Question

A cell , Ag | Ag₊ | | Cu , initially contains 1 M Ag+ and 1 M Cu+ ions . Calculate the changes in cell potential after the passage of 9.65 A ₊of current for 1h .

Answer 1

cSol.
Quantity of electricity passed
= 9.65 x 60 x 60 C
= 34740 C

Ag+ ions deposited
= 34740 / 96500
=0. 36 mole

Cu2+ ions deposited
= 34740 / ( 2 x 95600 )
= 0.18 mole

[ Ag + ] left = 1 - 0.36
= 0.64 M

[ Cu + ] left = 1 - 0.18
= 0.82 M

Cell reaction
Cu+ + 2Ag+ ----> Cu 2+ + 2Ag

Check calculation of the dE
in Pic

Answer 2

Explanation:

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