Q21.117 CP - Chemistry: Molecular Nature Of Matter And Change

Question

(a) How many minutes does it take to form 10.0 L of ${\text{O}}_{\text{2}}$ measured at  a$\text{99.8 kPa}$and  ${\text{28}}^{\text{o}}\text{C}$from water if a current of  1.3 A passes through the electrolytic cell? 

(b) What mass of ${\text{H}}_{\text{2}}$ forms?

Step-by-Step Solution

Answer

(a) The time required to form  10.0 L of ${\text{O}}_{\text{2}}$ is 1974 mins.

(b) The mass of ${\text{H}}_{\text{2}}$  that form is $\text{1.609g}$.

Step 1: Concept Introduction

The process of dissolving ionic compounds into their constituent components by delivering a direct electric current through the complex in a fluid form is known as electrolysis. At the cathode, cations are reduced, whereas anions are oxidised.

Step 2: Information Provided

• The reaction is – ${\text{Au}}^{\text{3 + }}{\text{(aq) + 3e}}^{\text{ - }}\rightleftharpoons \text{Au(s)}$
• Oxygen that needs to be formed: $\text{10.0 L}$ at  $\text{99.8 kPa}$and ${\text{28}}^{\text{o}}\text{C(28 + 273 K = 301 K)}$
• The current is $\text{0.013A(A = C/s)}$
• The ideal gas constant is $\text{R = 0.0821 L atm/K mol}$
• Faraday constant: charge of  1 mole of electrons $\text{F = 96485 C/mole}$

Step 3: (a) Calculation for Time

Convert the pressure from $\text{kPa}$  to atm  –

$$\begin{gathered} \text{P = 99.8 kPa}·\frac{\text{1000Pa}}{\text{1 kPa}}·\frac{{\text{9.8692310}}^{\text{ - 6}}\text{ atm}}{\text{1 Pa}} \\ \text{ = 0.985 atm} \end{gathered}$$

Calculate the number of moles of oxygen that need to be formed –

$$\begin{gathered} \text{PV = nRT} \\ \text{n = }\frac{\text{PV}}{\text{RT}} \\ \text{ = }\frac{\text{0.985atm}·10.0\text{L}}{{\text{0.0821 L atm mol}}^{\text{ - 1}}{\text{K}}^{\text{ - 1}}·\text{301K}} \\ {\text{ = 0.399 mol O}}_{\text{2}} \end{gathered}$$

It can be seen that in the reaction above, for every  ${\text{O}}_{\text{2}}$ formed,  4 moles of electrons are involved. Therefore, the number of moles of electrons is –

$$\begin{gathered} {\text{n}}_{{\text{e}}^{\text{ - }}}{\text{ = 0.399mol O}}_{\text{2}}·\frac{{\text{4mol e}}^{\text{ - }}}{{\text{1mol O}}_{\text{2}}} \\ {\text{ = 1.596mol e}}^{\text{ - }} \end{gathered}$$

Calculate the charge using Faraday constant –

$$\begin{gathered} {\text{Charge = n}}_{{\text{e}}^{\text{ - }}}·\text{F} \\ {\text{ = 1.596 mol e}}^{\text{ - }}·{\text{96485 C/mol e}}^{\text{ - }} \\ \text{ = 1.54}·{\text{10}}^5\text{ C} \end{gathered}$$

The time required (in minutes) –

$$\begin{gathered} \text{Current = }\frac{\text{ Charge }}{\text{ Time }} \\ \text{Time = }\frac{\text{ Charge }}{\text{ Current }} \\ \text{ = }\frac{\text{1.54}·{\text{10}}^5\text{ C}}{\text{1.3}\frac{\text{C}}{\text{s}}} \\ \text{ = 1.185}·{\text{10}}^{\text{5}}\text{s} \\ \text{ = 1.185}·{\text{10}}^{\text{5}}\text{s}·\frac{\text{1min}}{\text{60s}} \\ \text{ = }1\text{974 min} \end{gathered}$$

 

Therefore, the value for time is obtained as $1\text{974 min}$.

Step 4: (b) Calculation for Mass

The reaction is –

${\text{2H}}^{\text{ + }}{\text{ + 2e}}^{\text{ - }}\to {\text{H}}_{\text{2}}$

The number of moles of electrons is $\text{1.596 mol}$.

Since  2 moles of electrons are required to produce 1 mole of ${\text{H}}_{\text{2}}$, the number of moles of  produced are –

$$\begin{gathered} {\text{n}}_{{\text{H}}_{\text{2}}}{\text{ = 1.596mol e}}^{\text{ - }}·\frac{{\text{1mol H}}_{\text{2}}}{{\text{1mol e}}^{\text{ - }}} \\ {\text{ = 0.798mol H}}_{\text{2}} \end{gathered}$$

Molar mass of  is , hence, the mass of  produced is –

 $$\begin{gathered} {\text{m}}_{{\text{H}}_{\text{2}}}{\text{ = 0.798mol H}}_2·\frac{\text{2.016 g}}{{\text{1 molH}}_2} \\ {\text{ = 1.609g H}}_{\text{2}} \end{gathered}$$

Therefore, the value for mass is obtained as $\text{1.609g}$ .