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)
Could it be that you have reported 4 significant figures and you are allowed only 3.
I did the 1.85 x 10^-25,,,,the ^-25 was right but not the 1.85
I see.
The answer was marked wrong because it did not have the correct number of significant figures. In the question, it specifically asked for the answer to be reported with the correct number of significant figures.
In part c), the concentration [Cu+2]eq is calculated to be 1.852E-25 M. However, this answer has too many significant figures. To report the answer with the correct number of significant figures, we need to consider the least precise measurement involved in the calculation.
In this case, the initial concentration [Cu+2]0 is given as 0 M, which implies that it is an exact value with infinite precision. Since an exact value does not contribute to the uncertainty, we need to consider the precision of the other value involved, which is 2 M.
The value 2 M has only one significant figure, so the final answer should also have only one significant figure. Therefore, the correct answer is [Cu+2]eq = 2E-25 M.