Three faraday of electricity is passed through molten solutions of $A g N O_{3}, N i S O_{4}$ and $C r C l_{3}$ kept in three vessels using inert electrodes. The ratio in mol in which the metals $A g, N i$ and $C r$ will be deposited in:

1 : 2 : 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3 : 2 : 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6 : 3 : 2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2 : 3 : 6

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $6 : 3 : 2$
(i) $A g (a q) + e^{-} \to A g (s)$
$1 m o l = 1 F$ $1 m o l$
$3 F$ $3 m o l$
(ii) $N i^{2 +} (a q) + 2 e^{-} \to N i (s)$
$2 m o l = 2 F$ $1 m o l$
$3 F 3 / s m o l$
(iii) $C r^{3} + (a q) + 3 e^{-} \to C r (s)$
$3 m o l = 3 F$ $1 m o l$
The required ratio of moles of $A g, N i$ and $C r$ is:
$3 m o l A g : 3 / 2 m o l N i : 1 m o l C r$
i.e. $6 : 3 : 2$

Suggest Corrections

Similar questions

Four faradays of electricity were passed through AgNO₃ (l), CdSO₄(l), AlCl₃(l), & PbCl₄(l) kept in four vessels using inert electrodes. The ratio of moles of Ag, Cd, Al,& Pb deposited will be

A. 12 : 4 : 6 : 3

B.12 : 6 : 4 : 3

Three faraday of electricity is passed through three electrolytic cells connected in series containing $A g^{+}, C a^{2 +}$ and $A l^{3 +}$ ions respectively. The molar ratio in which the three metal ions are liberated at the electrodes is: