`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) 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

Key-idea
Use Nernst equation, i.e.
E (emf) `= E_("cell")^(@) - (0.0591)/(2) "log" [[Zn^(2+)]]/([Cu^(2+)])`
For reaction (i) `E_(1) = E_("cell")^(@) - (0.0591)/(2) "log" (0.1)/(1)`
`E_(1) = E_(1)= E_(("cell"))^(@)+ (0.0591)/(2)`
For reaction (ii)
`E_(2) = E_(("cell"))^(@) =(0.0591)/(2) "log " (1)/(1) =0`
`E_(2) = E_(("cell"))^(@)`
For reaction (iii)
`E_(3) = E_(("cell"))^(@) -(0.0591)/(2) "log" (1)/(0.1)`
`E_(3) = E_(("cell"))^(@) - (0.591)/(2)`
Hence `E_(1) gt E_(2) gt E_(3)` .