Problem

# Calcium metal can be obtained by the direct electrolysis of molten CaCl2, at a voltage o...

Calcium metal can be obtained by the direct electrolysis of molten CaCl2, at a voltage of 3.2 V.

(a) How many joules of electrical energy are required to obtain 12.0 lb of calcium?

(b) What is the cost of the electrical energy obtained in (a) if electrical energy is sold at the rate of nine cents per kilowatt hour?

#### Step-by-Step Solution

Solution 1

(a)

Molten $$\mathrm{CaCl}_{2}$$ undergoes direct electrolysis to form calcium metal. The chemical reaction is as follows:

$$\mathrm{Ca}^{2+}(a q)+2 e^{-} \rightarrow \mathrm{Ca}(s)$$

For the production of one mole calcium, two moles of electrons are consumed.

$$\frac{2 \mathrm{~mol} e^{-}}{1 \mathrm{~mol} \mathrm{Ca}}$$

The conversion factor for the relation between pounds and grams is as follows:

$$\frac{453.6 \mathrm{~g}}{1 \mathrm{lb}}$$

The molar mass of calcium is $$40.08 \mathrm{~g} / \mathrm{mol}$$.

The conversion factor for the relation between number of moles of electrons and coulomb is as follows:

$$\frac{1 \mathrm{~mol} e^{-}}{9.648 \times 10^{4} \mathrm{C}}$$

Calculate the coulombs by using $$12.0 \mathrm{lb}$$ of calcium:

$$12.0 \mathrm{\not6} \times \frac{453.6 \mathrm{~g}}{1 \mathrm{bb}} \times \frac{1 \mathrm{metCa}}{40.08 \mathrm{~g}} \times \frac{2 \mathrm{mote}}{1 \mathrm{motCa}} \times \frac{9.648 \times 10^{4} \mathrm{C}}{1 \text { mele }}=2.62 \times 10^{7} \mathrm{C}$$

Find the electrical energy in joules:

\begin{aligned} \mathrm{J} &=\mathrm{C} \times \mathrm{V} \\ &=2.62 \times 10^{7} \mathrm{C} \times 3.2 \mathrm{~V} \\ &=8.4 \times 10^{7} \mathrm{~J} \end{aligned}

(b)

The conversion factor for the relation between joules and $$\mathrm{kWh}$$ is as follows:

$$\frac{1 \mathrm{kWh}}{3.600 \times 10^{6} \mathrm{~J}}$$

Find the kilowatt hour by using above conversion factor:

$$8.4 \times 10^{7} \not \beta \times \frac{1 \mathrm{kWh}}{3.600 \times 10^{6} \not \beta}=23.3 \mathrm{kWh}$$

The electric energy is sold at the rate of nine cents per one kilowatt hour. The required conversion factor is as follows:

$$\frac{9 \text { cents }}{1 \mathrm{kWh}}$$

The conversion factor for the relation between cents and dollars is as follows:

$$\frac{1 \}{100 \text { cents }}$$

Calculate the cost of the electrical energy by using above conversion factors:

$$23.3 \mathrm{kWh} \times \frac{9 \text { cents }}{1 \mathrm{k} \mathrm{Wh}} \times \frac{1 \mathrm{~s}}{100 \text { certs }}=2.1 \mathrm{~s}$$