Question

A solution of NaOH is electrolyzed, generating ${\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$at the anode. The current of 0.15 A is passed through the cell for 75 minutes. Calculate the volume of (wet) oxygen generated in this period if the temperature is held at 25°C and total pressure is 0.985 atm. (hint: use the vapor pressure of water at this temperature from table 0.3)

Short answer

The volume of wetO₂(g) produced at anode is 0.0449 L.

Step-by-step solution

Cathode and Anode reactions

$\text{1} \text{M} \text{NaOH}$present in the solution dissociate as following :

$\text{NaOH}\left(\text{aq}\right)\to {\text{Na}}^{\text{+}}\left(\text{aq}\right){\text{+ OH}}^{\text{-}}\left(\text{aq}\right)$

Cathode reaction ${\text{Na}}^{\text{+}}\text{+ e}\to \text{Na}$

Anode reaction ${\text{4OH}}^{\text{-}}\text{+ 4e}\to {\text{O}}_{\text{2}}{\text{+ 2H}}_{\text{2}}\text{O + 4e}$

Number of moles of O2 produced

No. of moles of electron (n) produced on passing current for a time period (t) is given by:

$\begin{array}{rcl}\text{Q}& =& \text{It}\\ \text{nF}& =& \text{It}\\ \text{n}& =& \frac{\text{It}}{\text{F}}\end{array}$

$\begin{array}{rcl}\text{n}& =& \frac{\text{0.15A×75×60s}}{\text{96485} \text{C}}\\ & =& 7\times {10}^{-3}\text{moles}\end{array}$

According to equation 3:

No of moles of ${\text{O}}_{\text{2}}$ produced with 4 moles of electron = 1

No of moles of ${\text{O}}_{\text{2}}$produced with $7\times {10}^{-3}$ moles of electron $\frac{7\times {10}^{-3}}{4}=1.75\times {10}^{-3}$.

Calculation of partial pressure of oxygen

Total pressure (P_T) is developed because of O₂and H₂O produced at anode.

Partial pressure of O₂ = ${\text{P}}_{{\text{O}}_{\text{2}}}$

Partial pressure of H₂O =${\text{P}}_{{\text{H}}_{\text{2}}\text{O}}\text{= 0.03126} \text{atm} \text{at} {\text{25}}^{\text{o}}\text{C}$

Total pressure ${\text{=P}}_{{\text{O}}_{\text{2}}}=\text{0.985} \text{atm} \text{at} {\text{25}}^{\text{o}}\text{C}$

Therefore,

$\begin{array}{rcl}{\text{P}}_{\text{T}}& =& {\text{P}}_{{\text{H}}_{\text{2}}\text{O}}{\text{+P}}_{{\text{O}}_{\text{2}}}\\ {\text{P}}_{{\text{O}}_{\text{2}}}& =& {\text{P}}_{\text{T}}{\text{-P}}_{{\text{H}}_{\text{2}}\text{O}}\\ & =& \text{0.985} \text{atm -} \text{0.03126} \text{atm}\\ & =& \text{0.95374} \text{atm}\end{array}$

So, partial pressure of O₂ (g) is 0.95374 atm.

calculation of volume of wet O2 (g)

The volume of the gas can be calculated from the ideal gas law as follows:

$\begin{array}{rcl}\text{PV}& =& \text{nRT}\\ \text{V}& =& \frac{\text{nRT}}{\text{P}}\\ & =& \frac{\text{1.75}\times {\text{10}}^{\text{- 3}}\text{mol}\times \text{0.0826} \text{Latm/Kmol}\times \text{298K}}{\text{0.95374} \text{atm}}\\ & =& \text{0.0449} \text{L}\end{array}$

• P= pressure
• V= volume
• n = number of mole
• R = universal gas constant
• T = temperature

Hence, the volume of wet O₂ (g) produced at anode is$0.0449 \text{L}$ .