What would be the electrode potential for the given half-cell reaction at $pH = 5 ? _ _ _ _ _ _ _ _ _ .<br />$ 2{H}_{2}O \to {O}_{2} + 4\overset{⊕ $}{\text{H}}$ + 4{e}^{-}; {E}_{red}^{0} = 1.23V
\left(R = 8.314J{mol}^{- 1}{K}^{- 1}; Temp = 298K; oxygenunderstandard . atm . pressureof1bar\right)

Question

What would be the electrode potential for the given half-cell reaction at

pH = 5 ? _ _ _ _ _ _ _ _ _ .<br />

2{H}_{2}O \to {O}_{2} + 4\overset{⊕

}{\text{H}}

+ 4{e}^{-}; {E}_{red}^{0} = 1.23V
\left(R = 8.314J{mol}^{- 1}{K}^{- 1}; Temp = 298K; oxygenunderstandard . atm . pressureof1bar\right)

Quantrex Academy · Accepted Answer

Correct answer: 0.93

On applyinf Nernst equation- E = E ° - (0 . 0591)/(n)logQE = - 1 .23 - (0 .0591)/(4)log [ H ^ + ] ^ 4 = 1.23 + 0.0591 × pH = - 1.23 + 0.0591 × 5 = - 1.23 + 0.2955 = - 0.9345V = - 0.93V The Nernst equation defines the relationship between cell potential to standard potential and to the activities of the electrically active (electroactive) species. It relates the effective concentrations (activities) of the components of a cell reaction to the standard cell potential.

What would be the electrode potential for the given half-cell...

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