Identify the correct statements with reference to the given reaction
$$\mathrm{P}_4+3 \mathrm{OH}^{-}+3 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{PH}_3+3 \mathrm{H}_2 \mathrm{PO}_2^{-}$$
Which of the following electrodes will act as anodes, which connected to Standard Hydrogen Electrode ?
The reaction $\mathrm{Cl}_2(g)+2 \mathrm{OH}^{-}(a q) \rightarrow \mathrm{ClO}^{-}(a q)+\mathrm{Cl}^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l)$ represents the process of bleaching. Identify and name the species that bleaches the substances due to its oxidising action.
$$\mathop {C{l_2}}\limits^0 (g) + \mathop {2O{H^ - }}\limits^{ - 2\,\,\, + 1} (aq) \to \mathop {Cl{O^ - }}\limits^{ + 1\,\,\, - 2} (aq) + \mathop {C{l^ - }}\limits^{ - 1} (aq) + \mathop {{H_2}O}\limits^{ + 1\,\,\,\, - 2} (l)$$
In this reaction, $\mathrm{O} . \mathrm{N}$. of Cl increases from $0\left(\right.$ in $\left.\mathrm{Cl}_2\right)$ to 1 (in $\left.\mathrm{ClO}^{-}\right)$as well as decreases from 0 (in $\mathrm{Cl}_2$ ) to -1 (in $\mathrm{Cl}^{-}$). So, it acts both reducing as well as oxidising agent. This is an example of disproportionation reaction. In this reaction, $\mathrm{ClO}^{-}$species bleaches the substances due to its oxidising action. [In hypochlorite ion $\left(\mathrm{ClO}^{-}\right) \mathrm{Cl}$ can decrease its oxidation number from +1 to 0 or -1 .]
$\mathrm{MnO}_4^{2-}$ undergoes disproportionation reaction in acidic medium but $\mathrm{MnO}_4^{-}$does not. Give reason.
In $\mathrm{MnO}_4^{2-}$, the oxidation number of Mn is +6 . It can increase its oxidation number (to +7 ) or decrease its oxidation number (to $+4,+3,+2,0$ ). Hence, it undergoes disproportionation reaction in acidic medium.
In $\mathrm{MnO}_4^{-}, \mathrm{Mn}$ is in its highest oxidation state, i.e., +7 . It can only decrease its oxidation number. Hence, it cannot undergo disproportionation reaction.
PbO and $\mathrm{PbO}_2$ react with HCl according to following chemical equations
$$\begin{aligned} & 2 \mathrm{PbO}+4 \mathrm{HCl} \longrightarrow 2 \mathrm{PbCl}_2+2 \mathrm{H}_2 \mathrm{O} \\ & \mathrm{PbO}_2+4 \mathrm{HCl} \longrightarrow \mathrm{PbCl}_2+\mathrm{Cl}_2+2 \mathrm{H}_2 \mathrm{O} \end{aligned}$$
Why do these compounds differ in their reactivity ?
Writing the oxidation number of each element above its symbol in the following reactions
(a) $$\mathop {2\mathop {PbO}\limits^{ + 2\,\,\,\,\, - 2} }\limits_{Basic\,oxide} + \mathop {4\mathop {HCl}\limits^{ + 1\,\,\,\,\, - 1} }\limits_{Acid} \buildrel {} \over \longrightarrow \mathop {2\mathop {PbC{l_2}}\limits^{ + 2\,\,\,\,\, - 1} }\limits_{} + \mathop {2\mathop {{H_2}O}\limits^{ + 1\,\,\,\,\, - 2} }\limits_{} $$
In this reaction, oxidation number of each element remains same hence, it is not a redox reaction. In fact, it is an example of acid-base reaction.
(b) $$\mathop {\mathop {Pb{O_2}}\limits^{ + 4\,\,\,\,\, - 2} }\limits_{} + \mathop {4\mathop {HCl}\limits^{ + 1\,\,\,\,\, - 1} }\limits_{} \buildrel {} \over \longrightarrow \mathop {\mathop {PbC{l_2}}\limits^{ + 2\,\,\,\,\, - 1} }\limits_{} + \mathop {\mathop {C{l_2}}\limits^0 + 2\mathop {{H_2}O}\limits^{ + 1\,\,\,\,\, - 2} }\limits_{} $$
In $\mathrm{PbO}_2, \mathrm{~Pb}$ is in +4 oxidation state. Due to inert pair effect Pb in +2 oxidation state is more stable. $\mathrm{So}, \mathrm{Pb}$ in +4 oxidation state $\left(\mathrm{PbO}_2\right)$ acts as an oxidising agent. It oxidises $\mathrm{Cl}^{-}$to $\mathrm{Cl}_2$ and itself gets reduced to $\mathrm{Pb}^{2+}.$
