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A copper silver cell is setup. The coppe...

A copper silver cell is setup. The copper ion concentration is 0.10 M.The concentration of silver ion is not known. The cell potential when measured was 0.422 V . Determine the concentration of silver ions in the cell .
[Given `E_(Ag^(+)//Ag)^(@) = 0.80 V, E_(cu^(2+)//Cu)^(@) = + 0.34 V`]

Text Solution

Verified by Experts

Cell reaction is
`Cu(s) + 2Ag^(+) (aq) to Cu^(2+) (aq) + 2A g(s)`
`E_("cell")^(@) = E_(R) - E_(L) = + 0.80 - (+ 0.34) =0.46 V `
Number of electrons taking part, n = 2
By using Nernst equation ,
`E_("cell") = E_("cell")^(@) - (0.0591V)/(n) " log " ([Cu^(2+)])/([Ag^(+)]^(2))`
`0.422 V = 0.46 V- (0.0591)/(2) " log " ([Cu^(2+)])/([Ag^(+)]^(2))`
`0.422 V = 0.46 V - 0.0295 " log " ([Cu^(2+)])/([Ag^(+)]^(2))`
`" log " ([Cu^(2+)])/([Ag^(+)]^(2)) = (0.46 - 0.422)/(0.0295) = 1.288`
`([Cu^(2+)])/([Ag^(+)]^(2))` = antilog 1.288
`([Cu^(2+)])/([Ag^(+)]^(2))` = 19.41
`(0.10)/([Ag^(+)]^(2))=19.41 `
`[Ag^(+)]^(2) = 0.00515`
`[Ag^(+)] = 0.0717 = 7.17 xx 10^(-2)` M
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