The correct option is B Emf of the cell gets reduced from 0.177 to 0.118 V
From cell representation,
[Ag+]=0.001 M⇒ Anode
[Ag+]=1 M⇒ Cathode
Cell reaction will be,
Ag+(aq, 1 M)→Ag+(aq, 0.001 M)
By nernst equation,
Ecell=E0cell−0.05911 log [0.001][1]
Ecell=E0cell+0.05911log[1][0.001]
For concentration cells, cathode and anode consist of same metal and its solution.
Thus,
E0=0
Ecell=0.05911log[1][0.001]
Ecell=0.05911 log 103
Ecell=0.05911×3
Ecell≈0.177 V
Now, concentration of Ag+ at cathode is reduced to 0.1 M
Cell reaction will be,
Ag+(aq, 0.1 M)→Ag+(aq, 0.001 M)
Ecell=E0cell−0.05911 log [0.001][0.1]
Ecell=E0cell+0.05911log[0.1][0.001]
E0=0
Ecell=0.05911log[0.1][0.001]
Ecell=0.05911 log 102
Ecell=0.05911×2
Ecell≈0.118 V