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 will use the Nernst equation. The half-cell reaction is:

\[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]

The standard reduction potential (\(E^\circ\)) for this reaction is given as 1.23 V.

### Step-by-step Solution:

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