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Question 132 of 132

Q.(i) Complete the following equations :

(a) 2MnO4−+5SO32−+6H+→2MnO_4^- + 5SO_3^{2-} + 6H^+ \rightarrow
(b) Cr2O72−+6Fe2++14H+→Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \rightarrow
(ii) Based on the data, arrange Fe2+Fe^{2+}, Mn2+Mn^{2+} and Cr2+Cr^{2+} in the increasing order of stability of +2 oxidation state.
E°Cr3+/Cr2+=−0.4E°_{Cr^{3+}/Cr^{2+}} = -0.4 V
E°Mn3+/Mn2+=+1.5E°_{Mn^{3+}/Mn^{2+}} = +1.5 V
E°Fe3+/Fe2+=+0.8E°_{Fe^{3+}/Fe^{2+}} = +0.8 V OR Write the preparation of following :
(i) KMnO4KMnO_4 from K2MnO4K_2MnO_4
(ii) Na2CrO4Na_2CrO_4 from FeCr2O4FeCr_2O_4
(iii) Cr2O72−Cr_2O_7^{2-} from CrO42−CrO_4^{2-}
Puducherry CbseCBSE Class XII Board 2018Subjective· 3mImportance★★★★★
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(i) balanced redox equations of MnO4−MnO_4^- and Cr2O72−Cr_2O_7^{2-}; (ii) Cr2+<Fe2+<Mn2+Cr^{2+} < Fe^{2+} < Mn^{2+} in stability of +2. OR: standard preparations of KMnO4KMnO_4, Na2CrO4Na_2CrO_4 and dichromate.

Concept. Redox chemistry and oxidation-state stability of transition metals — CBSE Class-12 the-d-and-f-block-elements.

(i) Balanced equations.

  1. 2MnO4−+5SO32−+6H+→2Mn2++5SO42−+3H2O2MnO_4^- + 5SO_3^{2-} + 6H^+ \rightarrow 2Mn^{2+} + 5SO_4^{2-} + 3H_2O (Mn: +7→+2+7\rightarrow+2; S: +4→+6+4\rightarrow+6).
  2. Cr2O72−+6Fe2++14H+→2Cr3++6Fe3++7H2OCr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O (Cr: +6→+3+6\rightarrow+3; Fe: +2→+3+2\rightarrow+3).

(ii) Stability of the +2 state. A more positive E∘(M3+/M2+)E^\circ(M^{3+}/M^{2+}) means M3+M^{3+} is readily reduced to M2+M^{2+} — i.e. M2+M^{2+} is more stable (harder to oxidise). Given E∘E^\circ: Cr =−0.4=-0.4 V, Fe =+0.8=+0.8 V, Mn =+1.5=+1.5 V. Hence increasing stability of the +2 state:

Cr2+<Fe2+<Mn2+.Cr^{2+} < Fe^{2+} < Mn^{2+}.

OR — Preparations.

(i) KMnO4KMnO_4 from K2MnO4K_2MnO_4 (oxidation of the manganate, e.g. with Cl2Cl_2 or electrolytically): 2K2MnO4+Cl2→2KMnO4+2KCl2K_2MnO_4 + Cl_2 \rightarrow 2KMnO_4 + 2KCl (industrially by electrolytic oxidation of manganate). …

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