Question

Molten ${\text{CaCl}}_{\text{2}}$is electrolyzed for $8.00h$to produce $\mathrm{Ca}\left(s\right)$and data-custom-editor="chemistry" ${\text{Cl}}_{\text{2}}\text{(g)}$

a. What current is needed to produce$5.52\mathrm{kg}$of calcium metal?

b. If$5.52\mathrm{kg}$calcium metal is produced, what mass (inkg of${\text{Cl}}_{\text{2}}$is produced?

Short answer

1. 924 A is the current needed to produce $5.52kg$ of calcium metal.
2. The${\text{MassofCl}}_2\text{is9660g}$.

Step-by-step solution

Concept Introduction 

Electrolysis is a which occurs when an electric current is conducted through a molten or aqueous solution of an electrolyte to produce the desired chemical change.

The charge (Q) and the time (t) are connected by the current (I) that passes through the solution as follows:

$|=\frac{q}{T}$

Write the Half-cell reaction

a)

${\text{Ca}}^{\text{2 +}}$is reduced to metallic calcium and deposited at the cathode during the electrolysis of ${\text{CaCl}}_{\text{2}}$. In contrast, at the anode,$\text{Cl -}$is oxidised to ${\text{Cl}}_{\text{2}}$gas.

The half-reactions are as follows:

${\text{Cathode:Ca}}^{\text{2 +}}{\text{(aq) + 2e}}^{\text{-}}\to \text{Ca(s)}$

${\text{Anode:2Cl}}^{\text{-}}\text{(aq)}\to {\text{Cl}}_{\text{2}}{\text{(g) + 2e}}^{\text{-}}$

First, we must determine the moles of calcium, which we can do using the calcium's mass.

Divide the mass of magnesium by the molar mass of calcium to get moles of calcium.

$n\left(\text{Ca}\right)=\frac{\text{mass}}{\text{Molarmass}}=\frac{\text{5520g}}{\text{40g/mol}}\text{=138mol}$

Now calculate the moles of calcium:

The moles of calcium can then be compared to the moles of electrons.

The mole ratio is 1 to 2 if you look at the half-reactions above.

$n\left(e^{-}\right)\text{=2×n}\left(\text{Ca}\right)\text{=2×138mol}$

$n\left(e^{-}\right)\text{=276mol}$

Let us find the value of current:

The Faraday charge on one mole of electrons is 96,485 coulombs of charge per mole of electrons.

As a result, q equals the number of electron moles multiplied by the charge per mole of electrons:

$\text{q=n}\left(e^{-}\right){\text{×F=276mole}}^{-}{\text{×96,485C/mole}}^{-}$

${\text{q=26.6×10}}^{6}C$

The charge (Q) and time (t) are connected to the current (I) that passes through the solution as follows:

$\text{I=}\frac{q}{T}=\frac{{\text{26.6×10}}^{6}C}{\text{28800s}}$

$|=924A$

Calculate of mass of Cl2

b)

The moles of chlorine gas can then be compared to the moles of electrons.

The mole ratio is 1 to 2 if you look at the half-reactions above.

$n\left({\text{Cl}}_2\right)=\frac{1}{2}\text{×n}\left(e^{-}\right)=\frac{1}{2}\text{×276mol}$

$n\left({\text{Cl}}_2\right)\text{=138mol}$

The moles of ${\mathrm{Cl}}_2$and its molar mass can be used to compute the mass of${\mathrm{Cl}}_2$ .

${\text{MassofCl}}_2\text{=138mol×70g/mol}$

${\text{MassofCl}}_2\text{=9660g}$

Therefore, the mass of${\mathrm{Cl}}_{2}is9660g$ .