# 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.

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.

answers:  (with the exception of Cu=2 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

Go = – n FE0cell

= – 2 x 96485 x 0.74

= -142798 J

= -142 .8 kJ

Go = -142 .8 kJ

Go   = – 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) 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 ,  and DeltaG= 0 by definition

but what is CU+2 at equilibrium?

I don’t need an explanation just the answer in scientific notation…note part of the answer is 10^-25

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