A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The...

A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The initial concentrations of Pb2+ and Cu2+ are 5.30

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Expert Solution

I am giving you an example calculation.

Pls. verify with your own values.

E ( Complete Cell ) = E ( reduction of Cathode ) - E ( reduction of Anode )

E ( reduction of Cathode )
= Eo ( Cu^2+ : Cu ) - ( 0.059 / 2 ) log ( 1 / [ Cu^2+ ] )
= ( + 0.34 ) - ( 0.059 / 2 ) log ( 1 / [ 0.250 ] )

E ( reduction of Anode )
= Eo ( Pb^2+ : Pb ) - ( 0.059 / 2 ) log ( 1 / [ Pb^2+ ]
= ( - 0.13 ) - ( 0.059 / 2 ) log ( 1 / [ 5.30 x 10^-2 ] )

E ( Complete Cell )
= ( + 0.34 ) - ( - 0.13 ) - { ( 0.059 / 2 ) log ( [ 5.30 x 10^-2 ] / [ 0.250 ] ) }

= ( + 0.47 ) - { ( 0.059 / 2 ) log ( [ 5.30 x 10^-2 ] / [ 2.50 x 10^-1 ] ) }

= ( + 0.47 ) - ( 0.059 / 2 ) { ( log 5.30 - 2 ) - ( log 2.50 - 1 ) }

= ( + 0.47 ) - ( 0.059 / 2 ) { ( 0.7243 ) - ( 0.3797 ) -1 }

= ( + 0.47 ) - { ( 0.059 / 2 ) x ( - 0.6554 ) }

= ( + 0.47 ) - ( - 0.0193343 )

= + 0.4893343 V

If one concentration falss ,othe increases.

If [Cu2+] falls, then [Pb2+] increases because it is being produced at the anode.

Molarity . . . . . .Pb + Cu2+ ==> Pb2+ + Cu
initial . . . . . . . --- . . 1.50 . . . . .0.0500 . ---
change . . . . . .--- . .-1.30 . . . . .+1.30 . .---
final . . . . . . . . --- . . 0.20 . . . . .1.35 . . ---

Q = 1.35 / 0.20 = 6.75

E cell = 0.47 - 0.059/2 log 6.75 = 0.45 V

C. Note that the sum of [Pb2+] and [Cu2+] = 1.55.

E cell = 0.47 - 0.059/2 log Q
0.35 = 0.47 - 0295 loq Q
-0.12 = -0.0295 log Q
4.07 = log Q
Q = 10^4.07 = 11,700

If x = [Pb2+] and y = [Cu2+], then x + y = 1.55 and x/y = 11,700 (or, x = 11,700y).
11,700y + y = 1.55
11.700y = 1.55
y = 0.00013 = [Cu2+] and x ([Pb2+]) = essentially 1.55 M

Pls. calculate the last part by using the given values.

For complete Background Visit

http://chemistry.about.com/od/electrochemistry/a/nernstequation.htm

queen_honey_blossom answered 11 months ago

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