Assertion (A) The decomposition of hydrogen peroxide to form water and oxygen is an example of disproportionation reaction.
Reason (R) The oxygen of peroxide is in -1 oxidation state and it is converted to zero oxidation state in $\mathrm{O}_2$ and -2 oxidation state in $\mathrm{H}_2 \mathrm{O}$.
Assertion (A) Redox couple is the combination of oxidised and reduced form of a substance involved in an oxidation or reduction half cell.
Reason (R) In the representation $E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\ominus}$ and $E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\ominus}, \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}$ and $\mathrm{Cu}^{2+} / \mathrm{Cu}$ are redox couples.
Explain redox reaction on the basis of electron transfer. Given suitable examples.
As we know that, the reactions
$$ \begin{aligned} & 2 \mathrm{Na}(\mathrm{s})+\mathrm{Cl}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{NaCl}(\mathrm{s}) \\ & 4 \mathrm{Na}(\mathrm{s})+\mathrm{O}_2(g) \longrightarrow 2 \mathrm{Na}_2 \mathrm{O}(\mathrm{s}) \end{aligned}$$
are redox reactions because in each of these reactions sodium is oxidised due to the addition of either oxygen or more electronegative element to sodium. Simultaneously, chlorine and oxygen are reduced because of each of these, the electropositive element sodium has been added.
From our knowledge of chemical bonding we also know that, sodium chloride and sodium oxide are ionic compounds and perhaps better written as $\mathrm{Na}^{+} \mathrm{Cl}^{-}(\mathrm{s})$ and $\left(\mathrm{Na}^{+}\right)_2 \mathrm{O}^{2-}(\mathrm{s})$. Development of charges on the species produced suggests us to rewrite the above reaction in the following manner
For convenience, each of the above processes can be considered as two separate steps, one involving the loss of electrons and other the gain of electrons. As an illustration, we may further elaborate one of these, say, the formation of sodium chloride.
$$\begin{aligned} 2 \mathrm{Na}(s) & \longrightarrow 2 \mathrm{Na}^{+}(g)+2 e^{-} \\ \mathrm{Cl}_2(g)+2 \mathrm{e}^{-} & \longrightarrow 2 \mathrm{Cl}^{-}(g) \end{aligned}$$
Each of the above steps is called a half reaction, which explicitly shows involvement of electrons. Sum of the half reactions gives the overall reaction:
$$2 \mathrm{Na}(s)+\mathrm{Cl}_2(g) \longrightarrow 2 \mathrm{Na}^{+} \mathrm{Cl}^{-}(s) \text { or } 2 \mathrm{NaCl}(s)$$
The given reactions suggest that half reactions that involved loss of electrons are oxidation reactions. Similarly, the half reactions that involve gain of electrons are called reduction reactions.
It may not be out of context to mention here that the new way of defining oxidation and reduction has been achieved only by establishing a correlation between the behaviour of species as per the classical idea and their interplay in electron-transfer change. In the given reactions, sodium, which is oxidised, acts as a reducing agent because it donates electron to each of the elements interacting with it and thus helps in reducing them. Chlorine and oxygen are reduced and act as oxidising agents because these accept electrons from sodium.
To summarise, we may mention that
$$\begin{aligned} & \text { Oxidation Loss of electron(s) by any species. } \\ & \text { Reduction Gain of electron(s) by any species. } \\ & \text { Oxidising agent Acceptor of electron(s). } \\ & \text { Reducing agent Donor of electron(s). } \end{aligned}$$
On the basis of standard electrode potential values, suggest which of the following reactions would take place? (Consult the book for $E^{\ominus}$ value)
(a) $\mathrm{Cu}+\mathrm{Zn}^{2+} \longrightarrow \mathrm{Cu}^{2+}+\mathrm{Zn}$
(b) $\mathrm{Mg}+\mathrm{Fe}^{2+} \longrightarrow \mathrm{Mg}^{2+}+\mathrm{Fe}$
(c) $\mathrm{Br}_2+2 \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_2+2 \mathrm{Br}^{-}$
(d) $\mathrm{Fe}+\mathrm{Cd}^{2+} \longrightarrow \mathrm{Cd}+\mathrm{Fe}^{2+}$
As we know that,
$\begin{aligned} E_{\mathrm{Cu}^{2+}}^{\circ} / \mathrm{Cu} & =0.34 \mathrm{~V}, E_{\mathrm{Zn}}^{\circ} \mathrm{Zn}^{2+} / \mathrm{Zn}=-0.76 \mathrm{~V}, \\ E_{\mathrm{Mg}^{2+} / \mathrm{Mg}}^{\circ} & =-2.37 \mathrm{~V}, E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\circ}=-0.74 \mathrm{~V}, \\ E_{\mathrm{Br}_2 / \mathrm{Br}^{-}}^{-} & =+1.08 \mathrm{~V}, \mathrm{E}_{\mathrm{Cl}_2 / \mathrm{Cl}^{-}}^{\circ}=+1.36 \mathrm{~V} \\ E_{\mathrm{Cd}^{2+} / \mathrm{Cd}}^{\circ} & =-0.44 \mathrm{~V}\end{aligned}$
(a) $E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=+0.34 \mathrm{~V}$ and $E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V}$
$$\mathrm{Cu}+\mathrm{Zn}^{2+} \longrightarrow \mathrm{Cu}^{2+}+\mathrm{Zn}$$
In the given cell reaction, Cu is oxidised to $\mathrm{Cu}^{2+}$, therefore, $\mathrm{Cu}^{2+} / \mathrm{Cu}$ couple acts as anode and $\mathrm{Zn}^{2+}$ is reduced to Zn , therefore, $\mathrm{Zn}^{2+} / \mathrm{Zn}$ couple acts as cathode.
$$\begin{aligned} & E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ} \\ & E_{\text {cell }}^{\circ}=-0.76-(+0.34)=-1.10 \mathrm{~V} \end{aligned}$$
Negative value of $E_{\text {cell }}^{\circ}$ indicates that the reaction will not occur.
$\begin{aligned} & \text{(b)}\quad \mathrm{Mg}+\mathrm{Fe}^{2+} \longrightarrow \mathrm{Mg}^{2+}+\mathrm{Fe} \\ & E_{\mathrm{Mg}^{2+} / \mathrm{Mg}}^{\circ}=-2.37 \mathrm{~V} \text { and } E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\circ}=-0.74 \mathrm{~V}\end{aligned}$
In the given cell reaction, Mg is oxidised to $\mathrm{Mg}^{2+}$, and $\mathrm{Fe}^{2+}$ is reduced to Fe hence, $\mathrm{Fe}^{2+} / \mathrm{Fe}$ couple acts as cathode.
$$\begin{aligned} & E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ} \\ & E_{\text {cell }}^{\circ}=-0.74-(-2.37)=+1.63 \mathrm{~V} \end{aligned}$$
Positive value of $E_{\text {cell }}^{\circ} 5$ indicates that the reaction will occur.
(c) $\begin{array}{r}\mathrm{Br}_2+2 \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_2+2 \mathrm{Br}^{-} \\ E_{\mathrm{Br}^{-} / \mathrm{Br}_2}^{\circ}=+1.08 \mathrm{~V} \text { and } E_{\mathrm{Cl}^{-} / \mathrm{Cl}_2}^{\circ}=+1.36 \mathrm{~V}\end{array}$
In the given cell reaction, $\mathrm{Cl}^{-}$is oxidised to $\mathrm{Cl}_2$ hence, $\mathrm{Cl}^{-} / \mathrm{Cl}_2$ couple acts as anode and $\mathrm{Br}_2$ is reduced to $\mathrm{Br}^{-}$hence; $\mathrm{Br}^{-} / \mathrm{Br}_2$ couple acts as cathode.
$$\begin{aligned} & E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ} \\ & E_{\text {cell }}^{\circ}=+1.08-(+1.36)=-0.28 \mathrm{~V} \end{aligned}$$
Negative value of $E_{\text {cell }}^{\circ}$ indicates that the reaction will occur.
(d) $\begin{array}{r}\mathrm{Fe}+\mathrm{Cd}^{2+} \longrightarrow \mathrm{Cd}+\mathrm{Fe}^{2+} \\ E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\circ}=-0.74 \mathrm{~V} \text {. and } E_{\mathrm{Cd}^{2+} / \mathrm{Cd}}^{\circ}=-0.44 \mathrm{~V}\end{array}$
In the given cell reaction, Fe is oxidised to $\mathrm{Fe}^{2+}$ hence, $\mathrm{Fe}^{2+} / \mathrm{Fe}$ couple acts as anode and $\mathrm{Cd}^{2+}$ is reduced to Cd hence, $\mathrm{Cd}^{2+} / \mathrm{Cd}$ couple acts as cathode.
$$\begin{aligned} & E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ} \\ & E_{\text {cell }}^{\circ}=-0.44-(-0.74)=+0.30 \mathrm{~V} \end{aligned}$$
Positive value $E_{\text {cell }}^{\circ}$ indicates that the reaction will occur.
Why does fluorine not show disproportionation reaction?
In a disproportionation reaction, the same species is simultaneously oxidised as well as reduced. Therefore, for such a redox reaction to occur, the reacting species must contain an element which has atleast three oxidation states.
The element, in reacting species, is present in an intermediate state while lower and higher oxidation states are available for reduction and oxidation to occur (respectively).
Fluorine is the strongest oxidising agent. It does not show positive oxidation state. That's why fluorine does not show disproportionation reaction.