The solutions at the two Pb electrodes of a concentration cell were prepared as
ID: 1057184 • Letter: T
Question
The solutions at the two Pb electrodes of a concentration cell were prepared as follows:
Cell A: A mixture of 1.00 mL of 0.0500 M Pb(NO3)2 with 4.00 mL of 0.0500 M KX (the soluble potassium salt of an unspecified monovalent ion X-).
Some PbX2(s) precipitates.
Cell B: 5.00 mL of 0.0500 M Pb(NO3)2.
The cell potential was measured to be 0.05600 V at 25 °C.
1.By use of the Nernst equation, determine the concentration (M) of Pb2+ in the solution of Cell A.
2.In Cell A, how many moles of X- have reacted with Pb2+?
3.What is the concentration (M) of X- in the solution of cell A.
4.Calculate Ksp of PbX2.
Explanation / Answer
with help of nernst equation
cell equation a is
anode : Pb(A) = Pb+2(a) + 2e- E0(OXI) = 0.13v
Cathode : Pb(B)+2 + 2e- = Pb(B) E0(RED)= -0.13v
E0 = E0(OXI) +E0(RED)
= 0.13-0.13
= 0
now from nernst equation
E= E0-0.0591/n log[Pb(A)+2/Pb(B)+2]
0.067=-0.0591 log[Pb(A)+2/Pb(B)+2]
0.0054=Pb(A)+2/Pb(B)+2
Pb(A)+2=0.0054(0.05M)
Pb(A)+2=0.00027M --------------------------ANS (1)
Pb+2 + 2X- = PbX2
When we prepared cell A
Dissolved Pb+2 is
(0.00100L)(0.0500M Pb(NO3)2=5*10-5 moles of Pb+2 dissolved
when Pb(NO3)2 is dissolved in water we findout
Pb(A)+2 = 0.00027 As per Ans 1
so at equilibrium poimt moles of Pb+2 in solution is
=0.00027M(0.00500L)-1.35*10-6
4.865*10-5 moles of PbX2 precipitate
but moles of x is double than pb so
9.73*10-5 moles reacted-----------------------ans(2)
From the second calculation moles of x=9.73*10-5 moles
now prepare cell A
=(0.00400L)(0.0500M KX)
=2.00 * 10-4 moles
so 2.00 * 10-4 -9.73*10-5
=1.027*10-4 moles available in solution
so now concentration of solution
Cell A =1.027*10 / 0.00500L
=0.02054 M ------------------ans (3)
PbX2 = Pb+2 + 2x-
dissociation constant
KSp= [Pb+2][x-]2
=0.00027M [0.002054M]2
= 5.5 * 10-6 ----------------------ans (4)