Question

A voltaic cell has one half-cell with a Cu bar in a 1.00 M Cu2+ salt and the other half-cell with a Cd bar in the same volume of a 1.00 M Cd2+ salt. Report your answers to the correct number of significant figures.

a)Find E degree cell,

delta G degree ,

and K.

b)As the cell operates (Cd2+) increases;

Find E cell when (Cd2+) is 1.95M

c)Find E cell, Delta G, and (Cu2+) at equilibrium.

Note: Delta G in Joules and

(Cu2+) ........in scientific notation

(Please note all the answers are correct EXCEPT the CU2= at equilibrium which is what I need)

a)anode reaction: oxidation takes place

Cd(s) -------------------------> Cd+2 (aq) + 2e- , E0Cd+2/Cd = - 0.403 V

cathode reaction : reduction takes palce

Cu+2(aq) + 2e- -----------------------------> Cu(s) , E0Cu+2/Cu = + 0.34V

--------------------------------------------------------------------------------

net reaction: Cd(s) +Cu+2(aq) -------------------------> Cd+2 (aq) + Cu(s)

E0cell= E0cathode- E0anode

E0cell= E0Cu+2/Cu - E0Cd+2/Cd

= 0.34 - (-0.403)

= 0.74 V

E0cell= 0.74 V

\DeltaGo = - n FE0cell

= - 2 x 96485 x 0.74

= -142798 J

= -142 .8 kJ

\DeltaGo = -142 .8 kJ

\DeltaGo = - R T ln K

-142 . 8 = -8.314 x 10^-3 x 298 x ln K

lnK = 57.64

K = 1.08 x 10^25

b) Cd(s) +Cu+2(aq) -------------------------> Cd+2 (aq) + Cu(s)

t= 0 1 M 1M

t=t 1-0.95 =0.05 1+0.95 M

E cell = E0cell - 0.0591/2log( Cd+2/Cu+2)

= 0.74 V - 0.0591/2log( 1.95/0.05) = 0.692 V

c) At equlibrium, Ecell = 0 ,

\DeltaG= 0 by definition

02_img-avatar-gry-40x40.png.....


From (a) Equilibrium constant, K = [Cd+2]eq/[Cu+2]eq = 1.08E25
Cd(s) +Cu+2(aq) -----------------------> Cd+2 (aq) + Cu(s)
As equilibrium constant is very high we will solve the backward reaction assuming
all the Cu+2 consumed and similar Cd+2 produced initially

So Cd+2 (aq) + Cu(s) -----------------------> Cd(s) +Cu+2(aq)
K = [Cu+2]eq/[Cd+2]eq = 1/1.08E25 = 9.26E-26

Initially at t= 0
[Cu+2]0 = 1-1=0M
[Cd+2]0 = 1+1=2M


t=equillibrium
[Cu+2]eq = x
[Cd+2]eq = 2-x

K = x/(2-x) = 9.26E-26

as x is very small, x << 2, so we can assume 2-x = 2
so x/2 = 9.26E-26
[Cu+2]eq = x = 2*9.26E-26 = 1.852E-25 M

Dr Bob was is Cu2+ wrong? Everything else was right.. Thank you.

Answer 1

1.852 x 10^-25 was marked wrong. Why?