`underset(1 L solution)(Ag|Ag^(o+)(1M))||underset(1 L solution)(Ag^(o+)||Ag)`
`0.5F` electricity in the `LHS(` anode `)` and `1F` of electricity in the `RHS(` cathode`)` is first passed making them independent electrolytic cells at `298K`. `EMF` of the cell after electrolysis will be

Options

A. Increased
B. Decreased
C. No change
D. Time is also required

Answer

c

Explanation

`E=E^(c-)+(0.0591)/(1)log.([Ag^(o+)]_(R))/([Ag^(o+)]_(L))`
`=0+0.0591log 2=0.0591xx0.301V`
After current is passed `[Ag^(o+)]_(R)=1M`
`[Ag^(o+)]_(L)=0.5M`
Hence, no change in `EMF`.