Question

$XO_2^- \rightleftharpoons XO^- \quad E^° = p + q \text{ volts}$

$XO^- \rightleftharpoons \frac{1}{2}X_2 \quad E^° = p - 2q \text{ volts}$

(Here p, q are positive values)

If $XO_2^-, XO^-$ and $X_2$ are are placed together, then

• A. Concentration of $XO_2^-$ decreases
• B. Concentration of $X_2$ decreases
• C. Mass of $X_2$ will not change
• D. Concentration of $XO^-$ decreases

Answer 1

Answer: (A), (B)

A. Concentration of $XO_2^-$ decreases

Answer 2

The given standard electrode potentials are $E(XO_2^-/XO^-) = p+q$ and $E(XO^-/X_2) = p-2q$. Since $p, q > 0$, we have $p+q > p-2q$, meaning $E(XO_2^-/XO^-) > E(XO^-/X_2)$. This indicates that $XO_2^-$ is a stronger oxidizing agent than $XO^-$ (meaning $XO_2^-$ readily gets reduced to $XO^-$), and $X_2$ is a stronger reducing agent than $XO^-$ (meaning $X_2$ readily gets oxidized to $XO^-$). When $XO_2^-$, $XO^-$, and $X_2$ are placed together, the strongest oxidizing agent ($XO_2^-$) will react with the strongest reducing agent ($X_2$). The spontaneous reaction will be $XO_2^- + X_2 \rightarrow XO^-$ (after balancing). In this reaction, $XO_2^-$ is a reactant, so its concentration will decrease. $X_2$ is also a reactant, so its concentration will decrease. $XO^-$ is a product, so its concentration will increase. Therefore, options A and B are correct. In a single-choice question format, if multiple options are correct, it implies a flaw in the question. However, if a single answer must be chosen, A is a correct statement.