq) + 2e^------> Sn E⁰_Sn²⁺/Sn = -0.14V (Cathode)

Oxidation half :-

Pb ----> Pb²⁺(aq) + 2e^- E⁰_Pb²⁺/Pb = -0.13V (Anode)

now, E⁰_cell = E_cathode - E_anode

  = -0.14V -(-0.13V)

= -0.01V

Now,

[Sn²⁺] = 1.50M

[Pb²⁺] = ?

E⁰_cell = -0.01V

E_cell = 0.22V

no. of electrons transferred, n = 2

According to Nernst equation-

E_cell = E⁰_cell -( 0.0591V / n)*log[Pb²⁺]/[Sn2+]

Keeping all the values-

0.22V = -0.01V -(0.0591V/2)*log[Pb²⁺]/1.50M

0.22V+0.01V = -(0.0591V /2)*log[Pb²⁺]/1.50M

0.23V = -0.02955Vlog[Pb²⁺]/1.50M

- 0.23V/0.02955V = log[Pb²⁺] / 1.50M

-7.783 = log[Pb²⁺]/1.50M

[Pb²⁺]/1.50M = 10^-7.783

[Pb²⁺]/1.50M = 1.65 *10^-8

[Pb²⁺] = 1.50M * 1.65* 10^-8

[Pb²⁺] = 2.47 * 10^-8 M

part B :-

PbSO4(s) <----> Pb²⁺ + SO₄^2-

K_sp = [Pb²⁺][SO₄^2-]

[Pb²⁺] = 2.47*10^-8M

[SO₄^2-] = 1.50M

K_sp = (2.47*10^-8M)*(1.50M)

K_sp = 3.71*10^-8