Question

Which of the following reagent can not oxidize $H_2O(l)$ to $O_2(g)$ under standard state?

[Given: $E_{Cu^{2+}/Cu}^{\circ}$ = 0.34 volt

$E_{H_2O/O_2}^{\circ}$ = -0.83 volt

$E_{Pb^{+2}/Pb}^{\circ}$ = -0.13 volt

$E_{Cl_2/Cl^-}^{\circ}$ = 1.36 volt

$E_{H^+/H}^{\circ}$ = 0 volt

• A. $H^+(aq)$
• B. $Cu^{2+}(aq)$
• C. $Pb^{2+}(aq)$
• D. $Cl_2(aq)$

Answer 1

Answer: (A), (B), (C)

A, B, C (Multiple correct options based on calculation)

Answer 2

For a reagent to oxidize $H_2O(l)$ to $O_2(g)$, its standard reduction potential ($E_{red}^{\circ}$) must be greater than the standard reduction potential of the $O_2/H_2O$ couple. The oxidation potential of $H_2O$ to $O_2$ is given as $E_{H_2O/O_2}^{\circ}$ = -0.83 V. Therefore, the standard reduction potential for $O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l)$ is $E_{O_2/H_2O}^{\circ} = +0.83 \text{ V}$.

A reagent cannot oxidize $H_2O$ if its $E_{red}^{\circ} < +0.83 \text{ V}$.

1. For $H^+(aq)$, $E_{H^+/H_2}^{\circ}$ = 0.00 V. Since $0.00 < 0.83$, $H^+$ cannot oxidize $H_2O$.

2. For $Cu^{2+}(aq)$, $E_{Cu^{2+}/Cu}^{\circ}$ = 0.34 V. Since $0.34 < 0.83$, $Cu^{2+}$ cannot oxidize $H_2O$.

3. For $Pb^{2+}(aq)$, $E_{Pb^{2+}/Pb}^{\circ}$ = -0.13 V. Since $-0.13 < 0.83$, $Pb^{2+}$ cannot oxidize $H_2O$.

4. For $Cl_2(aq)$, $E_{Cl_2/Cl^-}^{\circ}$ = 1.36 V. Since $1.36 > 0.83$, $Cl_2$ can oxidize $H_2O$.

Thus, $H^+(aq)$, $Cu^{2+}(aq)$, and $Pb^{2+}(aq)$ cannot oxidize $H_2O(l)$ to $O_2(g)$.