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A copper-silver cell is set up. The copp...

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

Text Solution

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The given cell may be represented as
`Cu_(s)|Cu ^(2+) (0.10M)\\Ag^(+) (C )|Ag_(s)`
`E_(cell)^(@) = E_("cathode")^(@)-("anode") = 0.80V - 0.34V = 0.46 V`
`E_(cell)^(@) = E_(cell)^(@)(0.0591)/(2)log([Cu^2+])/([Ag^(+)]^(2)`
or `0.422V-0.46V- (0.0591)/(2) log (0.1)/([Ag^(+)]^(2)`
`-0.038V = -0.0295 log (0.1)/([Ag^(+) ]^(2)` or , log`(0.1)/([Ag^(+)]^(2) = (-0.038)/(-0.0295) = 1.288`
or log`(0.1)/([Ag^(+)]^(2) = ("antilog")1.288 = 19.41` `therefore ([Ag^(+))]^(2)= (0.1)/(19.41) = 5.1519xx10^(-3)`
`([Ag^(+)] = 7.1 xx10^(-2) M`
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