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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

`1.1V`

D

`0.72`

Text Solution

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The correct Answer is:
C
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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) Find emf at cell of 200K if E^(@) values are independent of temperature [log 2= 0.3]

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) At what concentration of Cu^(+2) emf of the cell will be zero (at 298K) and concentration of Zn^(+2) is remains same:

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