Question

The molar ratio of $B^{2+}$ and $C^{3+}$ ions discharged at the electrode when 0.2 faradays of electricity is passed throught the solution is:

• A. 5:2
• B. 4:5
• C. 3:2
• D. 2:1

Answer 1

The correct option is C 3:2

Amount of ion reduced, $m\propto Q$

For one mol $B^{2+}$, 2 mol of electrons are required.

So $Q=2F$

With $0.2F$, only 0.1 mol is obtained

For one mol $C^{3+}$, 3 moles of electrons required

so $Q=3F$

with $0.2F$, $\frac{0.2}{3}$ mol is obtained

molar ratio of $B^{2+}$:$C^{3+}$::$0.1:\frac{0.2}{3}$

i.e. $3:2$

C is correct.