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Given: E(Zn^(+2)//Zn)^(@) =- 0.76V E...


Given:
`E_(Zn^(+2)//Zn)^(@) =- 0.76V`
`E_(Cu^(+2)//Cu)^(@) = +0.34V`
`K_(f)[Cu(NH_(3))_(4)]^(2+) = 4 xx 10^(11)`
`(2.303R)/(F) = 2 xx 10^(-4)`
When 1 mole of `NH_(3)` added to cathode compartement than emf of cell is (at 298K)`:

A

0.81 V

B

1.91 V

C

1.1 V

D

0.72 V

Text Solution

Verified by Experts

The correct Answer is:
A
`undersetunderset(x)(0.2)(Cu^(+2))+undersetunderset(1-0.8)(1)(4NH_(3))hArrundersetunderset(0.2)(0)([Cu(CH_(3))_(4)]^(+2))`
`K_(f)=4.0xx10^(11)=(0.2)/(x xx(0.2)^(4))=(1)/(x xx(0.2)^(3))`
`x=(10^(-11))/((0.2)^(3)xx4)`
`x=3.125xx10^(-10)`
`[Cu^(+2)]=3.125xx10^(-10)`
`E_(cell)=0.75+0.34-(0.0591)/(2)log((2)/(3.125xx10^(-10))`
`=1.1-(0.0591)/(2)(10-0.194)=1.1-0.29=0.81"volt"`
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