An electric charge of 5 Faradays is passed through three electrolytes `AgNO_3, CuSO_4` and `FeCl_3` solution. The grams of each metal liberted at cathode will be

Options

A. Ag=10.8 g,Cu = 12.7 g, Fe = 1.11 g
B. Ag = 540 g, Cu = 367.5 g , Fe = 325 g
C. Ag =108 g , Cu =63.5 g, Fe = 56g
D. Ag= 540 g, Cu = 158.8 g, Fe = 93.3 g

Answer

D

Explanation

`underset(1 mol -= 1F)(Ag^(+) + e^(-) to Ag)`
`underset(2 "mol "-=2F)(Cu^(2+) + 2e^(-) to Cu)`
`underset(3 " mol "-= 3F)(Fe^(3+) + 3e^(-) to Fe)`
Grams of liberated `=(108)/(1)xx5=540 g`
Grams of Cu liberated `=(63.5)/(2)xx5=158.75 g`
Grams of liberated `=(56)/(3)xx5=93.3 g`