Question

A 1.000 g sample of metallic lead is immersed in 100.0 mL of a 0.1 M solution of Sn2+ .

(a) Calculate the equilibrium concentrations of Sn2+ and Pb2+ .

(b) What are the masses of metallic lead and tin in the container once the reaction reaches equilibrium?

Answer 1

A 1.000 g sample of metallic lead is immersed in 100.0 mL of a 0.1 M solution of Sn2+ .

(a) Calculate the equilibrium concentrations of Sn2+ and Pb2+ .

(b) What are the masses of metallic lead and tin in the container once the reaction reaches equilibrium?

Moles of Pb (s) = amount in g/ molar mass

= 1.00 g /207.2 g/ mole

= 4.83*10^-3 moles Pb

And moles of Sn 2+

Number of moles = molarity * volume in L

= 0.1 * 0.100

= 0.01 Moles Sn2+

Pb2+ + 2e- ==> Pb . . .Eo = -0.13 V
Sn2+ + 2e- ==> Sn . . .Eo = -0.14 V

We are told that the cathode (reduction) compartment contains Sn2+ and that the anode (oxidation) compartment contains Pb2+. So the overall reaction must be

Pb ==> Pb2+ + 2e- (oxidation) . . .Eo = +0.13 V
Sn2+ + 2e- ==> Sn (reduction) . . Eo = -0.14 V
======================================...
Pb + Sn2+ ==> Pb2+ + Sn . . . . . .Eo cell = -0.01 V

Now calculate the equilibrium constant;

E cell = 0.0592 /n log K

-0.01 = 0.0592/2 log K

Log K = -0.33

K = 0.467

Pb + Sn2+ ==> Pb2+ + Sn . . . . . .Eo cell = -0.01 V

at equilibrium ]Sn2+]= [0.1-x] and [Pb2+]= 4.83*10^-3 + x
K = [product ] / [reactants ]
0.467 = [Pb2+] /[Sn2+] = 4.83*10^-3 + x /[0.1-x]

0.0467 – 0.467 x = 0.00483+x

X = 0.0285

at equilibrium [Sn2+]= [0.1-x] = 0.1-0.0285= 0.071

and [Pb2+]= 4.83*10^-3 + x

= 0.033

(b) What are the masses of metallic lead and tin in the container once the reaction reaches equilibrium?

at equilibrium [Sn] =x= 0.0285 moles

amount of Sn = 0.0285 moles * 118.71 g/ moles

= 3.38 g

[Pb2+]= 4.83*10^-3 + x

= 0.033

Amount of Pb = 0.033*207.2 g/ mole

= 6.9 g