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d-and f-Block Elements - Potassium Dichromate and Potassium Permanganate

Grade 12CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Potassium Dichromate (K2Cr2O7K_2Cr_2O_7) is prepared from chromite ore (FeCr2O4FeCr_2O_4) via sodium chromate (Na2CrO4Na_2CrO_4) and sodium dichromate (Na2Cr2O7Na_2Cr_2O_7).

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In aqueous solution, chromate (CrO42−CrO_4^{2-}) and dichromate (Cr2O72−Cr_2O_7^{2-}) ions are interconvertible depending on the pH: 2CrO42−+2H+⇌Cr2O72−+H2O2CrO_4^{2-} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} + H_2O. Chromate is yellow (stable in alkaline pH) and dichromate is orange (stable in acidic pH).

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Potassium Dichromate acts as a powerful oxidizing agent in acidic medium, where chromium is reduced from +6+6 to +3+3 oxidation state.

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Potassium Permanganate (KMnO4KMnO_4) is prepared from pyrolusite ore (MnO2MnO_2) by fusion with alkali metal hydroxide and an oxidizing agent to form K2MnO4K_2MnO_4 (green), which then disproportionates in acidic/neutral solution to KMnO4KMnO_4 (purple).

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The manganate (MnO42−MnO_4^{2-}) ion is paramagnetic with one unpaired electron, while the permanganate (MnO4−MnO_4^-) ion is diamagnetic. Both have tetrahedral geometry.

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Potassium Permanganate is a strong oxidizing agent in acidic, neutral, and alkaline media. Its equivalent weight varies depending on the medium because the number of electrons transferred changes.

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In acidic medium, MnO4−MnO_4^- is reduced to Mn2+Mn^{2+}. In neutral or weakly alkaline medium, it is reduced to MnO2MnO_2.

📐Formulae

Cr2O72−+14H++6e−→2Cr3++7H2O(E∘=+1.33V)Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \quad (E^\circ = +1.33V)

MnO4−+8H++5e−→Mn2++4H2O(E∘=+1.51V)MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O \quad (E^\circ = +1.51V)

MnO4−+2H2O+3e−→MnO2+4OH−(Neutral/Alkaline)MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^- \quad (\text{Neutral/Alkaline})

2MnO4−+16H++10I−→2Mn2++8H2O+5I2(Oxidation of Iodide)2MnO_4^- + 16H^+ + 10I^- \rightarrow 2Mn^{2+} + 8H_2O + 5I_2 \quad (\text{Oxidation of Iodide})

Cr2O72−+6Fe2++14H+→2Cr3++6Fe3++7H2O(Oxidation of Ferrous)Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O \quad (\text{Oxidation of Ferrous})

💡Examples

Problem 1:

Explain the action of acidified K2Cr2O7K_2Cr_2O_7 on KIKI solution and write the balanced ionic equation.

Solution:

Cr2O72−+14H++6I−→2Cr3++7H2O+3I2Cr_2O_7^{2-} + 14H^+ + 6I^- \rightarrow 2Cr^{3+} + 7H_2O + 3I_2

Explanation:

In acidic medium, the dichromate ion acts as an oxidizing agent (Cr+6Cr^{+6} is reduced to Cr+3Cr^{+3}). It oxidizes iodide ions (I−I^-) to molecular iodine (I2I_2).

Problem 2:

How does KMnO4KMnO_4 react with H2SH_2S in acidic medium?

Solution:

2MnO4−+16H++5H2S→2Mn2++8H2O+5S2MnO_4^- + 16H^+ + 5H_2S \rightarrow 2Mn^{2+} + 8H_2O + 5S

Explanation:

Acidified potassium permanganate oxidizes hydrogen sulfide (H2SH_2S) to elemental sulfur (SS). The Mn7+Mn^{7+} in MnO4−MnO_4^- is reduced to Mn2+Mn^{2+}, which is nearly colorless in dilute solutions.

Problem 3:

What happens when KMnO4KMnO_4 reacts with I−I^- in a neutral or weakly alkaline medium?

Solution:

2MnO4−+H2O+I−→2MnO2+2OH−+IO3−2MnO_4^- + H_2O + I^- \rightarrow 2MnO_2 + 2OH^- + IO_3^-

Explanation:

In neutral or weakly alkaline solutions, permanganate oxidizes iodide (I−I^-) to iodate (IO3−IO_3^-), while the permanganate itself is reduced to manganese dioxide (MnO2MnO_2).