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
1.852 x 10^-25 was marked wrong. Why?
(everything else was right)
Thank you
See my response above.
The value 1.852 x 10^-25 M for [Cu2+]eq was marked wrong because it is not expressed in scientific notation as requested in the question. When expressing the concentration of [Cu2+]eq in scientific notation, it should be written as 1.85E-25 M.
In scientific notation, the coefficient is typically written as a number between 1 and 10, multiplied by 10 raised to a certain power (exponent). In this case, the coefficient is 1.85, and the exponent is -25, indicating 10 raised to the power of -25. Thus, the concentration of [Cu2+]eq in scientific notation is 1.85E-25 M.