The standard reduction potential of the reaction, H2 O + e- arrow (1/2) H2 + OH at 298 K is :

Question

The standard reduction potential of the reaction, H2 O + e- arrow (1/2) H2 + OH at 298 K is : (A) E° = (RT/F) ln Kw (B) E° = (RT/F) ln [PH2]1/2 [OH-] (C) E° = (RT/F) ln ([PH2]1/2/ [H+]) (D) E°=-(R T/F) ln Kw

Tardigrade Admin · Accepted Answer

E text cell =0 at eq m 0=E text cell °+(R T/F) ln [ P H2]1 / 2[ OH -] E text cell °=-(R T/F) ln [ P H2]1 / 2[ OH -]

Q. The standard reduction potential of the reaction, $H_{2} O + e^{-} \to \frac{1}{2} H_{2} + O H$ at $298 K$ is :

$E^{\circ} = \frac{RT}{F} l n K_{w}$

$E^{\circ} = \frac{RT}{F} l n [P_{H_{2}}]^{1/2} [O H^{-}]$

$E^{\circ} = \frac{RT}{F} l n \frac{[ P _{H_{2}} ] ^{1/2}}{[ H ^{+} ]}$

$E^{\circ} = - \frac{RT}{F} ln K_{w}$

$E_{cell} = 0$ at eq $^{m}$
$0 = E_{cell}^{\circ} + \frac{RT}{F} ln [P_{H_{2}}]^{1/2} [O H^{-}]$
$E_{cell}^{\circ} = - \frac{RT}{F} ln [P_{H_{2}}]^{1/2} [O H^{-}]$

Q. The standard reduction potential of the reaction, H2​O+e−→21​H2​+OH at 298K is :

E∘=FRT​lnKw​

E∘=FRT​ln[PH2​​]1/2[OH−]

E∘=FRT​ln[H+][PH2​​]1/2​

E∘=−FRT​lnKw​