Animated Solution for Chemistry - Electrochemistry: What would be the electrode potential for the given half-cell reaction at $\text{pH}=5$ ? …… . $2{\text{H}}_2\text{O}\to {\text{O}}_2+4{\text{H}}^{\oplus }+4{\text{e}}^{-}$; $E_{\text{red}}^{\circ }=1.23\text{ V}$ ($R=8.314{\text{ J mol}}^{-1}{\text{ K}}^{-1}$; Temp $=298\text{ K}$; oxygen under std. atm. pressure of $1\text{ bar}$)

Imagine a half-cell where water is being oxidized to oxygen gas at a platinum electrode. The pH of the water is 5, and the oxygen gas is bubbling out at a standard pressure of 1 bar. We are tasked with finding the electrode potential for this half-cell reaction.

The problem provides us with the oxidation half-reaction:

However, there is a massive conceptual trap here! According to the strict IUPAC conventions, the term 'electrode potential' always refers to the reduction potential, unless 'oxidation potential' is explicitly requested. Therefore, to find the correct electrode potential, we must first reverse the given reaction to represent a reduction process:

For this reduction reaction, the standard reduction potential is given as $E_{\text{red}}^{\circ }=1.23\text{ V}$.

To find the electrode potential at non-standard conditions (since the pH is 5, not 0), we deploy the Nernst equation:

Let's identify the components of our reaction quotient, $Q$. It is the ratio of the concentration of products to reactants. For our reduction reaction, water is a pure liquid, so its activity is taken as 1. The reactants are oxygen gas and hydrogen ions. By looking at the balanced equation, we can clearly see that 4 electrons are being transferred, so $n=4$.

We are given that the pH is 5. The pH is defined as the negative base-10 logarithm of the hydrogen ion concentration. This means the concentration of hydrogen ions is:

The pressure of oxygen is standard, $p_{{\text{O}}_2}=1\text{ bar}$. Let's substitute these values into our Nernst equation:

Now, let's simplify the denominator. Ten to the power of minus five, raised to the power of four, gives us $10^{-20}$. Bringing it to the numerator, it becomes $10^{20}$.

Using the property of logarithms, $\log (10^{20})$ is simply 20. So, we multiply our fraction by 20. Four goes into twenty exactly five times.

Multiplying $0.0591$ by 5 gives $0.2955$. Subtracting this from $1.23$, we get:

Rounding this off to two decimal places, we arrive at our final answer:

Notice how some reference solutions might calculate the oxidation potential ($E_{\text{oxi}}$) as $-0.94\text{ V}$. But since the official answer key states positive $0.94\text{ V}$, it confirms the universal rule: 'Electrode potential' always implies reduction potential unless explicitly stated otherwise. Always keep this convention in mind to avoid losing easy marks!