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Zn+Cu^(2+)(aq)hArrCu+Zn^(2+)(aq). Reac...

`Zn+Cu^(2+)(aq)hArrCu+Zn^(2+)(aq).`
Reaction quotient is `Q=([Zn^(2+)])/([Cu^(2+)])` . Variation of `E_(cell)` with log `Q` is of the type with `OA=1.10` `V.E_(cell) ` will be `1.1591V` when

A

`([Cu^(2+)])/([Zn^(2+)])=0.01`

B

`([Zn^(2+)])/([Cu^(2+)])= 0.01`

C

`([Zn^(2+)])/([Cu^(2+)])= 0.1`

D

`([ZN^(2+)])/([Cu^(2+)])= 1`

Text Solution

Verified by Experts

The correct Answer is:
B
`Zn(s) + Cu^(2+) (aq) Zn^(2) (aq) + Cu(s)`
`E_(Cell) = E_(Cell)^(@)- (0.0591)/(2)log_(10) Q`
At O, `log _(10) Q= 0 Rightarrow E _ (cell) = E_(cell)^(@) = 1.1V`
When `E_(cell) 1.1591VRightarrow 1.1591 = 1.1 -(0.0591) /(2) log _(10) Q`
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