The reaction quotient (Q) for the reduction of O2 to H2O in acid solution, O2(g)+4H+(aq)+4e- arrow 2H2O(l) is [Where , αH+ =activity of H+, pO2 = pressure of O2 when it is present in the reaction, p° = pressure of O2 at standard state ]

Question

The reaction quotient (Q) for the reduction of O2 to H2O in acid solution, O2(g)+4H+(aq)+4e- arrow 2H2O(l) is [Where , αH+ =activity of H+, pO2 = pressure of O2 when it is present in the reaction, p° = pressure of O2 at standard state ] (A)  Q=(ρ°/αH+4+ρO2)  (B)  Q=(ρO2/αH+4+ρ°)  (C)  Q=(αH+4+ρO2/ρ°)  (D)  Q=(αH+4+ρ°/ρO2)

Tardigrade Admin · Accepted Answer

The half reaction for the reduction is given as

O_{2} (g) + 4 H^{+} (a q) + 4 e^{-} \to 2 H_{2} O (l)

We have, activity for

H_{2} O

(or any pure liquid or solid)

\propto H_{2} O = 1

Activity for hydrogen ion

(H^{+}) = α_{H^{+}}

Activity for oxygen gas

(O_{2}) = α_{O_{2}} = \frac{p _{O_{2}}}{p _{o}}

Thus, reaction quotient

(Q)

is given as

Q = \frac{α _{H_{2} O}^{2}}{α _{H^{+}}^{4} \cdot α _{O_{2}}} = \frac{1 ^{2}}{α _{H^{+}}^{4} \cdot p _{O_{2}} p ^{o}}

= \frac{p ^{o}}{α _{H^{+}} \cdot p _{O_{2}}}

$Q = \frac{ρ ^{\circ}}{α _{H^{+}}^{4} + ρ _{O_{2}}}$

$Q = \frac{ρ _{O_{2}}}{α _{H^{+}}^{4} + ρ ^{\circ}}$

$Q = \frac{α _{H^{+}}^{4} + ρ _{O_{2}}}{ρ ^{\circ}}$

$Q = \frac{α _{H^{+}}^{4} + ρ ^{\circ}}{ρ _{O_{2}}}$

Q. The reaction quotient (Q) for the reduction of O2​ to H2​O in acid solution, O2​(g)+4H+(aq)+4e−→2H2​O(l) is [Where , αH+​ =activity of H+,pO2​​ = pressure of O2​ when it is present in the reaction, p∘ = pressure of O2​ at standard state ]

Q=αH+4​+ρO2​​ρ∘​

Q=αH+4​+ρ∘ρO2​​​

Q=ρ∘αH+4​+ρO2​​​

Q=ρO2​​αH+4​+ρ∘​

$Q = \frac{ρ ^{\circ}}{α _{H^{+}}^{4} + ρ _{O_{2}}}$

$Q = \frac{ρ _{O_{2}}}{α _{H^{+}}^{4} + ρ ^{\circ}}$

$Q = \frac{α _{H^{+}}^{4} + ρ _{O_{2}}}{ρ ^{\circ}}$

$Q = \frac{α _{H^{+}}^{4} + ρ ^{\circ}}{ρ _{O_{2}}}$