1.

(a) Calculate the equilibrium constant for the reaction : Cd^(2+) (aq) + Zn(s)

Answer»

Solution :(a) `E_("cell")^(@) = E_(Cd^(2+)//Cd)^(@) -E_(Zn^(2+)//Zn)^(@) = -0.403 V - (-0.763 V) = + 0.36 V`
`nFE^(@) =2.303 RT ln K`
or `log K = (nFE^(@))/(2.303 RT) =(nE^(@))/((2.303 Rt)/F) =(2 xx 0.36)/0.0591 = 12.2033`
or `K = 1.5969 xx 10^(12)`
(b) By FARADAY.s law.
Quantity of electricity = 25 x 60 x 0.75 coulombs = 1125 C = 0.01166 Faraday 0.01166 Faraday of charge deposits = 0.369 G of copper.
2 Faraday of charge deposits `=0.369/0.1166 xx 2g` of copper = 63.3 g
At. mass of copper = 63.3 u
( c) We want to see if the reaction:
`Ag_(2)S (s) + Zn to 2Ag(s) + ZnS`
can take place
`E_("cell") = E_(Ag_(2)S//Ag)^(@) -E_(Al^(3+)//Al)^(@)`
`=-0.71 -(-1.66)`
`=+0.95 V`
As the cell emf is POSITIVE, the reaction will take place i.e., tarnish can.be removed.


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