Find the amount of silver liberated at cathode if `0.500 A` of current is passed through `AgNO_3` electrolyte for 1 hour. Atomic weight of silver s `107.9 g mol^(-1)`

Answer

A, B

Explanation

At wt Ag = 107.9 g/mole `
`i=0.5 A, `
`E_(Ag) = 107.9 g [As Ag is monoatomic] Z_(Ag) = (E )/(f) = (107.9)/(965000)`
` =0.001118 So, amount of silver liberated, M = Zit = 0.00118xx0.500xx3600 =2.01g