What would be the electrode potential for the given half-cell reaction at $p H = 5 ?_________.$
$2 H_{2} O \to O_{2} + 4 \overset{\oplus}{H} + 4 e^{-}; E_{r e d}^{0} = 1.23 V$
$(R = 8.314 J {m o l}^{- 1} K^{- 1}; T e m p = 298 K; o x y g e n u n d e r s \tan dard . a t m . p r e s s u r e o f 1 b a r)$

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TestHub · Accepted Answer

On applyinf Nernst equation- E = E ° - 0 . 0591 n logQ E = - 1 .23 - 0 .0591 4 log [ H + ] 4 = 1.23 + 0.0591 × p H = - 1.23 + 0.0591 × 5 = - 1.23 + 0.2955 = - 0.9345 V = - 0.93 V 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.

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