I get 1.069 for Ecell.
In 8 hrs, coulombs = A x seconds.
C = 5A x 8 hrs x (60 min/1 hr) x (60sec/1 min) = about 144,000 coulombs.
That is how much of a Faraday?
144,000/96,485 = about 1.49 F.
1.49 F will dissolve 65.38/2 g Zn and deposit 63.54/2 g Cu. Subtract g Zn from 250 and add g Cu to 250 g to find amount electrodes weigh after 8 hours operation at 5 A.
How long before the battery will go dead delivering 5A? That is done this way.
96,485 C will dissolve 65.38/2 g = 32.69 g Zn. How many seconds will it take to dissolve 250 g (all of the Zn electrode) Zn . That will be
96,485 x (250/32.69 = ?C
5A x seconds = ?C. Solve for seconds and convert to min or hours as desired.
A zinc-copper battery is constructed as follows at 25 degree C:
Zn | Zn2+(0.15 M) || Cu2+(1.70 M) | Cu
The mass of each electrode is 250. g.
I found the cell potential to be 1.13V and the cell potential after 5.00 A of current has flowed for 8.00 h is 1.11V.
but how do I Calculate the mass of each electrode after 8.00 h. Zn =___g and Cu= ___g
AND How long can this battery deliver a current of 5.00 A before it goes dead?___hrs.
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