For the cell Cu ( s ) Cu 2 + ( aq ) ( 0 . 1 M ) ‖ Ag + ( aq ) ( 0 . 01 M ) Ag ( s ) the cell potential E 1 = 0 . 3095   V . For the cell Cu ( s ) Cu 2 + ( aq ) ( 0 . 01 M ) ‖ Ag + ( aq ) ( 0 . 001 M ) Ag ( s ) the cell potential = x × 10 - 2   V . Find value of x (Round off the Nearest Integer). [ Use : 2 . 303 RT F = 0 . 059   J ]

Step-by-step solution

Cell reaction is : Cu ( s ) + 2 Ag + ( aq ) → Cu 2 + ( aq ) + 2 Ag ( s ) Now, E cell = E Cell o - 0 . 059 2 log Cu 2 + Ag + 2 E 1 = 0 . 3095 = E Cell o - 0 . 059 2 · log 0 . 1 ( 0 . 01 ) 2 ....... 1 E 2 = E Cell o - 0 . 059 2 · log 0 . 01 ( 0 . 001 ) 2 . . . . . . . . . . ( 2 ) From ( 1 ) and ( 2 ) , E 2 = 0 . 28   V = 28 × 10 - 2   V