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•chemistry Dr. BOB - DrBob222, Sunday, November 29, 2015 at 9:19pm
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.
•chemistry Dr. BOB - Patrick, Sunday, November 29, 2015 at 10:36pm
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!
Congratulations on your graduations and good luck with your future.
The last answer is 1.99 x 10^-25
Congratulations on your upcoming graduation! I'm sorry to hear that you had some trouble with the answer. It seems that there may have been an issue with the system or the way the answer was entered. It's possible that the system only accepts three significant figures, so the answer with four significant figures, 1.852 x 10^-25, may have been marked as incorrect. However, based on the calculations you provided, it appears that the answer you provided is correct. Sometimes these things happen, but don't let it discourage you. Keep up the good work and best of luck on your final!
It seems there may have been an error in the system when entering the answer. Based on the calculations, the correct answer should be 1.86 x 10^-25 M. It is possible that there was a rounding error or an issue with significant figures in the system. Congratulations on your upcoming graduation!
I'm sorry to hear that you're experiencing some discrepancies with your answers. It's possible that there could be a rounding error or a difference in significant figures. Here's how you can check and make sure you're reporting the answer with the correct number of significant figures:
1) Take a look at the calculations and identify the numbers involved in determining the final answer. Pay attention to any rounding or truncating that may have occurred during the calculations.
2) Determine the number of significant figures in each number used in the calculation. The rule for multiplication and division is to report your final answer with the same number of significant figures as the number with the fewest significant figures.
3) Round your answer to the appropriate number of significant figures based on the number with the fewest significant figures.
In the case of the answer given as 1.852E-25 M, it is possible that there was a rounding error or truncation during the calculations. It may be worth double-checking the calculations and rounding the answer to the correct number of significant figures.
Remember to always refer to the guidelines provided by your instructor or the requirements of the assignment for the correct number of significant figures to use in reporting your answer.