1.

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

Answer»

`0.195`
`0.097`
`0.053`
`0.145`

Solution :`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`


Discussion

No Comment Found

Related InterviewSolutions