Redox Reactions and Electrochemistry MCQs with Answers – Part 1 (NEET Chemistry)

Explanation: Oxidation, according to the electronic concept, is the loss of one or more electrons by a chemical species. In the selected half-reaction, an iron(II) ion loses one electron and becomes an iron(III) ion. The transformation is: \[ \mathrm{Fe^{2+}\rightarrow Fe^{3+}+e^-} \] The oxidation number of iron increases from \(+2\) to \(+3\). This increase is consistent with the loss of one negatively charged electron. The electron appears on the product side because it is released during oxidation. The total electric charge is conserved: the reactant has charge \(+2\), while the combined charge on the product side is \(+3-1=+2\). In the remaining half-reactions, electrons appear on the reactant side and are gained by the reacting species. Those transformations therefore represent reduction rather than oxidation. A useful distinction is that oxidation does not necessarily require oxygen; electron loss is sufficient to identify the process.

Explanation: The reaction contains no oxygen, but it is still a redox reaction because electrons are transferred between reacting species. Zinc initially exists as an elemental metal with oxidation number zero. It forms \(\mathrm{Zn^{2+}}\), in which its oxidation number is \(+2\). The corresponding oxidation half-reaction is: \[ \mathrm{Zn\rightarrow Zn^{2+}+2e^-} \] Zinc therefore loses two electrons per atom. Silver ions accept these electrons and undergo reduction. Their reduction half-reaction is: \[ \mathrm{2Ag^++2e^-\rightarrow 2Ag} \] The two electrons released by zinc are exactly equal to the two electrons accepted by the silver ions. Zinc is consequently the electron donor and the reducing agent. The silver ions are electron acceptors and constitute the oxidising agent. The essential point is that the electronic definition of oxidation applies even when oxygen does not participate in the chemical reaction.

Explanation: \(\textbf{Oxidation-state change:}\) Each iron ion changes from oxidation state \(+2\) to \(+3\). The increase of one unit means that each iron(II) ion loses one electron. For a single ion: \[ \mathrm{Fe^{2+}\rightarrow Fe^{3+}+e^-} \] \(\textbf{Account for both ions:}\) Two iron(II) ions undergo the same oxidation. Therefore, the total number of electrons released is: \[ n(e^-)=2\times1=2 \] The resulting half-reaction is: \[ \mathrm{2Fe^{2+}\rightarrow 2Fe^{3+}+2e^-} \] \(\textbf{Verify electric charge:}\) The total charge on the reactant side is: \[ Q_{\mathrm{left}}=2(+2)=+4 \] The total charge on the product side is: \[ Q_{\mathrm{right}}=2(+3)+2(-1)=+4 \] Both sides contain two iron atoms, and the total electric charges are equal. Option A accounts for only one released electron, while option D assigns too many electrons. Option C places electrons on the reactant side despite the increase in iron's oxidation number. \(\textbf{Final answer:}\) \[ \boxed{2e^-\text{ released by two }\mathrm{Fe^{2+}}\text{ ions}} \]

Explanation: \(\textbf{Electron demand of sample P:}\) Aluminium changes from oxidation number \(0\) to \(+3\). Its oxidation half-reaction is: \[ \mathrm{Al\rightarrow Al^{3+}+3e^-} \] Thus, one mole of aluminium atoms releases three moles of electrons on complete oxidation. For \(0.10\ \mathrm{mol}\) of aluminium: \[ n(e^-)_{\mathrm{P}}=0.10\times3 \] \[ n(e^-)_{\mathrm{P}}=0.30\ \mathrm{mol} \] \(\textbf{Electron demand of sample Q:}\) Iron(II) undergoes oxidation according to: \[ \mathrm{Fe^{2+}\rightarrow Fe^{3+}+e^-} \] One mole of iron(II) ions releases one mole of electrons. Therefore: \[ n(e^-)_{\mathrm{Q}}=0.20\times1 \] \[ n(e^-)_{\mathrm{Q}}=0.20\ \mathrm{mol} \] \(\textbf{Compare with the electron-accepting capacity:}\) The oxidising agent in each separate experiment can accept only \(0.25\ \mathrm{mol}\) of electrons. Sample P requires \(0.30\ \mathrm{mol}\), so its complete oxidation is not possible with the available oxidising agent. Sample Q requires \(0.20\ \mathrm{mol}\), which lies within the available capacity. Consequently, all the iron(II) ions in sample Q can be oxidised, whereas some aluminium must remain unoxidised in sample P. This demonstrates why the amount of a reducing species cannot be compared with oxidant capacity without first accounting for electrons transferred per particle. \(\textbf{Final answer:}\) Only sample Q can be oxidised completely.

Explanation: Reduction is defined electronically as the gain of electrons by a chemical species. In molecular iodine, each iodine atom has oxidation number zero. The formation of an iodide ion changes its oxidation number to \(-1\). This decrease in oxidation number indicates reduction. Because \(\mathrm{I_2}\) contains two iodine atoms, the molecule accepts two electrons in the displayed half-reaction. Electrons are written on the reactant side because they are consumed during the reduction process. The final iodide ions carry negative charges as a consequence of electron gain. The reduction is not identified merely because the product is negatively charged; it is identified by the change from oxidation state \(0\) to \(-1\). Reduction also does not require oxygen removal, as no oxygen-containing species appears in this transformation. Thus, the electronic and oxidation-number interpretations give the same conclusion.

Explanation: \(\textbf{Determine the initial chromium oxidation number:}\) Let the oxidation number of each chromium atom in \(\mathrm{Cr_2O_7^{2-}}\) be \(x\). Oxygen has oxidation number \(-2\) in this ion. The sum of oxidation numbers must equal the ionic charge: \[ 2x+7(-2)=-2 \] \[ 2x-14=-2 \] \[ 2x=12 \] \[ x=+6 \] \(\textbf{Identify the reduction:}\) Chromium changes from \(+6\) in dichromate to \(+3\) in \(\mathrm{Cr^{3+}}\). Each chromium atom therefore accepts: \[ 6-3=3e^- \] There are two chromium atoms in one dichromate ion. The total electrons accepted are: \[ n(e^-)=2\times3=6 \] The complete acidic-medium reduction half-reaction is: \[ \mathrm{Cr_2O_7^{2-}+14H^++6e^-\rightarrow 2Cr^{3+}+7H_2O} \] \(\textbf{Conservation check:}\) Both sides contain two chromium atoms, seven oxygen atoms and fourteen hydrogen atoms. The total charge on the left is \(-2+14-6=+6\). The total charge on the right is \(2(+3)=+6\). Therefore, the six-electron requirement is consistent with both oxidation-number change and charge conservation. \(\textbf{Final answer:}\) \[ \boxed{6e^-\text{ accepted per }\mathrm{Cr_2O_7^{2-}}\text{ ion}} \]

Explanation: \(\textbf{Initial oxidation number:}\) Nitrogen has oxidation number \(+5\) in \(\mathrm{NO_3^-}\). This follows from: \[ x+3(-2)=-1 \] \[ x=+5 \] \(\textbf{Process P — formation of nitric oxide:}\) Oxygen has oxidation number \(-2\) in \(\mathrm{NO}\). Therefore, nitrogen has oxidation number \(+2\). Its oxidation-number change is: \[ +5\rightarrow+2 \] The number of electrons accepted per nitrogen atom is: \[ n_{\mathrm{P}}(e^-)=5-2=3 \] \(\textbf{Process Q — formation of nitrogen dioxide:}\) In \(\mathrm{NO_2}\), the two oxygen atoms contribute a total oxidation number of \(-4\). Nitrogen consequently has oxidation number \(+4\). The change is: \[ +5\rightarrow+4 \] Thus: \[ n_{\mathrm{Q}}(e^-)=5-4=1 \] \(\textbf{Compare the two reduction pathways:}\) Each nitrate ion contains one nitrogen atom, so equal numbers of nitrate ions correspond to equal numbers of nitrogen atoms being reduced. The required ratio is: \[ \frac{n_{\mathrm{P}}(e^-)}{n_{\mathrm{Q}}(e^-)}=\frac{3}{1} \] Both transformations are reductions, but formation of \(\mathrm{NO}\) involves a larger decrease in nitrogen's oxidation number. The oxidising species therefore accepts more electrons when its reduction product is \(\mathrm{NO}\) rather than \(\mathrm{NO_2}\). \(\textbf{Final answer:}\) \[ \boxed{3:1} \]

Explanation: The Assertion is true because iron(III) ions accept electrons to become iron(II) ions. The relevant half-reaction is: \[ \mathrm{Fe^{3+}+e^-\rightarrow Fe^{2+}} \] During this transformation, the oxidation number of iron decreases from \(+3\) to \(+2\). This decrease establishes that reduction has occurred. The resulting ion remains positively charged because gaining one electron does not completely neutralise the original \(+3\) charge. The Reason is also true because reduction concerns electron gain and oxidation-number decrease, not necessarily the formation of a neutral or negatively charged product. For comparison, \(\mathrm{Cu^{2+}}\) may be reduced to neutral copper by accepting two electrons. Alternatively, \(\mathrm{Fe^{3+}}\) can be reduced to \(\mathrm{Fe^{2+}}\) by accepting only one electron. Both transformations qualify as reduction despite producing species with different final charges. The Reason directly explains why a positively charged product does not contradict the reduction described in the Assertion. Hence, both statements are true, and the Reason correctly explains the Assertion.

Explanation: \(\textbf{Identify the oxidation process:}\) Tin changes from oxidation number \(+2\) to \(+4\). An increase in oxidation number indicates the loss of electrons. The oxidation half-reaction is: \[ \mathrm{Sn^{2+}\rightarrow Sn^{4+}+2e^-} \] Thus, each tin(II) ion releases two electrons. Tin(II) is the reducing agent because it supplies electrons to another species. \(\textbf{Identify the reduction process:}\) Iron changes from oxidation number \(+3\) to \(+2\). The reduction half-reaction for one iron ion is: \[ \mathrm{Fe^{3+}+e^-\rightarrow Fe^{2+}} \] Each iron(III) ion accepts one electron. Two iron(III) ions are therefore required to accept the two electrons released by one tin(II) ion. \(\textbf{Verify electron conservation:}\) \[ \text{Electrons lost}=1\times2=2 \] \[ \text{Electrons gained}=2\times1=2 \] The total numbers of electrons lost and gained are equal. This is also consistent with the balanced equation, whose total ionic charge is \(+8\) on each side. Tin(II) is the electron donor and reducing agent, whereas iron(III) is the electron acceptor and oxidising agent. The electron roles must be identified from their actual oxidation-state changes rather than from the fact that both initial species are positively charged.

Explanation: A reaction is classified as redox when oxidation and reduction occur together, as indicated by changes in oxidation numbers. Reaction I is a precipitation reaction forming silver chloride. Silver remains in oxidation state \(+1\), sodium remains \(+1\), chloride remains \(-1\), and the oxidation numbers of nitrogen and oxygen in nitrate remain unchanged. Therefore, Reaction I is not redox. In Reaction II, magnesium changes from oxidation number \(0\) to \(+2\), while oxygen changes from \(0\) to \(-2\). Magnesium undergoes oxidation, and oxygen undergoes reduction. Reaction II is consequently a redox reaction. Reaction III involves decomposition of calcium carbonate, but calcium remains \(+2\), carbon remains \(+4\), and oxygen remains \(-2\). Decomposition by itself does not establish that an electron-transfer process has occurred. Therefore, Reaction III is not redox. In Reaction IV, mercury changes from \(+2\) in \(\mathrm{HgO}\) to \(0\) in elemental mercury. At the same time, oxygen changes from \(-2\) in \(\mathrm{HgO}\) to \(0\) in molecular oxygen. Mercury is reduced, while oxygen is oxidised. Reaction IV is therefore redox. The comparison establishes that reaction classification must be based on oxidation-number changes rather than merely on precipitation, combination or decomposition labels.

Explanation: \(\textbf{Identify the electron-transfer changes:}\) Aluminium undergoes oxidation from oxidation number \(0\) to \(+3\). Copper undergoes reduction from \(+2\) to \(0\). The oxidation half-reaction is: \[ \mathrm{Al\rightarrow Al^{3+}+3e^-} \] The reduction half-reaction is: \[ \mathrm{Cu^{2+}+2e^-\rightarrow Cu} \] \(\textbf{Determine the available electrons:}\) One mole of aluminium releases three moles of electrons. For \(0.12\ \mathrm{mol}\) of aluminium: \[ n(e^-)=0.12\times3 \] \[ n(e^-)=0.36\ \mathrm{mol} \] \(\textbf{Calculate copper(II) consumption:}\) Each copper(II) ion accepts two electrons. Therefore, the amount of copper(II) that can be reduced using the electrons from all the aluminium is: \[ n(\mathrm{Cu^{2+}})_{\mathrm{consumed}}=\frac{0.36}{2} \] \[ n(\mathrm{Cu^{2+}})_{\mathrm{consumed}}=0.18\ \mathrm{mol} \] Only \(0.18\ \mathrm{mol}\) of copper(II) is required, whereas \(0.25\ \mathrm{mol}\) is initially available. Thus, aluminium is the limiting reactant and copper(II) ions are in excess. \(\textbf{Determine the residual amount:}\) \[ n(\mathrm{Cu^{2+}})_{\mathrm{remaining}}=n_{\mathrm{initial}}-n_{\mathrm{consumed}} \] \[ n(\mathrm{Cu^{2+}})_{\mathrm{remaining}}=0.25-0.18 \] \[ n(\mathrm{Cu^{2+}})_{\mathrm{remaining}}=0.07\ \mathrm{mol} \] \(\textbf{Stoichiometric check:}\) The balanced reaction requires \(3\) moles of copper(II) ions for every \(2\) moles of aluminium. Hence, \(0.12\times(3/2)=0.18\ \mathrm{mol}\) of copper(II) is consumed, confirming the electron-accounting result. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{Cu^{2+}})_{\mathrm{remaining}}=0.07\ \mathrm{mol}} \]

Explanation: \(\textbf{Statement I: Correct.}\) Chlorine changes from oxidation number \(0\) in \(\mathrm{Cl_2}\) to \(-1\) in \(\mathrm{Cl^-}\). The decrease in oxidation number indicates reduction. Because chlorine accepts electrons and causes oxidation of another species, it acts as the oxidising agent. \(\textbf{Statement II: Correct.}\) Bromine changes from oxidation number \(-1\) in \(\mathrm{Br^-}\) to \(0\) in \(\mathrm{Br_2}\). This increase in oxidation number indicates oxidation. Bromide ions release electrons and therefore act as reducing agents. \(\textbf{Statement III: Incorrect.}\) A negative charge does not itself establish that oxidation has occurred. In this reaction, chloride ions are products of the reduction of molecular chlorine. Chlorine gains electrons during chloride formation rather than losing them. \(\textbf{Statement IV: Correct.}\) The reduction half-reaction is: \[ \mathrm{Cl_2+2e^-\rightarrow 2Cl^-} \] Therefore, each molecule of chlorine accepts two electrons. The corresponding oxidation half-reaction is: \[ \mathrm{2Br^-\rightarrow Br_2+2e^-} \] Both half-reactions involve two electrons, giving a balanced overall reaction. The correct combination is I, II and IV only.

Explanation: \(\textbf{Initial oxidation number of oxygen:}\) In hydrogen peroxide, each hydrogen atom has oxidation number \(+1\). Let the oxidation number of oxygen be \(x\). For the neutral molecule: \[ 2(+1)+2x=0 \] \[ x=-1 \] Thus, oxygen initially has oxidation number \(-1\). \(\textbf{Oxidation pathway:}\) Oxygen forming molecular \(\mathrm{O_2}\) changes from oxidation number \(-1\) to \(0\). Each oxygen atom undergoing this change loses one electron. Two oxygen atoms enter the \(\mathrm{O_2}\) product. Therefore: \[ n(e^-)_{\mathrm{lost}}=2\times[0-(-1)] \] \[ n(e^-)_{\mathrm{lost}}=2 \] \(\textbf{Reduction pathway:}\) Oxygen forming water changes from oxidation number \(-1\) to \(-2\). Each oxygen atom undergoing this change gains one electron. The two molecules of water contain two oxygen atoms in total. Therefore: \[ n(e^-)_{\mathrm{gained}}=2\times[(-1)-(-2)] \] \[ n(e^-)_{\mathrm{gained}}=2 \] \(\textbf{Interpret the reaction:}\) The same element, oxygen, initially present in one oxidation state, undergoes both oxidation and reduction. This is a disproportionation reaction. Hydrogen peroxide therefore participates as both an oxidising and a reducing agent in its decomposition. The two electrons lost by oxygen atoms forming \(\mathrm{O_2}\) are balanced by the two electrons gained by oxygen atoms forming \(\mathrm{H_2O}\). \(\textbf{Final answer:}\) \[ \boxed{(e^-_{\mathrm{lost}},e^-_{\mathrm{gained}})=(2,2)} \]

Explanation: \(\textbf{Determine the oxidation numbers in potassium chlorate:}\) Potassium has oxidation number \(+1\), while oxygen has oxidation number \(-2\). Let the oxidation number of chlorine in \(\mathrm{KClO_3}\) be \(x\). The oxidation-number sum gives: \[ (+1)+x+3(-2)=0 \] \[ x=+5 \] \(\textbf{Follow chlorine into potassium chloride:}\) Chlorine has oxidation number \(-1\) in \(\mathrm{KCl}\). Therefore: \[ \mathrm{Cl:\ +5\rightarrow-1} \] The decrease of six oxidation-number units indicates reduction of chlorine. \(\textbf{Follow oxygen into molecular oxygen:}\) Oxygen changes from \(-2\) in potassium chlorate to \(0\) in \(\mathrm{O_2}\). Therefore: \[ \mathrm{O:\ -2\rightarrow0} \] The increase in oxidation number indicates oxidation of oxygen. \(\textbf{Verify electron conservation:}\) The two chlorine atoms collectively gain: \[ 2\times[5-(-1)]=12e^- \] Six oxygen atoms collectively lose: \[ 6\times[0-(-2)]=12e^- \] The electron loss and gain are equal. This establishes that decomposition of potassium chlorate is a redox reaction. However, it is not disproportionation because chlorine undergoes only reduction and oxygen undergoes only oxidation. In disproportionation, the same element initially present in one oxidation state must undergo both oxidation and reduction. The reaction therefore illustrates why decomposition of a single compound is not automatically classified as disproportionation.

Explanation: \(\textbf{Analyse Stage I:}\) Copper initially appears as \(\mathrm{Cu^{2+}}\) in the aqueous solution. It changes to elemental copper, whose oxidation number is zero. The corresponding half-reaction is: \[ \mathrm{Cu^{2+}+2e^-\rightarrow Cu} \] Copper(II) ions accept two electrons from zinc. Thus, copper undergoes reduction in Stage I. Copper(II) also acts as the oxidising agent because it causes zinc to lose electrons. \(\textbf{Analyse Stage II:}\) The deposited copper is now present as elemental metal with oxidation number zero. On reaction with silver(I) ions, copper changes into \(\mathrm{Cu^{2+}}\). The relevant half-reaction is: \[ \mathrm{Cu\rightarrow Cu^{2+}+2e^-} \] Copper therefore loses two electrons and undergoes oxidation. The released electrons are accepted by two silver(I) ions: \[ \mathrm{2Ag^++2e^-\rightarrow 2Ag} \] Copper consequently acts as the reducing agent in Stage II. \(\textbf{Connect the two transformations:}\) Across the successive reactions, the oxidation-number sequence of copper is: \[ \mathrm{Cu:\ +2\rightarrow0\rightarrow+2} \] The first change represents reduction, whereas the second represents oxidation. This is not disproportionation because the two changes occur in separate reactions with different reactants. The redox role of a chemical species must always be determined from its transformation in the particular reaction, rather than being assigned permanently to that element.

Explanation: \(\textbf{Identify the initial oxidation state:}\) Oxygen has oxidation number \(-1\) in hydrogen peroxide. Hydrogen peroxide can undergo either oxidation or reduction depending on the other reacting species. Its redox role must therefore be determined independently for each reaction. \(\textbf{Reaction P — oxidation of iodide:}\) Iodine changes from oxidation number \(-1\) in \(\mathrm{I^-}\) to \(0\) in \(\mathrm{I_2}\). Thus, iodide ions lose electrons. Oxygen in hydrogen peroxide changes from \(-1\) to \(-2\) in water. Hydrogen peroxide consequently accepts electrons and undergoes reduction. The reduction half-reaction is: \[ \mathrm{H_2O_2+2H^++2e^-\rightarrow 2H_2O} \] Since hydrogen peroxide causes iodide oxidation while itself undergoing reduction, it acts as the oxidising agent. \(\textbf{Reaction Q — reduction of permanganate:}\) Manganese has oxidation number \(+7\) in \(\mathrm{MnO_4^-}\). It changes to oxidation number \(+2\) in \(\mathrm{Mn^{2+}}\). Manganese therefore undergoes reduction. Meanwhile, oxygen in hydrogen peroxide changes from \(-1\) to \(0\) in molecular oxygen. Hydrogen peroxide loses electrons according to: \[ \mathrm{H_2O_2\rightarrow O_2+2H^++2e^-} \] Therefore, hydrogen peroxide undergoes oxidation and acts as the reducing agent in Reaction Q. \(\textbf{Check the electron balance in Reaction Q:}\) Each permanganate ion accepts five electrons during the change from manganese(VII) to manganese(II). Two permanganate ions collectively accept: \[ 2\times5=10e^- \] Each hydrogen peroxide molecule releases two electrons when converted into molecular oxygen. Five hydrogen peroxide molecules collectively release: \[ 5\times2=10e^- \] The electron counts agree with the balanced overall reaction. \(\textbf{Compare the roles:}\) In Reaction P, hydrogen peroxide is reduced and behaves as an oxidising agent. In Reaction Q, hydrogen peroxide is oxidised and behaves as a reducing agent. The comparison demonstrates that oxidising and reducing behaviour depends on the actual reaction and the oxidation-state changes involved, not solely on the identity of a chemical species.

Explanation: \(\textbf{Oxidation-number interpretation:}\) The average oxidation number of sulphur in thiosulphate is obtained using the ionic charge. \[ 2x+3(-2)=-2 \] \[ 2x=4 \] \[ x=+2 \] For tetrathionate: \[ 4y+6(-2)=-2 \] \[ 4y=10 \] \[ y=+2.5 \] These are average oxidation numbers; they do not imply that every sulphur atom possesses the same individual oxidation state. \(\textbf{Electron balance:}\) Two thiosulphate ions contain four sulphur atoms, which become the four sulphur atoms of one tetrathionate ion. The total increase in oxidation-number sum is: \[ \Delta N=4(2.5)-4(2)=2 \] Therefore, the correctly balanced oxidation half-reaction is: \[ \mathrm{2S_2O_3^{2-}\rightarrow S_4O_6^{2-}+2e^-} \] Two moles of electrons are released per mole of tetrathionate formed. \(\textbf{Determine the limiting capacity:}\) Complete conversion of the available thiosulphate would release: \[ n(e^-)_{\mathrm{required}}=0.40\times\frac{2}{2} \] \[ n(e^-)_{\mathrm{required}}=0.40\ \mathrm{mol} \] However, the oxidising agent accepts only \(0.25\ \mathrm{mol}\) of electrons. Thus, the oxidising agent is limiting. \(\textbf{Calculate tetrathionate formation:}\) \[ n(\mathrm{S_4O_6^{2-}})=\frac{0.25}{2} \] \[ n(\mathrm{S_4O_6^{2-}})=0.125\ \mathrm{mol} \] The thiosulphate consumed is: \[ n(\mathrm{S_2O_3^{2-}})_{\mathrm{consumed}}=2(0.125) \] \[ =0.250\ \mathrm{mol} \] Hence: \[ n(\mathrm{S_2O_3^{2-}})_{\mathrm{remaining}}=0.400-0.250 \] \[ =0.150\ \mathrm{mol} \] The electron balance, sulphur-atom balance and total ionic charge are consistent with the half-reaction. \(\textbf{Final answer:}\) \[ \boxed{\left(n_{\mathrm{tetrathionate}},n_{\mathrm{thiosulphate\ remaining}}\right)=(0.125,\ 0.150)\ \mathrm{mol}} \]

Explanation: The oxidation number of oxygen in the superoxide ion is obtained from its total charge. \[ 2x=-1 \] \[ x=-\frac12 \] Each superoxide ion contains two oxygen atoms with an average oxidation number of \(-\frac12\). In molecular oxygen, the oxidation number of oxygen is zero. Consequently, the superoxide ion converted into \(\mathrm{O_2}\) undergoes oxidation. Its total oxidation-number increase is: \[ 2\left[0-\left(-\frac12\right)\right]=1 \] Thus, this superoxide ion loses one electron. In the peroxide ion, \(\mathrm{O_2^{2-}}\), each oxygen atom has an average oxidation number of \(-1\). The other superoxide ion therefore undergoes reduction. Its total oxidation-number decrease has a magnitude of: \[ 2\left[\left(-\frac12\right)-(-1)\right]=1 \] Hence, it accepts one electron. The corresponding formal half-reactions are: \[ \mathrm{O_2^-\rightarrow O_2+e^-} \] \[ \mathrm{O_2^-+e^-\rightarrow O_2^{2-}} \] Adding these half-reactions cancels the electron and reproduces the given equation. The fractional oxidation number of oxygen does not mean that a fraction of an electron is transferred. It is an average value, and the complete superoxide ions exchange one electron. The reaction is disproportionation because oxygen initially in one oxidation state undergoes both oxidation and reduction.

Explanation: \(\textbf{Statement I: Correct.}\) Potassium has oxidation number \(+1\) in potassium superoxide. The \(\mathrm{O_2}\) unit consequently carries a charge of \(-1\). Its two oxygen atoms have an average oxidation number of \(-\frac12\), not \(-1\). This distinguishes a superoxide from an ordinary peroxide. \(\textbf{Statement II: Correct.}\) Fluorine has oxidation number \(-1\) in its compounds. Let the oxidation number of each oxygen atom in \(\mathrm{O_2F_2}\) be \(x\). \[ 2x+2(-1)=0 \] \[ x=+1 \] Oxygen is assigned a positive oxidation number because fluorine is more electronegative. \(\textbf{Statement III: Incorrect.}\) In \(\mathrm{OF_2}\), the two fluorine atoms together contribute \(-2\). Oxygen must therefore have oxidation number \(+2\), not \(-2\). The familiar oxygen value of \(-2\) cannot be applied indiscriminately to oxygen-fluorine compounds. \(\textbf{Statement IV: Correct.}\) Calcium is a Group 2 metal and has oxidation number \(+2\) in calcium hydride. The two hydrogen atoms must together contribute \(-2\). Each hydrogen atom consequently has oxidation number \(-1\). This is the characteristic assignment for hydrogen in metal hydrides. The correct combination is I, II and IV, demonstrating why compound identity and exception rules must be examined before assigning oxidation numbers.

Explanation: \(\textbf{Identify the two initial oxidation states:}\) Sodium hydride is an ionic hydride in which sodium has oxidation number \(+1\). Hydrogen in \(\mathrm{NaH}\) therefore has oxidation number \(-1\). Hydrogen in water has oxidation number \(+1\), since oxygen ordinarily has oxidation number \(-2\). In molecular hydrogen, \(\mathrm{H_2}\), the oxidation number of each hydrogen atom is zero. \(\textbf{Hydride hydrogen:}\) The formal oxidation-number change is: \[ \mathrm{H:\ -1\rightarrow0} \] This increase of one unit represents the loss of one electron. Hydride hydrogen consequently undergoes oxidation. \(\textbf{Water hydrogen:}\) The corresponding change for hydrogen from water is: \[ \mathrm{H:\ +1\rightarrow0} \] This decrease of one unit represents the gain of one electron. Water hydrogen consequently undergoes reduction. \(\textbf{Verify the electron accounting:}\) For one formula unit of sodium hydride reacting according to the equation, the formal electron loss and gain are: \[ n(e^-)_{\mathrm{lost}}=1 \] \[ n(e^-)_{\mathrm{gained}}=1 \] The sodium oxidation number remains \(+1\), and oxygen remains at \(-2\). The two oxidation-state changes of hydrogen therefore account for the redox process. Sodium hydride acts as the reducing agent through its hydride ion, whereas water supplies the hydrogen undergoing reduction. Although both hydrogen contributions ultimately have oxidation number zero, their starting oxidation states are different. This is why a product consisting of elemental hydrogen can arise through simultaneous oxidation and reduction of hydrogen originally present in different chemical environments.