A 1.0 M solution of Cd^(+2) is added to ...

A 1.0 M solution of `Cd^(+2)` is added to excess iron and the system is allowed to reach equilibrium . What is the concentration of `Cd^(+2) `? `Cd^(2+)(aq) + Fe(s) to Cd(s) + Fe^(2+) (aq) , E^(0) = 0.037`

`0.195`

`0.097`

`0.053`

`0.145`

Text Solution

Verified by Experts

The correct Answer is:

`Cd^(+) + Fe to Fe^(+2) + Cd , E=E^(0) -(0.0591)/(n) log Q`
` 0=0.037 - (0.0591)/(2) log""([Fe^(+2)])/([Cd^(+2)]) ,log""([Fe^(+2)])/([Cd^(+2)]) = (0.037 xx 2)/( 0.0591) = 1.25`
`log""((y)/(1-y)) = 1.252 , ((y)/(1-y)) = 10 xx 1.79 , (y)/(1-y) =17.9`
`y = 17.9 - 17.9 y , y + 17.9 y = 17.9 , y=(17.9)/(18.9) = 0.947 , (1-y) = 1- 0.947 = 0.0529 =0.053`

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