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Calculate equilibrium constant for the r...

Calculate equilibrium constant for the reaction :
`Zn+Cd^(2+) hArr Zn^(2+)+Cd`,
(Given `E_(cell)^(@)=0.36 "V "`)

Text Solution

Verified by Experts

The correct Answer is:
`1.596 xx 10^12`
`E_(cell)^(Theta) = E_((Cd^(2+) |Cd))^(Theta) - E_((Zn^(2+)|Zn))^(Theta)`
`=0.403 - (- 0.763) = 0.360 V `
`log K_(c) = (2 E_(cell)^(@))/(0.059)`
`( 2 xx 0.36)/(0.059) = 12.203`
`K_c = 1.596 xx 10^(12)`
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