Question

Calculate the reversible potential of oxygen electrode in a solution of \[\text{ pH = 1 }\], when the partial pressure of ${{\text{O}}_{\text{2}}}$ is \[\text{1}{{\text{0}}^{\text{-2}}}\text{atm}\] [Assume\[\text{E}_{{{\text{O}}_{\text{2}}}\text{/}{{\text{H}}_{\text{2}}}\text{O}}^{\text{0}}\text{= 1}\text{.23V}\] SHE at 1 atm and \[\text{2}{{\text{5}}^{\text{0}}}\text{C}\]
(A)\[\text{ 1}\text{.19 V }\]
(B)\[\text{ 1}\text{.71 V }\]
(C)\[\text{ 1}\text{.97 V }\]
(D)\[\text{ 1}\text{.112 V }\]

Answer 1

Hint: The hydrogen ion concentration is found to be related to the voltage of the hydrogen electrode. The concentration is expressed in the form of pH and pH is related to the electrode potential by the Nernst equation.

Complete step by step answer:
The pH or pOH of the solution is related to the hydrogen or hydroxide ion concentration. The pH is generally represented as the hydrogen potential.
$\text{pH=-log}\left[ {{\text{H}}^{\text{+}}} \right]$
But the pH and pOH are related as,
$\text{14 = pH + pOH}$
The Nernst equation can be modified for the concentration of ions in the solution and with the partial pressure of the gas at the electrode. The Nernst equation is
\[{{\text{E}}_{\text{Cell}}}\text{=E}_{\text{cell}}^{\text{0}}\text{-}\dfrac{\text{0}\text{.0591}}{\text{n}}\text{lo}{{\text{g}}_{\text{10}}}\dfrac{\text{Concentration of ion}}{\text{partial pressure at electrode}}\]
We are given the following data,
The pH of the solution is 1.0
The partial pressure of oxygen at the electrode (${{p}_{{{o}_{2}}}}$) is \[\text{1}{{\text{0}}^{\text{-2}}}\text{atm}\]
The standard electrode potential $\text{E}_{{{\text{O}}_{\text{2}}}\text{/}{{\text{H}}_{\text{2}}}\text{O}}^{\text{0}}$ is given as 1.23V at 1 atm and ${{25}^{0}}\text{C}$
The reaction at the oxygen electrode between the ${{\text{O}}_{\text{2}}}$ and ${{\text{H}}_{\text{2}}}\text{O}$is,
\[{{\text{O}}_{\text{2}}}\text{+2}{{\text{H}}_{\text{2}}}\text{O+4}{{\text{e}}^{\text{-}}}\to \text{4O}{{\text{H}}^{\text{-}}}\]
The pH of the solution is 1.
Since we know that,
$\text{pOH = 14}-\text{pH=14}-\text{1=13}$
Since, $\text{pOH=-log}\left[ \text{O}{{\text{H}}^{-}} \right]$
Therefore, $\left[ \text{O}{{\text{H}}^{-}} \right]={{10}^{\text{-poH}}}$
Here, $\left[ \text{O}{{\text{H}}^{-}} \right]={{10}^{\text{-13}}}\text{M}$
Let’s use the Nernst equation. The Nernst equation for partial pressure of oxygen and concentration of hydroxide can be written as,
\[{{\text{E}}_{{{\text{O}}_{\text{2}}}\text{/}{{\text{H}}_{\text{2}}}\text{O}}}\text{=E}_{{{\text{O}}_{\text{2}}}\text{/}{{\text{H}}_{\text{2}}}\text{O}}^{\text{0}}\text{-}\dfrac{\text{0}\text{.0591}}{\text{n}}\text{lo}{{\text{g}}_{\text{10}}}\dfrac{\left[ \text{concentration of hydroxide} \right]}{\text{partial pressure of oxygen}}\]
Let's substitute the values.
\[{{\text{E}}_{{{\text{O}}_{\text{2}}}\text{/}{{\text{H}}_{\text{2}}}\text{O}}}\text{=E}_{{{\text{O}}_{\text{2}}}\text{/}{{\text{H}}_{\text{2}}}\text{O}}^{\text{0}}\text{-}\dfrac{\text{0}\text{.0591}}{\text{4}}\text{lo}{{\text{g}}_{\text{10}}}\dfrac{{{\left[ \text{O}{{\text{H}}^{-}} \right]}^{4}}}{p{{o}_{2}}}\]
Since no.of electrons involved in the reaction is 4.
\[\begin{align}
& {{\text{E}}_{{{\text{O}}_{\text{2}}}\text{/}{{\text{H}}_{\text{2}}}\text{O}}}\text{= 1}\text{.23}-\dfrac{\text{0}\text{.0591}}{\text{4}}\text{lo}{{\text{g}}_{\text{10}}}\dfrac{{{({{10}^{-13}})}^{4}}}{{{10}^{-2}}} \\
& \text{ = 1}\text{.23}-\dfrac{\text{0}\text{.0591}}{\text{4}}\text{lo}{{\text{g}}_{\text{10}}}{{10}^{-50}} \\
 & \text{ = 1}\text{.23}-\dfrac{\text{0}\text{.0591}}{\text{4}}\times (-50) \\
 & \text{ = 1}\text{.23+}\dfrac{2.95}{4} \\
 & \text{ = 1}\text{.23+0}\text{.7375} \\
& \therefore {{\text{E}}_{{{\text{O}}_{\text{2}}}\text{/}{{\text{H}}_{\text{2}}}\text{O}}}=1.967\simeq 1.97\text{ } \\
\end{align}\]
Therefore , the reversible potential of oxygen is equal to \[1.97\text{ V}\].

Hence, (C) is the correct option.

Note: The $\dfrac{\text{0}\text{.0591}}{\text{n}}$value in the Nernst equation is obtained from the Nernst equation apply to the system at the standard condition. It is equal to the $\dfrac{\text{2}\text{.303 }\!\!\times\!\!\text{ R }\!\!\times\!\!\text{ T}}{\text{nF}}$. Here, R is the gas constant is room temperature $298K$ and F is Faraday's constant $\text{ }96500\text{C }$.