# Question e42d6

May 26, 2016

Choice D.

#### Explanation:

Molten calcium chloride $C a C {l}_{2}$ will dissociate into its corresponding ions according to the following equation:

$C a C {l}_{2} \to C {a}^{2 +} + 2 C {l}^{-}$

The reduction reaction of calcium ion $C {a}^{2 +}$ that happens on the cathode is:

$C {a}^{2 +} + 2 {e}^{-} \to C a \left(s\right)$

Therefore, for $1 m o l$ of $C {a}^{2 +}$ to be reduced, $2 m o l$ of electrons will be needed ($2 {n}_{c {a}^{2 +}} = {n}_{{e}^{-}}$).

The number of mole of calcium $C {a}^{2 +}$ could be found from its mass ($m = 60 g$):

$n = \frac{m}{M M} = \frac{60 \cancel{g}}{40 \frac{\cancel{g}}{m o l}} = 1.5 m o l C {a}^{2 +}$

$\implies {n}_{{e}^{-}} = 2 \times 1.5 = 3.0 \text{mol} {e}^{-}$

The charge $q$ applied could be calculated from the current $I = 5 A$ and the time $t$ in seconds as follows:

$q = I \times t = {n}_{{e}^{-}} \times F$ where $F = 96485 \frac{C}{\text{mol} {e}^{-}}$ is Faraday's constant.

=>t=(n_(e^(-))xxF)/I=(3.0cancel("mol"e^(-))xx(96485 C)/(1cancel("mol"e^(-))))/(5A)=57891s#

$\implies t = \left(57891 \cancel{s}\right) \times \frac{1 h}{3600 \cancel{s}} = 16 h$

Here is video that explains further this topic:
Electrochemistry | Electrolysis, Electrolytic Cell & Electroplating.