What would be the electrode potential for the given half cell reaction at pH = 5 ?
`2H_(2)Oto O_(2) + 4H^(+) + 4e^(-) , E_(red)^(@) = 1.23V`
(`R=8.314Jmol^(-1)K^(-1) Temp = 298K, ` oxygen under std. atm. Pressure of 1 bar )

Explanation

To find the electrode potential for the given half-cell reaction at pH = 5, we can follow these steps:

### Step 1: Understand the Reaction
The half-cell reaction provided is:
\[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]
The standard reduction potential (\(E_{red}^\circ\)) for this reaction is given as \(1.23 \, V\). However, since this reaction is an oxidation reaction (loss of electrons), we need to reverse the sign for the oxidation potential.

### Step 2: Calculate the Concentration of \(H^+\)
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