Question

Consider the electrochemical cell below, which was initially constructed of two 1.0 L solutions of equal concentration, one of Pb⁺² and the other with Sn⁺². The electrochemical cell reached equilibrium after 0.40 moles of electrons had been transferred. If the K_c value for the reaction is 2.2, what was the initial concentration of the lead and tin solutions?

Sn (s) + Pb⁺² (aq) --> Sn⁺² (aq) + Pb (s)

Answer 1

Sn(s) + Pb2+ (aq) <---> Sn2+ (aq) + Pb (s)

Given 0.4 moles of electrons transferred

For the reaction 2 electrons transferred per 1 atom   ( Pb2+ reduction to Pb takes 2electrons)

since 0.4 moles electrons transferred the number of moles of Pb2+ reacted = 0.4/2 = 0.2

Hence Sn2+ moles formed = 0.2

Let Pb2+ initial moles = m = Sn2+ initial

at equilibrium Pb2+ moles = m-0.2 , Sn2+ moles = m+0.2

[Pb2+] = ( m-0.2) /1 , [Sn2+] = (m+0.2)/1

Kc = [Sn2+] /[Pb2+]

2.2 = ( m+0.2) / ( m-0.2)

2.2m-0.44 = m+0.2

m= 0.533

Initially

[Pb2+] = [Sn2+] = 0.533 /1 = 0.533 M