Question

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

• A. $1:2:3$
• B. $2:3:1$
• C. $6:3:2$
• D. $3:4:2$

Answer 1

The correct option is C $6:3:2$

For all the solution $Q=i\times t$ is same,

$n_{Ag}=\frac{Q}{nF}=\frac{Q}{F}({Ag}^{+}+e^{-}⟶Ag)$

$n_{Ca}=\frac{Q}{2F}({Ca}^{2+}+2e^{-}\to Ca)$

$n_{Al}=\frac{Q}{3F}({Al}^{3+}+3e^{-}\to Al)$

$\therefore$ $n_{Ag}:n_{Ca}:n_{Al}=1:\frac{1}{2}:\frac{1}{3}=6:3:2$

$\therefore$ Molar ratio$=6:3:2$.