A cell Ag|`Ag^+||Cu^(++)|Cu` initially contains `2M Ag^+` and `2M Cu^(++)` ions. The change in cell potential after the passage of 10 amp current for 4825 sec is:

Options

A. `-0.0074 V`
B. `-1.00738 V`
C. `-0.0038 V`
D. none

Answer

A

Explanation

Q=10x4825
=48250 C
no. of mole of `Ag^+ =48250/96500=0.5`
`Ag+1/2 Cu^(++)toAg^+ +1/2 Cu`
`{:(2.00,2.00),(2-0.25,2+0.50):}`
`E_"cell"=E_"cell"^@-0.0591/1 "log" ([Ag^+])/([Cu^(++)]^(1//2)) implies E_1=E_"cell"^@-0.0591/1"log"2.00/((2.00)^1//2)`
`E_2=E_"cell"^@-0.0591/1 "log" (2.50)/(1.75)^((1//2)) implies DeltaE=E_2-E_1=0.0591/1["log"sqrt2-"log"2.50/sqrt(1.75)]`
`=0.0591/1["log"1.41-"log"1.88]=0.0591/1[0.1492-0.2742]=-0.0591/2xx0.125=-0074 V`