Question

At what pH, oxidation potential of water is \[-0.81V\]?

Answer 1

Hint:At first write the half reaction of water, and write the standard oxidation potential of that half reaction.
Then we will use the nest equation to find out the concentration of H+ ions.
Then we will use that value to calculate the pH of the solution.

Complete step by step answer:
In the field of electrochemistry, the Nernst equation is an equation which relates the reduction potential of a reaction which is electrochemical in nature (half-cell or full cell reaction depending on the conditions) to the standard electrode potential, activities and temperature, (often approximated by concentrations) of the chemical species undergoing reduction and oxidation.
For a complete electrochemical reaction (full cell), the equation can be written as
\[{{E}_{cell}}=E{{{}^\circ }_{red}}-\dfrac{RT}{zF}lnQr=E{{{}^\circ }_{red}}-\dfrac{RT}{zF}ln\dfrac{{{a}_{red}}}{{{a}_{ox}}}\]
Which gives \[{{E}_{cell}}=E{{{}^\circ }_{cell}}-\dfrac{0.059}{z}logQ\]
Here, half-cell reduction potential is denoted by \[{{E}_{red}}\]at the temperature under observation,
Standard half-cell reduction potential is denoted by \[E{{{}^\circ }_{red}}\],
Cell potential also called the emf of the cell is denoted by\[~{{E}_{cell}}\] at the same temperature of consideration
Standard cell potential is denoted by \[E{{{}^\circ }_{cell}}\],
Universal gas constant is denoted by \[R\] which has the value:\[R\text{ }=\text{ }8.31446261815324\text{ }J\text{ }{{K}^{-1}}mo{{l}^{-1}}\],
Temperature is denoted by \[T\] which is in kelvins,
The number of electrons transferred in the half-reaction or full cell reaction is denoted by \[~z\] is,
Faraday constant is denoted by \[F\], it can be defined as the number of coulombs present per mole of electrons which has the value: \[F\text{ }=\text{ }96485.3321233100184\text{ }C\text{ }mo{{l}^{-1}}\],
Or in other words it is the reaction quotient of the cell reaction, and finally
The chemical activity for the significant species under observation is denoted by a is, where activity of the reduced form is \[{{a}_{Red}}\] and activity of the oxidized form is \[{{a}_{Ox.}}\]
The modified version of the equation is used to solve this equation, as the value of constants are known to us already, and since the hydrogen is getting oxidised in the reaction, we need to find out the only concentration of proton in the solution.
So in the question, oxidation potential of water is \[-0.81V\]
The half reaction for the oxidation of water can be written as,
\[2{{H}_{2}}O\to {{O}_{2}}+4{{H}^{+}}+4{{e}^{-}}\]
The value of standard electrode potential is \[-1.23~V\]
Now substituting the values of \[E{{{}^\circ }_{cell}}\] and \[~{{E}_{cell}}\]we get,
\[-0.81=-1.23-\dfrac{0.059}{4}log{{[{{H}^{+}}]}^{4}}\]
Hence the concentration of proton is,
 \[[{{H}^{+}}]=1\times {{10}^{-7}}M\]
Sp the pH of the solution is
\[-log\left[ {{H}^{+}} \right]=-\left( log\text{ }1\times {{10}^{-7}} \right)=7\]
So the pH is \[7\].


Note:
The pH and pOH of a water solution at \[25{}^\circ C\] are related by the following equation.\[pH\text{ }+\text{ }pOH\text{ }=\text{ }14\]
If either the pH or the pOH of a solution is known, the other can be quickly calculated.