What volume of `0.2M FeSO_(4)` can be oxidized by a current of `0.965` ampere hour?

Options

A. 0.07 L
B. 0.08 L
C. 0.09L
D. 0.1 L

Answer

C

Explanation

`c. 1 F=96500 C =1` equivalent of `Fe ^(2+)`
`0.965xx 3600C =(0.965xx3600)/(96500C)=(36)/(1000)`equivalent
`V ( i n L)xxN=` Equivalent `(n` factor `Fe^(2+)=2)`
`V( i n L)xx2xx0.2=(36)/(1000)`
`V (i n L)=(36)/(1000xx2xx0.2)=0.09L`