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)

Question

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) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

2H2O arrow O2+4H⊕+4e-;ESRP0=1.23 E=-1.23-(0 . 059/4)log(H+)4 E=-1.23+0.059pH E=-1.23+0.059× 5 E=-1.23+0.295 E=-0.935

Q. What would be the electrode potential for the given half cell reaction at pH=5? 2H2​O→O2​+4H⊕+4e−;Ered0​=1.23V R=8.314Jmol−1K−1; temp =298K; oxygen under std. atm. pressure of 1 bar) (Round off answer to the magnitude of nearest integer value)

2H2​O→O2​+4H⊕+4e−;ESRP0​=1.23 E=−1.23−40.059​log(H+)4 E=−1.23+0.059pH E=−1.23+0.059×5 E=−1.23+0.295 E=−0.935

$2 H_{2} O \to O_{2} + 4 H^{\oplus} + 4 e^{-}; E_{SRP}^{0} = 1.23$
$E = - 1.23 - \frac{0.059}{4} l o g (H^{+})^{4}$
$E = - 1.23 + 0.059 p H$
$E = - 1.23 + 0.059 \times 5$
$E = - 1.23 + 0.295$
$E = - 0.935$