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  4. What would be the electrode potential for the given half cell reaction at pH=5? 2H2O arrow O2+4H⊕+4e-;Ered0=1.23V R = 8 . 314 J mol- 1 K- 1 ; temp =298K; oxygen under std. atm. pressure of 1 bar) (Round off answer to the magnitude of nearest integer value)

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Q. What would be the electrode potential for the given half cell reaction at $pH=5?$
$2H_{2}O \rightarrow O_{2}+4H^{\oplus}+4e^{-};E_{red}^{0}=1.23V$
$R = 8 . 314 J mol^{- 1} K^{- 1} ;$ temp $=298K;$ oxygen under std. atm. pressure of $1$ bar)
(Round off answer to the magnitude of nearest integer value)

NTA AbhyasNTA Abhyas 2022

Solution:

$2H_{2}O \rightarrow O_{2}+4H^{\oplus}+4e^{-};E_{SRP}^{0}=1.23$
$E=-1.23-\frac{0 . 059}{4}log\left(H^{+}\right)^{4}$
$E=-1.23+0.059pH$
$E=-1.23+0.059\times 5$
$E=-1.23+0.295$
$E=-0.935$