`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)`

Options

A. `E_(2) gt E_(1) gt E_(3)`
B. `E_(1) gt E_(2) gt E_(3)`
C. `E_(3) gt E_(1) gt E_(2)`
D. `E_(3) gt E_(2) gt E_(1)`

Answer

B

Explanation

`E_(cell) = E_(cell)^(@) - (0.059)/(n) log ([Zn^(2+)])/([Cu^(2+)])`
(i) `Zn (s) |Zn^(2+) (0.1 M) ||Cu^(2+) ( 1M) |Cu (s) `
`E_1 = E_(cell)^(@) - (0.059)/(2) "log" ((0.1)/(1))`
`= E_(cell)^(@) + (0.059)/(2)`
(ii) `Zn(s) |Zn^(2+) (1M) ||Cu^(2+) (1 M) |Cu (s)`
`E_(2) = E_(cell)^(@) - (0.059)/(2) "log" ((1.0)/(1.0))`
`E_(cell)^(@)`
(iii) `Zn (s) |Zn^(2+) (1M) ||Cu^(2+) (0.1M) |Cu`
`E_(3) = E_(cell)^(@) - (0.059)/(2) "log" (1.0)/(0.1)`
` = E_(cell)^(@) - (0.059)/(2)` .
`therefore E_(1) gt E_(2) gt E_(3)`