Given that,

$$\begin{array}{r} \mathrm{Cu}+\mathrm{Zn}^{2+} \longrightarrow \mathrm{Cu}^{2+}+\mathrm{Zn} \\ \mathrm{Mg}+\mathrm{Fe}^{2+} \longrightarrow \mathrm{Mg}^{2+}+\mathrm{Fe} \\ \mathrm{Br}_2+2 \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_2+2 \mathrm{Br} \\ \mathrm{Fe}+\mathrm{Cd}^{2+} \longrightarrow \mathrm{Cd}+\mathrm{Fe}^{2+} \end{array}$$

(a) $\mathrm{Cu}^{2+} / \mathrm{Cu}$ and $\mathrm{Zn}^{2+} / \mathrm{Zn}$

(b) $\mathrm{Mg}^{2+} / \mathrm{Mg}$ and $\mathrm{Fe}^{2+} / \mathrm{Fe}$

(c) $\mathrm{Br}_2 / \mathrm{Br}^{-}$and $\mathrm{Cl}_2 / \mathrm{Cl}^{-}$

(d) $\mathrm{Fe}^{2+} / \mathrm{Fe}$ and $\mathrm{Cd}^{2+} / \mathrm{Cd}$

Suppose that the oxidation number of chlorine in these compounds be x.

$$ \begin{array}{lll} \text { O.N. of } \mathrm{Cl} \text { in } \mathrm{NaClO}_4 & \therefore & +1+x+4(-2)=0 \text { or, } x=+7 \\ \text { O.N. of } \mathrm{Cl} \text { in } \mathrm{NaClO}_3 & \therefore & +1+x+3(-2)=0 \text { or, } x=+5 \\ \text { O.N. of } \mathrm{Cl} \text { in } \mathrm{NaClO}_2 & \therefore & +1+x+1(-2)=0 \text { or, } x=+1 \\ \text { O.N. of } \mathrm{Cl} \text { in } \mathrm{KClO}_2 & \therefore & +1+x+2(-2)=0 \text { or, } x=+3 \\ \text { O.N. of } \mathrm{Cl} \text { in } \mathrm{Cl}_2 \mathrm{O}_7 & \therefore & +2 x+7(-2)=0 \text { or, } x=+7 \\ \text { O.N. of } \mathrm{Cl} \text { in } \mathrm{ClO}_3 & \therefore & x+3(-2)=0 \text { or, } x=+6 \\ \text { O.N. of } \mathrm{Cl} \text { in } \mathrm{Cl}_2 \mathrm{O} & \therefore & 2 x+1(-2)=0 \text { or, } x=+1 \\ \text { O. } \mathrm{N} \text {. of } \mathrm{Cl} \text { in } \mathrm{NaCl}_2 & \therefore & +1+x=0 \text { or, } x=-1 \\ \text { O. N. of } \mathrm{Cl} \text { in } \mathrm{Cl}_2 & \therefore & 2 x=0 \text { or, } x=0 \\ \text { O. } \mathrm{N} \text {. of } \mathrm{Cl} \text { in } \mathrm{ClO}_2 & \therefore & x+2(-2)=0 \text { or, } x=+4 \end{array}$$

None of these compounds have an oxidation number of +2.

Increasing order of oxidation number of chlorine is : $-1,0,+1,+3,+4,+5,+6,+7$ Therefore, the increasing order of oxidation number of Cl in compounds is

$$\mathrm{NaCl}<\mathrm{Cl}_2<\mathrm{NaClO}<\mathrm{KClO}_2<\mathrm{ClO}_2<\mathrm{NaClO}_3<\mathrm{ClO}_3<\mathrm{Cl}_2 \mathrm{O}_7$$

Measure the electrode potential of the given species by connecting the redox couple of the given species with standard hydrogen electrode. If it is positive, the electrode of the given species acts as reductant and if it is negative, it acts as an oxidant.

Find the electrode potentials of the other given species in the same way, compare the values and determine their comparative strength as an reductant or oxidant.

e.g., measurement of standard electrode potential of $\mathrm{Zn}^{2+} / \mathrm{Zn}$ electrode using SHE as a reference electrode.

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The EMF of the cell comes out to be 0.76 V . (reading of voltmeter is 0.76 V ). $\mathrm{Zn}^{2+}$ / Zn couple acts as anode and SHE acts as cathode.

$$\begin{aligned} \therefore \quad E_{\text {cell }}^{\circ} & =0.76=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ} \\ 0.76 & =0-E_{\text {anode }}^{\circ} \\ E_{\text {anode }}^{\circ} & =-0.76 \mathrm{~V} \\ E_{\mathrm{Zn^{2+}/Zn}}^{o} & =-0.76 \mathrm{~V} . \end{aligned}$$