Can someone help me find Cu2+ at equilibrium? All parts to this problem are corr

Question

Can someone help me find Cu2+ at equilibrium?

All parts to this problem are correct except the CU2+ at equilibrium. (I know that part of that answer is 10^-25) (It's not 2.14, it's not 1.478, nor 1.9, nor 1.08)

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)

Answers: 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.......Again, I need the answer to Cu2+ at equilibrium which is.....

..(blank) .....x 10^-25..............................what is blank?

Explanation / Answer

question is true

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