Chapter 18, Problem 18.50QP

How many faradays of electricity are required to produce (a) 0.84 L of O₂ at exactly 1 atm and 25°C from aqueous H₂SO₄ solution, (b) 1.50 L of Cl₂ at 750 mmHg and 20°C from molten NaCl, and (c) 6.0 g of Sn from molten SnCl₂?

Interpretation Introduction

Interpretation:

Need to calculate the Faraday of electricity needed for the production of 0.84L of O₂ with 1 atm pressure upon electrolysis of aqueous H₂SO₄ at 25^oC.

Concept introduction:

Electrolysis of aqueous sulfuric acid i.e. acidified water resulted in the production of oxygen and hydrogen gas, which will be liberated at the anode and cathode respectively. The presence of H⁺ and SO^-₄ made the solution to be more conductivity.  SO^-₄ is more stable to be inert at the anode.

Further the cell reaction can be written as shown below

$\begin{array}{l}\text{Anode }{\text{2H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{ O}}_{\text{2}}{\text{+4H}}^{\text{+}}\text{+4e-}\\ \text{Cathode}{\text{4H}}^{\text{+}}{}_{\text{(aq)}}{\text{+4e}}^{\text{-}}\stackrel{}{\to }{\text{2H}}_{\text{2(g)}}\\ \text{Overall reaction}{\text{2H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{ O}}_{\text{2}}{\text{+2H}}_{\text{2(g)}}\end{array}$

Since volume and pressure of oxygen produced was given, by applying it into ideal gas equation we can calculate the number of mole of oxygen produced

The ideal gas equation can be given as follows.

$\begin{array}{l}\text{PV}\text{=}\text{nRT}\\ \text{n}\text{=}\frac{\text{PV}}{\text{RT}}\\ \text{n}\text{=}\text{number}\text{of}\text{moles}\text{of}\text{gas}\\ \text{P}\text{=}\text{pressure}\text{of}\text{gas}\\ \text{V}\text{=}\text{Volume}\\ \text{R}\text{=}\text{universal}\text{gas}\text{constant (0}\text{.08206L}\text{.atm/K}\text{mol)}\\ \text{T}\text{=}\text{Temperature}\end{array}$

On applying the number of moles of oxygen produced into stoichiometry of the reaction, the number of moles electron involved in the reaction can be calculated. In case of the given reaction 4 mole of electron was liberated during the production of one mole of oxygen. Since one Faraday is equal to one mole of electron, so 4 Faraday of electricity will be needed to produce one mole of oxygen.

 Finally the Faraday of electricity utilized to produce the required amount of oxygen can be calculated according the formula

$\text{Faraday}\text{=}\text{n×}\frac{\text{4F}}{\text{1}\text{mole}{\text{O}}_{\text{2}}}$

To find: Amount of Faraday of electricity need to produce 0.076L of O₂ with pressure 755mmHg, at 298K, through electrolysis of water.

Answer to Problem 18.50QP

Ideal gas equation can be used to calculate the number of moles of oxygen produced, from that faraday of electricity needed will be calculated in successive steps (a)

$\begin{array}{l}\text{PV}\text{=}\text{nRT}\\ \text{n}\text{=}\frac{\text{PV}}{\text{RT}}\\ \text{n}\text{=}\frac{\text{(1atm)(0}\text{.84L)}}{\text{(0}\text{.08206L}\text{.atm/K}\text{mol)(298K)}}\\ \text{=}\frac{\text{(0}\text{.84L)}}{\text{(24}\text{.4539)}}\\ \text{=}\text{3}{\text{.4350×10}}^{\text{-2}}\text{mol}{\text{O}}_{\text{2}}\end{array}$

Since one mole of oxygen need 4 Faraday of electricity, so the Faraday of electricity needed to produce $\text{3}{\text{.4350×10}}^{\text{-2}}\text{mol}{\text{O}}_{\text{2}}$ can be calculated according to given formula.

$\begin{array}{l}\text{Faraday}\text{=}\text{n×}\frac{\text{4F}}{\text{1}\text{mole}{\text{O}}_{\text{2}}}\\ \text{=}\text{3}{\text{.4350×10}}^{\text{-2}}\text{mole}{\text{O}}_{\text{2}}\text{×}\frac{\text{4F}}{\text{1mole}{\text{O}}_{\text{2}}}\\ =1.3740{\text{×10}}^{\text{-1}}\text{F}\end{array}$

Faraday of electricity need to produce 0.84L of oxygen with pressure 1 atm was calculated as $1.3740{\text{×10}}^{\text{-1}}\text{F}$.

Explanation of Solution

Ideal gas equation can be used to calculate the number of moles of oxygen produced

$\begin{array}{l}\text{PV}\text{=nRT}\\ \text{n}\text{=}\frac{\text{PV}}{\text{RT}}\\ \text{n}\text{=number}\text{of}\text{moles}\text{of}\text{gas}\\ \text{P}\text{= 1atm}\\ \text{V}\text{=0}\text{.076L}\\ \text{R}\text{=0}\text{.08206L}\text{.atm/K}\text{mol}\\ \text{T}\text{=298K}\\ \text{n}\text{=}\frac{\text{(1atm)(0}\text{.84L)}}{\text{(0}\text{.08206L}\text{.atm/K}\text{mol)(298K)}}\\ \text{=}\frac{\text{(0}\text{.84L)}}{\text{(24}\text{.4539)}}\\ \text{=3}{\text{.4350×10}}^{\text{-2}}\text{mol}{\text{O}}_{\text{2}}\end{array}$

Since one mole of oxygen need 4 Faraday of electricity, so the Faraday of electricity needed to produce $\text{3}{\text{.4350×10}}^{\text{-2}}\text{mol}{\text{O}}_{\text{2}}$ can be calculated according to given formula.

$\begin{array}{l}\text{Faraday=}\text{n×}\frac{\text{4F}}{\text{1}\text{mole}{\text{O}}_{\text{2}}}\\ \text{=3}{\text{.4350×10}}^{\text{-2}}\text{mole}{\text{O}}_{\text{2}}\text{×}\frac{\text{4F}}{\text{1mole}{\text{O}}_{\text{2}}}\\ =1.3740{\text{×10}}^{\text{-1}}\text{F}\end{array}$

Faraday of electricity need to produce 0.84L of oxygen with pressure 1 atm was calculated as $1.3740{\text{×10}}^{\text{-1}}\text{F}$.

At first the number of moles of oxygen produced through electrolysis was calculated using ideal gas equation, from the given volume and pressure. Thus it was calculated as $\text{3}{\text{.4350×10}}^{\text{-2}}\text{mol}{\text{O}}_{\text{2}}$. Further, from the stoichiometry of the reaction it was found that four mole of electron was liberated to produce one mole of oxygen, since one mole of electron is equal to one Faraday, so it was under stood that four Faraday of electricity will be needed to produce one mole of oxygen. Therefore it was calculated that $1.3740{\text{×10}}^{\text{-1}}\text{F}$ of electricity was needed to produce 0.84L of oxygen with pressure 1 atm.

Conclusion

The amount of electricity needed to produce 0.84L of oxygen with pressure of 1atm was determined to be $1.3740{\text{×10}}^{\text{-1}}\text{F}$. The calculation for the above result was well explained in steps.

Interpretation Introduction

Interpretation:

Need to calculate the Faraday of electricity needed for the production of 1.50L of Cl₂ with pressure 750 mmHg at 20^oC by electrolysis of molten NaCl.

Concept introduction:

Electrolysis of molten sodium chloride was represented by the below equation. By using ideal gas equation, the number of moles of chlorine liberated at the anode can be calculated, provided the pressure and volume are known.

$\begin{array}{l}\text{Anode (oxidation)}{\text{2Cl}}^{{}^{\text{\_}}}{}_{\text{(l)}}\stackrel{}{\to }{\text{Cl}}_{\text{2(g)}}{\text{+2e}}^{\text{-}}\\ \text{Cathode (Reduction)}2{\text{Na}}^{+}{}_{\text{(l)}}{\text{+ 2e}}^{\text{-}}\stackrel{}{\to }{\text{2Na}}_{\text{(s)}}\\ \text{Overallreaction}2{\text{Na}}^{\text{+}}\text{+}{\text{2Cl}}^{{}^{\text{\_}}}{}_{\text{(l)}}\stackrel{}{\to }{\text{2Na}}_{\text{(s)}}{\text{+Cl}}_{\text{2(g)}}\end{array}$

The ideal gas equation can be given as follows.

$\begin{array}{l}\text{PV}\text{=}\text{nRT}\\ \text{n}\text{=}\frac{\text{PV}}{\text{RT}}\\ \text{n}\text{=}\text{number}\text{of}\text{moles}\text{of}\text{gas}\\ \text{P}\text{=}\text{pressure}\text{of}\text{gas}\\ \text{V}\text{=}\text{Volume}\\ \text{R}\text{=}\text{universal}\text{gas}\text{constant (0}\text{.08206L}\text{.atm/K}\text{mol)}\\ \text{T}\text{=}\text{Temperature}\end{array}$

The equation we find that two mole of electron is needed to produce one mole of chlorine gas , Since one Faraday is equal to one mole of electron, The Faraday of electricity utilized to produce given amount of chlorine can be calculated according the formula

$\text{Faraday=}\text{n×}\frac{\text{2F}}{\text{1}\text{mole}{\text{Cl}}_{\text{2}}}$

To find: Faraday of electricity need to produce 1.50 L of Cl₂ with pressure 750mmHg, at 293K, through electrolysis of molten NaCl.

Answer to Problem 18.50QP

Ideal gas equation can be used to calculate the number of moles of chlorine gas produced, Further from the number of moles of chlorine; the faraday of electricity utilized will be calculated in successive steps (b)

$\begin{array}{l}\text{PV}\text{=}\text{nRT}\\ \text{n}\text{=}\frac{\text{PV}}{\text{RT}}\\ \text{n}\text{=}\frac{\left(\text{750mmHg×}\frac{\text{1}\text{atm}}{\text{760mmHg}}\right)\text{(1}\text{.50L)}}{\text{(0}\text{.08206L}\text{.atm/K}\text{mol)(293K)}}\\ \text{=}\frac{\text{(0}\text{.9868}\text{atm)(1}\text{.50L)}}{\text{(0}\text{.08206L}\text{.atm/K}\text{mol)(293K)}}\\ =\frac{1.4802}{24.4539}\\ \text{=}\text{6}{\text{.0530×10}}^{\text{-2}}\text{mol}{\text{Cl}}_{\text{2}}\end{array}$

Since one mole of chlorine need two Faraday of electricity, so the Faraday of electricity needed to produce $\text{6}{\text{.0530×10}}^{\text{-2}}\text{mol}$ can be calculated according to the given formula.

$\begin{array}{l}\text{Faraday}\text{=}\text{n×}\frac{\text{2F}}{\text{1}\text{mole}{\text{Cl}}_{\text{2}}}\\ \text{=}\text{6}{\text{.0530×10}}^{\text{-2}}\text{mole}{\text{Cl}}_{\text{2}}\text{×}\frac{\text{2F}}{\text{1mole}{\text{Cl}}_{\text{2}}}\\ =1.2106{\text{×10}}^{\text{-1}}\text{F}\end{array}$

Faraday of electricity need to produce 1.50L of chlorine with pressure 750mm Hg was calculated as $1.2106{\text{×10}}^{\text{-1}}\text{F}$.

Explanation of Solution

Ideal gas equation can be used to calculate the number of moles of oxygen produced

$\begin{array}{l}\text{PV}\text{=}\text{nRT}\\ \text{n}\text{=}\frac{\text{PV}}{\text{RT}}\\ \text{n}\text{=}\text{number}\text{of}\text{moles}\text{of}\text{gas}\\ \text{P}\text{=}\text{750 mmHg (760 mm Hg= 1atm)}\\ \text{V}\text{=}\text{1}\text{.50L}\\ \text{R}\text{=}\text{0}\text{.08206L}\text{.atm/K}\text{mol}\\ \text{T}\text{=}\text{293K}\\ \text{n}\text{=}\frac{\left(\text{750mmHg×}\frac{\text{1}\text{atm}}{\text{760mmHg}}\right)\text{(1}\text{.50L)}}{\text{(0}\text{.08206L}\text{.atm/K}\text{mol)(293K)}}\\ \text{=}\frac{\text{(0}\text{.9868}\text{atm)(1}\text{.50L)}}{\text{(0}\text{.08206L}\text{.atm/K}\text{mol)(293K)}}\\ =\frac{1.4802}{24.4539}\\ \text{=}\text{6}{\text{.0530×10}}^{\text{-2}}\text{mol}{\text{Cl}}_{\text{2}}\end{array}$

Since one mole of chlorine need two Faraday of electricity, so the Faraday of electricity needed to produce $\text{6}{\text{.0530×10}}^{\text{-2}}\text{mol}$ can be calculated according to given formula.

$\begin{array}{l}\text{Faraday of electricity =}\text{6}{\text{.0530×10}}^{\text{-2}}\text{mole}{\text{Cl}}_{\text{2}}\text{×}\frac{\text{2F}}{\text{1mole}{\text{Cl}}_{\text{2}}}\\ =1.2106{\text{×10}}^{\text{-1}}\text{F}\end{array}$

The number of moles of chlorine produced through electrolysis was calculated using ideal gas equation, from the given volume and pressure it was calculated as $\text{6}{\text{.0530×10}}^{\text{-2}}\text{mol}$. Further, from the stoichiometry of the reaction it was found that two mole of electrons were liberated to produce one mole of chlorine, since one mole of electron is equal to one Faraday, it was under stood that two Faraday of electricity will be needed to produce one mole of chlorine. Therefore it was calculated that $1.2106{\text{×10}}^{\text{-1}}\text{F}$ of electricity was need to produce 0.076L of chlorine with pressure 755mm Hg.

Conclusion

The amount of electricity needed to produce 1.5 L of chlorine with pressure of 750 mmHg was determined to be $1.2106{\text{×10}}^{\text{-1}}\text{F}$.

Interpretation Introduction

Interpretation:

Need to calculate the Faraday of electricity needed for the production of 6g of Sn though electrolysis of molten SnCl₂.

Concept introduction:

Electrolysis of molten stannous chloride will result in the formation of Tin (Sn) and chlorine gas and half-cell reaction at the anode and cathode was given below. In this two electrons were released by chloride ion at the anode and get liberated as chlorine gas, further stannous ion accept two electron and for tin metal

$\begin{array}{l}\text{Anode (oxidation)}{\text{2Cl}}^{{}^{\text{\_}}}{}_{\text{(l)}}\stackrel{}{\to }{\text{Cl}}_{\text{2(g)}}{\text{+2e}}^{\text{-}}\\ {\text{Cathode (Reduction) Sn}}^{\text{+}}{}_{\text{(l)}}{\text{+ 2e}}^{\text{-}}\stackrel{}{\to }{\text{Sn}}_{\text{(s)}}\\ \text{Overallreaction}{\text{Sn}}^{\text{+}}{}_{\text{(l)}}\text{+}{\text{2Cl}}^{{}^{\text{\_}}}{}_{\text{(l)}}\stackrel{}{\to }{\text{Sn}}_{\text{(s)}}{\text{+Cl}}_{\text{2(g)}}\end{array}$

The ideal gas equation can be given as follows.

Since the mass of the metal produced was given, from that number of moles of Sn can be calculated. Further from the number of moles of Sn produced the Faraday of electricity can be calculated by the formula given below, since one Faraday is equal to one mole of electron. In the present case 2 moles of electrons are needed to reduce one mole of Sn²⁺

$\text{Number}\text{of}\text{moles (n)}\text{=}\frac{\text{Weight}}{\text{Atomic}\text{Mass}}$

So

$\text{Faraday=}\text{n×}\frac{\text{2F}}{\text{1}\text{mole}\text{Sn}}$

To find: Faraday of electricity need to produce 6 g of Tin, by electrolysis of molten SnCl₂.

Answer to Problem 18.50QP

From the mass of tin produced during the electrolysis, the Faraday of electricity needed for reaction can be calculated in the following steps (c).

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{=}\frac{\text{Weight}}{\text{Atomic}\text{Mass}}\\ \text{=}\frac{\text{6}\text{.0}}{\text{118}\text{.7}}\\ \text{=}\text{5}{\text{.0548×10}}^{\text{-2}}\text{mole}\text{Sn}\end{array}$

Since one mole of Sn²⁺ ion need two Faraday of electricity to, so the Faraday of electricity needed to produce $\text{5}{\text{.0548×10}}^{\text{-2}}\text{mole}\text{Sn}$ can be calculated according to given formula.

$\begin{array}{l}\text{Faraday of electricity}\text{=5}{\text{.0548×10}}^{\text{-2}}\text{mole}\text{Sn×}\frac{\text{2F}}{\text{1mole}\text{Sn}}\\ =1.011{\text{×10}}^{\text{-1}}\text{F}\end{array}$

Explanation of Solution

From the molar mass calculation

$\text{Number}\text{of}\text{moles}\text{=}\frac{\text{Weight}}{\text{Atomic}\text{Mass}}$

Weight = 6g

Atomic Mass = 118.7g

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{=}\frac{\text{6}\text{.0}}{\text{118}\text{.7}}\\ \text{=}\text{5}{\text{.0548×10}}^{\text{-2}}\text{mole}\text{Sn}\end{array}$

Since one mole of Sn²⁺ ion need two Faraday of electricity, so the Faraday of electricity needed to produce $\text{5}{\text{.0548×10}}^{\text{-2}}\text{mole}\text{Sn}$ can be calculated according to given formula.

$\begin{array}{l}\text{Faraday=}\text{n×}\frac{\text{2F}}{\text{1}\text{mole}\text{Sn}}\\ \text{=5}{\text{.0548×10}}^{\text{-2}}\text{mole}\text{Sn×}\frac{\text{2F}}{\text{1mole}\text{Sn}}\\ =1.011{\text{×10}}^{\text{-1}}\text{F}\end{array}$

The number of moles of tin produced through electrolysis was calculated as $\text{5}{\text{.0548×10}}^{\text{-2}}\text{mole}$. From the stoichiometry of the reaction it was found that two mole of electron was needed to produce one mole of tin, since one mole of electron is equal to one Faraday, so it was under stood that two Faraday of electricity will be needed to produce one mole of tin. Therefore it was calculated that $1.011{\text{×10}}^{\text{-1}}\text{F}$ of electricity will be needed to produce 6g of tin.

Conclusion

Conclusion

The Faraday of electricity need to produce 6g of Tin from by electrolysis of molten SnCl₂ was identified as $1.011{\text{×10}}^{\text{-1}}\text{F}$. The detailed steps involved in the calculation were well explained.