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E(1), E(2) and E(3) are the emfs of the ...

`E_(1)`, `E_(2)` and `E_(3)` are the emfs of the following three galvanic cells respectively
I. `Zn((s))`|`Zn^(2+) (0.1M)`| |`CU^(2+) (1M)`| `Cu((s))`
II. `Zn((s)) ZN^(2+) (1M)` ||`Cu^(2+)(1M)`|`Cu(s)`
III. `Zn(s)` | `Zn^(2+) (1M)`||`CU^(2+)(0.1M)CU(s)`

A

`E_(2)gtE_(3)gtE_(1)`

B

`E_(3)gtE_(2)gtE_(1)`

C

`E_(1)gtE_(2)gtE_(3)`

D

`E_(1)gtE_(3)gtE_(2)`.

Text Solution

Verified by Experts

The correct Answer is:
B

`Zn+Cu^(2+)rarrZn^(2+)+Cu`
`E_("cell")=E_("cell")^(@)-(0.0591)/(2)"log"([Zn^(2+)])/([Cu^(2+)])`
Lesser the `[Zn^(2+)]//[Cu^(2+)]` value more will be `E_("cell")`
for (i) `([Zn^(2+)])/([Cu^(2+)])=(1)/(01)=10`
for (ii) `([Zn^(2+)])/([Cu^(2+)])=(1)/(1)=1` and
for (iii) `([Zn^(2+)])/([Cu^(2+)])=(01)/(1)=0.1`
Hence `E_(3)gtE_(2)gtE_(1)`
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