1.

Calculate the e.m.f. of the cell in which the following reaction takes place. Ni_((s)) + 2Ag_((0.002M))^(+) to Ni_((0.160M))^(2+) + 2Ag_((s)) , "Given "E_("cell")^@ = 1.05 V

Answer»

Solution :(a) From the given cell reaction and Nernst equation
`E_(cell) = E^(circ)_(cell) - 0.0591/n log[NI^(2+)]/[Ag^(+)]^(2)`
`= 1.05V-0.0591/2 log[0.160]/[0.002]^(2)`
`1.05-0.0591/2 log(4 xx 10^(4))`
`1.05 - 0.0591/2 (4.6021)`
`1.05 - 0.14 = 0.91 V` = `E_(cell) = 0.91 V`
(b) The amount of substance deposited at an electrode is directly proportional to QUANTITY of current passed through its solution. MATHEMATICALLY,
`WQ`
`WIT`
`W = ZIT`
Where,
W = mass,
I = current in ampere,
t = time in second,
Z = electrochemical equivalent.


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