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)•Chemistry Dr Bob - DrBob222, Sunday, November 29, 2015 at 7:15pm
Could it be that you have reported 4 significant figures and you are allowed only 3.
•Chemistry Dr Bob - Patrick, Sunday, November 29, 2015 at 8:54pm
I did the 1.85 x 10^-25,,,,the ^-25 was right but not the 1.85
The only thing I might suggest is to go back and leave all of those numbers from above(start with 142798 J) in the calculator. If I do that I come up with 1.86E-25 but I wouldn't think a data base would count that wrong.
I am wondering if there is some sort of system error. Thank you for all your help this year , Dr Bob. This is my last assignment for chemistry, I graduate from college in 21 days, ( if I pass the final). Thanks again!
In this question, you are asked to calculate various values related to a voltaic cell.
a) To find the standard cell potential (E degree cell), you need to subtract the standard reduction potential of the anode from the standard reduction potential of the cathode.
E degree cell = E0Cu+2/Cu - E0Cd+2/Cd
E0Cu+2/Cu is given as +0.34 V, and E0Cd+2/Cd is given as -0.403 V.
Therefore, E degree cell = +0.34 - (-0.403) = +0.74 V.
To find the standard Gibbs free energy change (delta G degree), you can use the equation:
delta G degree = - n F E degree cell
Here, n represents the number of moles of electrons transferred, F is Faraday's constant (96485 C/mol), and E degree cell is the standard cell potential.
In this case, n = 2 (because 2 moles of electrons are transferred), and E degree cell is +0.74 V.
Therefore, delta G degree = - 2 x 96485 x 0.74 = -142798 J = -142.8 kJ.
To find the equilibrium constant (K), use the equation:
delta G degree = - R T ln K
Here, delta G degree is the standard Gibbs free energy change, R is the gas constant (8.314 J/(mol K)), T is the temperature in Kelvin, and ln represents the natural logarithm.
In this case, delta G degree is -142.8 kJ, R is 8.314 x 10^-3 kJ/(mol K), and T is 298 K.
Therefore, -142.8 = -8.314 x 10^-3 x 298 x ln K.
Solving for ln K, we get ln K = 57.64.
Finally, to find K, take the exponent of both sides: K = 1.08 x 10^25.
b) In this part, you need to find the cell potential (E cell) when the concentration of Cd2+ (Cd+2) is 1.95 M.
To calculate E cell, use the Nernst equation:
E cell = E degree cell - (0.0591/n)log(Cd+2/Cu+2)
Here, E degree cell is +0.74 V, n is 2 (because 2 moles of electrons are transferred), and Cd+2 and Cu+2 represent the concentrations of Cd2+ and Cu2+ ions, respectively.
In this case, Cd+2 is 1.95 M and Cu+2 is still 1 M (as given).
Plugging in these values, we have:
E cell = 0.74 V - (0.0591/2)log(1.95/1) = 0.692 V.
c) In this part, you need to find the cell potential (E cell), delta G, and the concentration of Cu2+ (Cu+2) at equilibrium.
At equilibrium, the cell potential (E cell) is 0, and by definition, the Gibbs free energy change (delta G) is also 0.
From part a), we know that:
E degree cell = +0.74 V.
Therefore, E cell = 0 implies that 0.74 - (0.0591/2)log(Cu+2/Cd+2) = 0.
Solving for Cu+2/Cd+2, we have:
log(Cu+2/Cd+2) = 2(0.74)/0.0591 = 24.85.
Taking the exponent of both sides, we get:
Cu+2/Cd+2 = 1.08 x 10^25.
Note that this is the same value of K obtained in part a), which confirms that it is at equilibrium.
However, in this part, you are specifically asked to find the concentration of Cu2+ (Cu+2) at equilibrium.
From the equation Cu+2/Cd+2 = 1.08 x 10^25, we can rewrite it as:
Cu+2 = (1.08 x 10^25)Cd+2.
Since Cd+2 is still 1 M (as given), we can substitute this value into the equation:
Cu+2 = (1.08 x 10^25) * 1 M = 1.08 x 10^25 M.
Therefore, the correct value for [Cu+2] at equilibrium is 1.08 x 10^25 M.