Question

On being placed in water, sodium peroxide not only produces an alkaline solution but also some bubbles. If we assume that the peroxide ion picks up two protons from water to produce a compound that can be seen as the dibasic conjugate acid of peroxide ion and then this compound undergoes a redox disproportionation.
$Na_{2}O_{2\left(s\right)}+H_{2}O_{\left(\ell\right)} \rightarrow\left(A\right)+\left(B\right) \left(A\right) \, $ and $ \, \left(B\right)$ are

• A. $H_{2}O_{2} \,$ and $\, NaOH$
• B. $H_{2}O \,$ and $O_{2}$
• C. $NaOH$ and $ O_{2}$
• D. $ Na_{2}O$ and $NaOH$

Answer 1

Answer: (C)

The reaction proceeds $\upsilon ia$ formation of $H_{2}O_{2}$ (which is dibasic conjugate acid of peroxide ion), $H_{2}O_{2}$ then disproportionates into water and oxygen.
$Na_{2}O_{2\left(s\right)}+2 H_{2}O_{\left(\ell\right)} \rightarrow 2 NaOH_{\left(a q\right)}+H_{2}O_{2\left(aq\right)}$
$H_{2}O_{2\left(aq\right)}\rightarrow H_{2}O_{\left(\ell\right)}+\frac{1}{2}O_{2\left(g\right)}$
Thus, overall reaction is
$Na_{2}O_{2\left(s\right)}+H_{2}O_{\left(\ell\right)}\rightarrow 2 NaOH_{\left(aq\right)}+\frac{1}{2} \,O_{2\left(g\right)}$