The standard reduction potential at 25° C of the reaction, 2H2 O + 2e- leftharpoons H2 + 2OH-, is -0.8277 V. Calculate the equilibrium constant for the reaction, 2H2 O leftharpoons H3 O+ + OH- at 25° C

Question

The standard reduction potential at 25° C of the reaction, 2H2 O + 2e- leftharpoons H2 + 2OH-, is -0.8277 V. Calculate the equilibrium constant for the reaction, 2H2 O leftharpoons H3 O+ + OH- at 25° C (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/92c169e9b376a60a246738a5a052aff3-.png" /> E° = 0.0592 log K ⇒ log K = (-.08277/0.0592)=-13.98 ⇒ K = 1.04 × 10-14

Q. The standard reduction potential at $2 5^{\circ} C$ of the reaction,
$2 H_{2} O + 2 e^{-} ⇌ H_{2} + 2 O H^{-},$ is $- 0.8277 V$ . Calculate the
equilibrium constant for the reaction,
$2 H_{2} O ⇌ H_{3} O^{+} + O H^{-}$ at $2 5^{\circ} C$

$E^{\circ} = 0.0592 l o g K$
$\Rightarrow lo g K = \frac{- .08277}{0.0592} = - 13.98$
$\Rightarrow K = 1.04 \times 1 0^{- 14}$

Q. The standard reduction potential at 25∘C of the reaction, 2H2​O+2e−⇌H2​+2OH−, is −0.8277V. Calculate the equilibrium constant for the reaction, 2H2​O⇌H3​O++OH− at 25∘C

E∘=0.0592logK ⇒logK=0.0592−.08277​=−13.98 ⇒K=1.04×10−14

Q. The standard reduction potential at $2 5^{\circ} C$ of the reaction,
$2 H_{2} O + 2 e^{-} ⇌ H_{2} + 2 O H^{-},$ is $- 0.8277 V$ . Calculate the
equilibrium constant for the reaction,
$2 H_{2} O ⇌ H_{3} O^{+} + O H^{-}$ at $2 5^{\circ} C$

$E^{\circ} = 0.0592 l o g K$
$\Rightarrow lo g K = \frac{- .08277}{0.0592} = - 13.98$
$\Rightarrow K = 1.04 \times 1 0^{- 14}$