Class 12 Chemistry MCQs | Chapter 4: The d-and f-Block Elements – Part 4

Explanation: In acidic medium, the manganese in permanganate is in the very high oxidation state \(+7\). It readily gains electrons and is reduced to \(Mn^{2+}\), where manganese is in the \(+2\) state. This large decrease in oxidation state is the main reason acidic permanganate is such a strong oxidising agent. The reduction product is not \(MnO_2\) in acidic medium; that belongs to neutral or weakly alkaline conditions.

Explanation: \(\textbf{Given:}\) Permanganate ion is reduced in acidic medium to \(Mn^{2+}\) \(\textbf{Required:}\) Balanced reduction half-reaction \(\textbf{Relevant Principle:}\) A balanced half-reaction must conserve both atoms and total charge \(\textbf{Why this principle applies:}\) Reduction of \(MnO_4^-\) in acid involves balancing oxygen with \(H_2O\), hydrogen with \(H^+\), and charge with electrons \(\textbf{Identify known values:}\) Manganese changes from \(+7\) in \(MnO_4^-\) to \(+2\) in \(Mn^{2+}\), so \(5\) electrons are gained There are \(4\) oxygen atoms, so \(4H_2O\) must appear on the product side That requires \(8H^+\) on the reactant side \(\textbf{Substitution:}\) \(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\) \(\textbf{Intermediate Simplification:}\) Atoms of Mn, O, and H are balanced \(\textbf{Final Simplification:}\) Charge on left \(= -1 + 8 - 5 = +2\), charge on right \(= +2\) Both mass and charge are balanced \(\textbf{Final Answer:}\) \(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\)

Explanation: \(\textbf{Given:}\) Ion \(= MnO_4^-\) Oxidation state of oxygen \(=-2\) \(\textbf{Required:}\) Oxidation state of manganese \(\textbf{Relevant Principle:}\) The algebraic sum of oxidation states in a polyatomic ion equals the ionic charge \(\textbf{Why this principle applies:}\) \(MnO_4^-\) is a polyatomic ion with overall charge \(-1\) \(\textbf{Identify known values:}\) Four oxygen atoms contribute \(4 \times (-2) = -8\) \(\textbf{Substitution:}\) Let oxidation state of manganese \(= x\) Then \(x - 8 = -1\) \(\textbf{Intermediate Simplification:}\) \(x = -1 + 8\) \(\textbf{Final Simplification:}\) \(x = +7\) \((+7) + 4(-2) = -1\) \(\textbf{Final Answer:}\) Oxidation state of manganese \(= +7\)

Explanation: \(\textbf{Given:}\) In acidic medium, one mole of \(MnO_4^-\) gains \(5\) electrons \(Fe^{2+}\) is oxidised to \(Fe^{3+}\) \(\textbf{Required:}\) Moles of \(Fe^{2+}\) oxidised by \(1\) mole of permanganate \(\textbf{Relevant Principle:}\) \(Fe^{2+} \rightarrow Fe^{3+} + e^-\) \(\textbf{Why this principle applies:}\) Each \(Fe^{2+}\) ion loses one electron on oxidation \(\textbf{Identify known values:}\) One mole of permanganate accepts \(5\) moles of electrons One mole of \(Fe^{2+}\) supplies \(1\) mole of electrons \(\textbf{Substitution:}\) Moles of \(Fe^{2+} = 5/1\) \(\textbf{Intermediate Simplification:}\) \(= 5\) \(\textbf{Final Simplification:}\) One mole of \(MnO_4^-\) oxidises \(5\) moles of \(Fe^{2+}\) The result is in moles of \(Fe^{2+}\) \(\textbf{Final Answer:}\) \(5\) moles

Explanation: \(\textbf{Given:}\) Amount of \(MnO_4^- = 0.20\) mol \(1\) mol \(MnO_4^-\) oxidises \(5\) mol \(Fe^{2+}\) \(\textbf{Required:}\) Moles of \(Fe^{2+}\) oxidised \(\textbf{Relevant Principle:}\) The balanced redox relation gives the stoichiometric ratio \(MnO_4^- : Fe^{2+} = 1:5\) \(\textbf{Why this principle applies:}\) Each permanganate ion gains \(5\) electrons, and each \(Fe^{2+}\) loses \(1\) electron \(\textbf{Identify known values:}\) Mole ratio \(= 1:5\) \(\textbf{Substitution:}\) Moles of \(Fe^{2+} = 0.20 \times 5\) \(\textbf{Intermediate Simplification:}\) \(= 1.00\) \(\textbf{Final Simplification:}\) \(0.20\) mol permanganate oxidises \(1.00\) mol \(Fe^{2+}\) The result is in moles \(\textbf{Final Answer:}\) \(1.00\) mol

Explanation: Oxalate ion, \(C_2O_4^{2-}\), acts as a reducing agent toward acidic permanganate. During oxidation, carbon in oxalate goes to the \(+4\) oxidation state found in carbon dioxide. Therefore the usual oxidation product is \(CO_2\). This reaction is a standard example used in redox titration chemistry.

Explanation: \(\textbf{Given:}\) Amount of \(MnO_4^- = 0.10\) mol Oxalate ion is oxidised to \(CO_2\) \(\textbf{Required:}\) Moles of \(C_2O_4^{2-}\) oxidised \(\textbf{Relevant Principle:}\) Each \(MnO_4^-\) gains \(5\) electrons in acidic medium Each \(C_2O_4^{2-}\) loses \(2\) electrons: \(C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-\) \(\textbf{Why this principle applies:}\) Electron gain by permanganate must equal electron loss by oxalate \(\textbf{Identify known values:}\) \(1\) mol \(MnO_4^-\) oxidises \(5/2 = 2.5\) mol \(C_2O_4^{2-}\) \(\textbf{Substitution:}\) Moles of oxalate \(= 0.10 \times 2.5\) \(\textbf{Intermediate Simplification:}\) \(= 0.25\) \(\textbf{Final Simplification:}\) \(0.10\) mol permanganate oxidises \(0.25\) mol oxalate The result is in moles of \(C_2O_4^{2-}\) \(\textbf{Final Answer:}\) \(0.25\) mol

Explanation: Iodide ion is a reducing agent and can lose electrons to strong oxidising agents such as acidic permanganate. In its usual acidic reaction with permanganate, the main oxidation product is iodine, \(I_2\). Meanwhile, permanganate is reduced to \(Mn^{2+}\). This makes iodine the expected product in acidic medium.

Explanation: \(\textbf{Given:}\) Amount of \(MnO_4^- = 0.20\) mol \(SO_2\) is oxidised to sulfate \(\textbf{Required:}\) Moles of \(SO_2\) oxidised \(\textbf{Relevant Principle:}\) Sulfur in \(SO_2\) changes from \(+4\) to \(+6\), so each mole of \(SO_2\) loses \(2\) electrons Each mole of \(MnO_4^-\) gains \(5\) electrons in acidic medium \(\textbf{Why this principle applies:}\) Total electrons lost by \(SO_2\) must equal total electrons gained by permanganate \(\textbf{Identify known values:}\) \(1\) mol \(MnO_4^-\) oxidises \(5/2 = 2.5\) mol \(SO_2\) \(\textbf{Substitution:}\) Moles of \(SO_2 = 0.20 \times 2.5\) \(\textbf{Intermediate Simplification:}\) \(= 0.50\) \(\textbf{Final Simplification:}\) \(0.20\) mol permanganate oxidises \(0.50\) mol \(SO_2\) The answer is in moles of \(SO_2\) \(\textbf{Final Answer:}\) \(0.50\) mol

Explanation: In acidic solution, permanganate undergoes a large fall in oxidation state from \(+7\) to \(+2\). This means each permanganate ion can accept five electrons. Such a large tendency to gain electrons makes it a powerful oxidising agent. The oxidising power is therefore centered on manganese, not on potassium.

Explanation: Permanganate ion, \(MnO_4^-\), is deeply purple. In acidic medium it is reduced to \(Mn^{2+}\), which is nearly colourless or only very pale pink in dilute solution. So the intense purple colour disappears as the reaction proceeds. This visual feature is closely linked with the acidic reduction product.

Explanation: \(\textbf{Given:}\) Permanganate oxidises \(Fe^{2+}\) to \(Fe^{3+}\) in acidic medium \(\textbf{Required:}\) Balanced redox equation \(\textbf{Relevant Principle:}\) Reduction half-reaction: \(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\) Oxidation half-reaction: \(Fe^{2+} \rightarrow Fe^{3+} + e^-\) \(\textbf{Why this principle applies:}\) The two half-reactions must be combined so that electrons cancel \(\textbf{Identify known values:}\) Multiply the iron half-reaction by \(5\) to balance electrons \(\textbf{Substitution:}\) \(MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}\) \(\textbf{Intermediate Simplification:}\) Electrons cancel out exactly \(\textbf{Final Simplification:}\) Atoms and charge are balanced Left charge \(= -1 + 8 + 10 = +17\), right charge \(= +2 + 15 = +17\) \(\textbf{Final Answer:}\) \(MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}\)

Explanation: The reduction product of permanganate depends strongly on the medium. In neutral or weakly alkaline solution, permanganate is reduced to manganese dioxide, \(MnO_2\). This product usually appears as a brown precipitate. It is different from acidic medium, where the reduction product is \(Mn^{2+}\).

Explanation: \(\textbf{Given:}\) Permanganate is reduced in neutral or weakly alkaline medium to \(MnO_2\) \(\textbf{Required:}\) Balanced reduction half-reaction \(\textbf{Relevant Principle:}\) In neutral or weakly alkaline medium, oxygen is balanced with \(H_2O\), hydrogen with \(OH^-\), and charge with electrons \(\textbf{Why this principle applies:}\) The reduction product here is \(MnO_2\), not \(Mn^{2+}\) \(\textbf{Identify known values:}\) Manganese changes from \(+7\) in \(MnO_4^-\) to \(+4\) in \(MnO_2\), so \(3\) electrons are gained \(\textbf{Substitution:}\) \(MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-\) \(\textbf{Intermediate Simplification:}\) Mn, O, and H atoms are balanced on both sides \(\textbf{Final Simplification:}\) Charge on left \(= -1 - 3 = -4\), charge on right \(= 4(-1) = -4\) Both atom balance and charge balance are satisfied \(\textbf{Final Answer:}\) \(MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-\)

Explanation: In strongly alkaline medium, permanganate is reduced only by one oxidation-state unit. Manganese goes from \(+7\) in \(MnO_4^-\) to \(+6\) in manganate, \(MnO_4^{2-}\). The manganate ion is green in colour. This behaviour contrasts with neutral medium, where \(MnO_2\) is formed, and acidic medium, where \(Mn^{2+}\) is formed.

Explanation: \(\textbf{Given:}\) In strongly alkaline medium, permanganate is reduced to manganate \(\textbf{Required:}\) Balanced reduction half-reaction \(\textbf{Relevant Principle:}\) Only the oxidation state of manganese changes from \(+7\) to \(+6\) \(\textbf{Why this principle applies:}\) No oxygen or hydrogen balancing is needed because the oxyanion framework remains the same \(\textbf{Identify known values:}\) The change \(+7 \rightarrow +6\) means gain of \(1\) electron \(\textbf{Substitution:}\) \(MnO_4^- + e^- \rightarrow MnO_4^{2-}\) \(\textbf{Intermediate Simplification:}\) All atoms are already balanced \(\textbf{Final Simplification:}\) Charge on left \(= -1 -1 = -2\), charge on right \(= -2\) Mass and charge are both balanced \(\textbf{Final Answer:}\) \(MnO_4^- + e^- \rightarrow MnO_4^{2-}\)

Explanation: The reduction product of permanganate depends on the reaction medium. In acidic solution, it is reduced to \(Mn^{2+}\). In neutral or weakly alkaline medium, it gives \(MnO_2\). In strongly alkaline medium, it is reduced only to manganate, \(MnO_4^{2-}\).

Explanation: \(\textbf{Given:}\) Amount of \(MnO_4^- = 0.40\) mol In neutral or weakly alkaline medium, each mole of \(MnO_4^-\) gains \(3\) electrons \(\textbf{Required:}\) Total moles of electrons gained \(\textbf{Relevant Principle:}\) \(Mn\) changes from \(+7\) in \(MnO_4^-\) to \(+4\) in \(MnO_2\) \(\textbf{Why this principle applies:}\) A decrease of \(3\) in oxidation state means gain of \(3\) electrons per mole of permanganate \(\textbf{Identify known values:}\) Electron factor \(= 3\) \(\textbf{Substitution:}\) Moles of electrons \(= 0.40 \times 3\) \(\textbf{Intermediate Simplification:}\) \(= 1.20\) \(\textbf{Final Simplification:}\) \(0.40\) mol \(MnO_4^-\) gains \(1.20\) mol electrons The answer is in moles of electrons \(\textbf{Final Answer:}\) \(1.20\) mol

Explanation: In strongly alkaline medium, the reduction product is manganate ion, \(MnO_4^{2-}\). In this ion, manganese is in the \(+6\) oxidation state. So the change is only by one unit, from \(+7\) to \(+6\). This is much smaller than the change observed in acidic medium.

Explanation: In acidic medium, permanganate is reduced to \(Mn^{2+}\), which is nearly colourless in dilute solution. As long as reducing agent is present, added permanganate loses its purple colour. Once the reducing agent is consumed, the next small excess of permanganate gives a faint permanent pink or purple tinge. This makes a separate indicator unnecessary.