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NCERT Exemplar · Q36

Q.Write redox couples involved in the reactions

(i) to
(iv) given in question 34.
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A redox couple is the oxidised and reduced form of a substance in a half reaction,

written oxidised form / reduced form. For the four reactions of question 34 the

couples are: (i) Cu²⁺/Cu and Zn²⁺/Zn · (ii) Mg²⁺/Mg and Fe²⁺/Fe ·

(iii) Br₂/Br⁻ and Cl₂/Cl⁻ · (iv) Fe²⁺/Fe and Cd²⁺/Cd.

What a redox couple is

Every redox reaction is the sum of two half reactions, and each half reaction connects

two forms of one substance that differ only by electrons. That pair — written with the

oxidised form first, a slash, then the reduced form — is the redox couple. The slash

stands for the interface between the two forms (metal/ion, or the two ions in solution).

The couples in each reaction of question 34

(i) Cu+ZnX2+→CuX2++Zn\ce{Cu + Zn^{2+} -> Cu^{2+} + Zn}

Half reactions: Cu→CuX2++2 eX−\ce{Cu -> Cu^{2+} + 2e^-} and ZnX2++2 eX−→Zn\ce{Zn^{2+} + 2e^- -> Zn}.

Couples: CuX2+/Cu\ce{Cu^{2+}/Cu} and ZnX2+/Zn\ce{Zn^{2+}/Zn}.

(ii) Mg+FeX2+→MgX2++Fe\ce{Mg + Fe^{2+} -> Mg^{2+} + Fe}

Half reactions: Mg→MgX2++2 eX−\ce{Mg -> Mg^{2+} + 2e^-} and FeX2++2 eX−→Fe\ce{Fe^{2+} + 2e^- -> Fe}.

Couples: MgX2+/Mg\ce{Mg^{2+}/Mg} and FeX2+/Fe\ce{Fe^{2+}/Fe}.

(iii) BrX2+2 ClX−→ClX2+2 BrX−\ce{Br2 + 2Cl^- -> Cl2 + 2Br^-}

Half reactions: BrX2+2 eX−→2 BrX−\ce{Br2 + 2e^- -> 2Br^-} and 2 ClX−→ClX2+2 eX−\ce{2Cl^- -> Cl2 + 2e^-}.

Couples: BrX2/BrX−\ce{Br2/Br^-} and ClX2/ClX−\ce{Cl2/Cl^-}.

(iv) Fe+CdX2+→Cd+FeX2+\ce{Fe + Cd^{2+} -> Cd + Fe^{2+}}

Half reactions: Fe→FeX2++2 eX−\ce{Fe -> Fe^{2+} + 2e^-} and CdX2++2 eX−→Cd\ce{Cd^{2+} + 2e^- -> Cd}. …

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