Question

The reaction quotient $(Q)$ for the reduction of $O_2$ to $H_{2}O$ in acid solution,
$O_2\left(g\right)+4H^{+}\left(aq\right)+4e^{-}\to 2H_{2}O\left(l\right)$ is
[Where , ${\alpha }_{H^{+}}$ =activity of $H^{+},p_{O_2}$ = pressure of $O_2$ when it is present in the reaction, $p^{\circ }$ = pressure of $O_2$ at standard state ]

• A. $Q=\frac{{\rho }^{\circ }}{{\alpha }_{H^{+}}^4+{\rho }_{O_2}}$
• B. $Q=\frac{{\rho }_{O_2}}{{\alpha }_{H^{+}}^4+{\rho }^{\circ }}$
• C. $Q=\frac{{\alpha }_{H^{+}}^4+{\rho }_{O_2}}{{\rho }^{\circ }}$
• D. $Q=\frac{{\alpha }_{H^{+}}^4+{\rho }^{\circ }}{{\rho }_{O_2}}$

Answer 1

Answer: (A)

The half reaction for the reduction is given as
$O_2(g)+4H^{+}(aq)+4e^{-}\to 2H_{2}O(l)$
We have, activity for $H_{2}O$ (or any pure liquid or solid) $\propto H_{2}O=1$
Activity for hydrogen ion $\left(H^{+}\right)={\alpha }_{H^{+}}$
Activity for oxygen gas $\left(O_2\right)={\alpha }_{O_2}=\frac{p_{O_2}}{p_o}$
Thus, reaction quotient $(Q)$ is given as
$Q=\frac{{\alpha }_{H_{2}O}^2}{{\alpha }_{H^{+}}^4\cdot {\alpha }_{O_2}}=\frac{1^2}{{\alpha }_{H^{+}}^4\cdot p_{O_2}{p}^o}$
$=\frac{p^o}{{\alpha }_{H^{+}}\cdot p_{O_2}}$