(a) How many minutes does it take to form 10.0 Lof ${\text{O}}_{\text{2}}$measured ata$\text{99.8 kPa}$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?

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.

• 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 electronsrole="math" localid="1663386701676" $\text{F = 96485 C/mole}$

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 –

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

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 –

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

Calculate the charge using Faraday constant –

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

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 $1\text{974 min}$.

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 –

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

Molar mass ofis, hence, the mass ofproduced is –

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

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