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Calculate the equilibrium constant for t...

Calculate the equilibrium constant for the reaction at 298K.
`Zn(s) +Cu^(2+)(aq) hArr Zn^(2+)(aq) +Cu(s)`
Given, `E_(Zn^(2+)//Zn)^(@) =- 0.76V` and `E_(Cu^(2+)//Cu)^(@) = +0.34 V`

Text Solution

Verified by Experts

We know that, `log K_(c) = (nE_(cell)^(@))/(0.0591)`
`E_(cell) = [E_("Cathode") - E_("Anode")]`
`=[(+0.34 V)-(-0.76V)] = 1.10 V, n =2`,
`:. log_(c)= (2xx (1.10V))/((0.0591V)) = 37.29`
`K_(c) = "Antilog" 37.29 = 1.95 xx 10^(37)`
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