College Chemistry Quiz: Redox Reactions And Oxidation States
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Redox Reactions And Oxidation StatesQuestion 1 of 20

In the reaction 2MnO4−+16H++10Cl−→2Mn2++5Cl2+8H2O2 MnO_4^- + 16 H^+ + 10 Cl^- \rightarrow 2 Mn^{2+} + 5 Cl_2 + 8 H_2O, what is the change in oxidation state of the chlorine atoms?

From -1 to 0, an increase of +1
From -1 to 0, an increase of +2
From 0 to -1, a decrease of -1
From +1 to 0, a decrease of -1
From -1 to +1, an increase of +2
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College Chemistry Quiz

College Chemistry Quiz: Redox Reactions And Oxidation States

Practice Redox Reactions And Oxidation States in College Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Redox Reactions And Oxidation States, giving you a quick way to practice the rules, question types, and explanations that matter most for College Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

In the reaction 2MnO4−+16H++10Cl−→2Mn2++5Cl2+8H2O2 MnO_4^- + 16 H^+ + 10 Cl^- \rightarrow 2 Mn^{2+} + 5 Cl_2 + 8 H_2O, what is the change in oxidation state of the chlorine atoms?

  1. From -1 to 0, an increase of +1 (correct answer)
  2. From -1 to 0, an increase of +2
  3. From 0 to -1, a decrease of -1
  4. From +1 to 0, a decrease of -1
  5. From -1 to +1, an increase of +2
Explanation: When you encounter redox reactions, focus on tracking how oxidation states change for each element involved. Oxidation state changes reveal which species are being oxidized (losing electrons) and which are being reduced (gaining electrons). To find chlorine's oxidation state change, examine its form on both sides of the equation. On the reactant side, chlorine exists as Cl−Cl^- ions with an oxidation state of -1 (the charge on a monatomic ion equals its oxidation state). On the product side, chlorine appears as Cl2Cl_2 molecules, where each chlorine atom has an oxidation state of 0 (elements in their pure form always have oxidation state 0). The change is from -1 to 0, which represents an increase of +1. This means chlorine is being oxidized—it's losing electrons. The reaction shows 10Cl−10 Cl^- converting to 5Cl25 Cl_2, confirming that chloride ions are losing electrons to form chlorine gas. Option A correctly identifies this +1 increase. Option B incorrectly states the increase is +2, which would only be true if chlorine went from -1 to +1. Option C has the right oxidation states but in reverse order—this would describe the opposite reaction. Option D assumes chlorine starts at +1, which doesn't match the Cl−Cl^- ions shown in the equation. Remember: always determine oxidation states from the actual chemical formulas given in the equation, and calculate the change as final state minus initial state. A positive change means oxidation (electron loss), while a negative change means reduction (electron gain).

Question 2

Which of the following compounds contains an element in its highest possible oxidation state?

  1. SO2SO_2
  2. ClO3−ClO_3^-
  3. MnO4−MnO_4^- (correct answer)
  4. CrO3CrO_3
  5. NO2−NO_2^-
Explanation: When determining oxidation states, you need to consider the maximum number of electrons an element can lose or the maximum positive charge it can achieve based on its electron configuration. Let's examine each compound by calculating the oxidation state of the central atom. In option C, MnO4−MnO_4^-, manganese has an oxidation state of +7. Since manganese has the electron configuration [Ar] 3d⁵ 4s², it can lose all seven of its outer electrons (five 3d and two 4s electrons) to achieve its maximum oxidation state of +7, which corresponds to having the same electron configuration as argon. Looking at the incorrect options: In A, SO2SO_2, sulfur has an oxidation state of +4, but sulfur can achieve a higher oxidation state of +6 (as seen in SO3SO_3 or H2SO4H_2SO_4). In B, ClO3−ClO_3^-, chlorine has an oxidation state of +5, but chlorine's maximum oxidation state is +7 (as in ClO4−ClO_4^-). In D, CrO3CrO_3, chromium has an oxidation state of +6, which is indeed chromium's highest possible oxidation state. However, this makes D a strong distractor since chromium is at its maximum oxidation state here. The key distinction is that manganese in MnO4−MnO_4^- represents the highest oxidation state (+7) among all the options presented, and manganese cannot exceed this oxidation state. Strategy tip: Memorize the maximum oxidation states for common transition metals and main group elements. The maximum oxidation state often equals the group number for main group elements and the total number of valence electrons for transition metals.

Question 3

In the reaction 3Cu+8HNO3→3Cu(NO3)2+2NO+4H2O3 Cu + 8 HNO_3 \rightarrow 3 Cu(NO_3)_2 + 2 NO + 4 H_2O, which element is reduced?

  1. Copper, because it loses electrons and forms Cu²⁺
  2. Nitrogen in HNO₃, because it goes from +5 to +2 in NO (correct answer)
  3. Hydrogen, because it goes from +1 to 0 in water
  4. Oxygen, because it forms water molecules in the products
  5. Nitrogen in NO, because it has a lower oxidation state
Explanation: When you encounter redox reactions, you need to track oxidation states to identify which elements are oxidized (lose electrons) and reduced (gain electrons). Reduction means gaining electrons and decreasing oxidation state. Let's examine the nitrogen in this reaction. In HNO3HNO_3, nitrogen has an oxidation state of +5 (since H is +1, O is -2, and the compound is neutral: +1 + N + 3(-2) = 0, so N = +5). In the product NONO, nitrogen has an oxidation state of +2 (since O is -2 and the compound is neutral: N + (-2) = 0, so N = +2). The nitrogen goes from +5 to +2, decreasing by 3, which means it gained electrons and was reduced. Choice A is incorrect because copper is actually oxidized, not reduced. It goes from 0 in elemental Cu to +2 in Cu(NO3)2Cu(NO_3)_2, losing electrons. Choice C misunderstands hydrogen's role—hydrogen remains +1 in both HNO3HNO_3 and H2OH_2O, so it's neither oxidized nor reduced. Choice D incorrectly focuses on oxygen forming water molecules, but oxygen maintains its -2 oxidation state throughout the reaction in both HNO3HNO_3 and H2OH_2O. The correct answer is B because nitrogen in HNO3HNO_3 is reduced when it forms NONO. Study tip: Always calculate oxidation states for each element before and after the reaction. The element whose oxidation state decreases is reduced, while the one whose oxidation state increases is oxidized. Remember: reduction = gain electrons = decrease in oxidation state.

Question 4

What is the oxidation state of chromium in the dichromate ion Cr2O72−Cr_2O_7^{2-}?

  1. +3
  2. +4
  3. +5
  4. +6 (correct answer)
  5. +7
Explanation: When you encounter questions about oxidation states in polyatomic ions, you need to use the known oxidation states of other elements and the overall charge to determine the unknown oxidation state. To find chromium's oxidation state in Cr2O72−Cr_2O_7^{2-}, start with what you know: oxygen typically has an oxidation state of -2, and the overall ion charge is -2. Set up an equation where the sum of all oxidation states equals the total charge. Let x = oxidation state of chromium. Since there are 2 chromium atoms and 7 oxygen atoms: 2x+7(−2)=−22x + 7(-2) = -2 2x−14=−22x - 14 = -2 2x=+122x = +12 x=+6x = +6 Therefore, chromium has an oxidation state of +6, making answer D correct. Looking at the wrong answers: A) +3 is a common oxidation state for chromium in compounds like Cr2O3Cr_2O_3, but applying this here would give a total charge of -8, not -2. B) +4 would result in a total charge of -6, which doesn't match the dichromate ion's actual charge. C) +5 would give a total charge of -4, again inconsistent with the -2 charge of dichromate. Remember this systematic approach: identify known oxidation states, set up an equation based on the overall charge, and solve algebraically. This method works for any polyatomic ion. Also note that transition metals like chromium can have multiple oxidation states, so don't assume the most common one without calculating.

Question 5

Which of the following compounds contains sulfur in the +4 oxidation state?

  1. H2SH_2S
  2. SO2SO_2 (correct answer)
  3. SO3SO_3
  4. H2SO4H_2SO_4
  5. S8S_8
Explanation: To find oxidation states, you need to apply the rules systematically: hydrogen is typically +1, oxygen is typically -2, and the sum of all oxidation states in a neutral compound must equal zero. Let's work through SO2SO_2: With two oxygen atoms each at -2, that's a total of -4 from oxygen. Since the compound is neutral, sulfur must be +4 to balance this out (+4 + (-4) = 0). This confirms that SO2SO_2 contains sulfur in the +4 oxidation state. Now let's check why the other options are incorrect. In choice A, H2SH_2S: two hydrogens at +1 each give +2 total, so sulfur must be -2 to make the sum zero. In choice C, SO3SO_3: three oxygens at -2 each give -6 total, so sulfur must be +6 to balance this. In choice D, H2SO4H_2SO_4: two hydrogens (+2) and four oxygens (-8) give a net of -6, so sulfur must be +6 to achieve neutrality. The key pattern here is that sulfur's oxidation state increases as it bonds with more oxygen atoms: -2 in H2SH_2S, +4 in SO2SO_2, and +6 in both SO3SO_3 and H2SO4H_2SO_4. When tackling oxidation state problems, always start by assigning the known values (usually H = +1, O = -2), then solve for the unknown element using the rule that all oxidation states must sum to the compound's overall charge.

Question 6

In the reaction Cl2+2KI→2KCl+I2Cl_2 + 2 KI \rightarrow 2 KCl + I_2, how many moles of electrons are transferred when 0.25 mol of Cl2Cl_2 reacts?

  1. 0.25 mol electrons
  2. 0.50 mol electrons (correct answer)
  3. 0.75 mol electrons
  4. 1.0 mol electrons
  5. 2.0 mol electrons
Explanation: When you encounter a question about electron transfer in chemical reactions, you're dealing with redox chemistry. The key is to identify which atoms are gaining or losing electrons by tracking oxidation state changes. In this reaction, chlorine starts as Cl2Cl_2 (oxidation state 0) and becomes Cl−Cl^- in KClKCl (oxidation state -1). Each chlorine atom gains one electron. Since Cl2Cl_2 contains two chlorine atoms, each molecule of Cl2Cl_2 gains 2 electrons total. With 0.25 mol of Cl2Cl_2 reacting, the electron transfer is: 0.25 mol Cl2Cl_2 × 2 electrons per Cl2Cl_2 = 0.50 mol electrons transferred. You can verify this by looking at iodine: I−I^- (oxidation state -1) becomes I2I_2 (oxidation state 0), so each iodide loses one electron. The balanced equation shows 2 mol KIKI react with 1 mol Cl2Cl_2, meaning 2 electrons are lost by iodine for every 2 electrons gained by chlorine. Answer choice (A) 0.25 mol incorrectly assumes only one electron transfers per Cl2Cl_2 molecule. Choice (C) 0.75 mol might result from incorrectly adding electrons gained and lost (0.25 + 0.50). Choice (D) 1.0 mol could come from mistakenly thinking each chlorine atom gains 2 electrons instead of 1. The correct answer is (B) 0.50 mol electrons. Study tip: Always determine the oxidation state change per atom first, then multiply by the number of atoms in the molecule and the moles present. Remember that electrons transferred = electrons gained = electrons lost in any redox reaction.

Question 7

Which of the following is the correct balanced half-reaction for the reduction of permanganate ion in acidic solution?

  1. MnO4−+4H++3e−→MnO2+2H2OMnO_4^- + 4 H^+ + 3 e^- \rightarrow MnO_2 + 2 H_2O
  2. MnO4−+8H++5e−→Mn2++4H2OMnO_4^- + 8 H^+ + 5 e^- \rightarrow Mn^{2+} + 4 H_2O (correct answer)
  3. MnO4−+2H2O+3e−→MnO2+4OH−MnO_4^- + 2 H_2O + 3 e^- \rightarrow MnO_2 + 4 OH^-
  4. MnO4−+4H++4e−→Mn2++2H2OMnO_4^- + 4 H^+ + 4 e^- \rightarrow Mn^{2+} + 2 H_2O
  5. MnO4−+8H++7e−→Mn2++4H2OMnO_4^- + 8 H^+ + 7 e^- \rightarrow Mn^{2+} + 4 H_2O
Explanation: When balancing redox half-reactions, you need to systematically account for atoms and charge while considering the solution conditions. In acidic solution, you use H+H^+ and H2OH_2O to balance hydrogen and oxygen. For permanganate reduction to Mn2+Mn^{2+} in acidic solution, start by balancing the manganese atoms (already balanced: 1 on each side). Next, balance oxygen atoms by adding water molecules. Since there are 4 oxygen atoms on the left in MnO4−MnO_4^- and none in Mn2+Mn^{2+}, you need 4 H2OH_2O molecules on the right. Then balance hydrogen by adding H+H^+ ions. With 4 water molecules contributing 8 hydrogen atoms, you need 8 H+H^+ on the left. Finally, balance charge: the left side has −1+8(+1)=+7-1 + 8(+1) = +7, while the right side has +2+2, requiring 5 electrons on the left to achieve charge balance. Choice B correctly shows this complete balancing: MnO4−+8H++5e−→Mn2++4H2OMnO_4^- + 8 H^+ + 5 e^- \rightarrow Mn^{2+} + 4 H_2O. Choice A has the wrong number of electrons (3 instead of 5) and reduces to MnO2MnO_2 instead of Mn2+Mn^{2+}. Choice C uses OH−OH^- ions, which indicates basic conditions, not acidic. Choice D has 4 electrons instead of 5 and only 2 water molecules instead of 4, making it unbalanced. Remember the systematic approach: balance atoms first (Mn, then O with H2OH_2O, then H with H+H^+), then balance charge with electrons. Always verify your final equation is balanced for both mass and charge.

Question 8

In the reaction 2Fe3++Sn2+→2Fe2++Sn4+2 Fe^{3+} + Sn^{2+} \rightarrow 2 Fe^{2+} + Sn^{4+}, which species acts as the oxidizing agent?

  1. Fe3+Fe^{3+}, because it gains electrons and is reduced to Fe2+Fe^{2+} (correct answer)
  2. Sn2+Sn^{2+}, because it loses electrons and is oxidized to Sn4+Sn^{4+}
  3. Fe2+Fe^{2+}, because it is the reduced form of iron
  4. Sn4+Sn^{4+}, because it has the highest oxidation state in the products
  5. Both Fe3+Fe^{3+} and Sn2+Sn^{2+}, because they both participate in electron transfer
Explanation: When you encounter redox reactions, you need to identify which species gains electrons (gets reduced) and which loses electrons (gets oxidized). The oxidizing agent is always the species that gets reduced—it causes oxidation in another species by accepting electrons from it. Let's trace the electron changes in this reaction. Iron goes from Fe3+Fe^{3+} to Fe2+Fe^{2+}, meaning it gained an electron (reduction). Tin goes from Sn2+Sn^{2+} to Sn4+Sn^{4+}, meaning it lost two electrons (oxidation). Since Fe3+Fe^{3+} accepts electrons and gets reduced, it's the oxidizing agent that drives the reaction forward. Choice A correctly identifies Fe3+Fe^{3+} as the oxidizing agent and gives the right reasoning—it gains electrons and is reduced. Choice B describes Sn2+Sn^{2+} accurately as losing electrons and being oxidized, but this makes it the reducing agent, not the oxidizing agent. Choice C incorrectly identifies Fe2+Fe^{2+} as the oxidizing agent; while it is indeed the reduced form of iron, it's a product of the reaction, not the species causing oxidation. Choice D picks Sn4+Sn^{4+}, but having the highest oxidation state doesn't make something an oxidizing agent—it's actually the oxidized product. Remember this key relationship: the oxidizing agent gets reduced (accepts electrons), while the reducing agent gets oxidized (donates electrons). Look for the species that decreases in oxidation state—that's your oxidizing agent.

Question 9

In the balanced equation Cr2O72−+14H++6Fe2+→2Cr3++6Fe3++7H2OCr_2O_7^{2-} + 14 H^+ + 6 Fe^{2+} \rightarrow 2 Cr^{3+} + 6 Fe^{3+} + 7 H_2O, which species undergoes oxidation?

  1. Cr2O72−Cr_2O_7^{2-}, because it contains chromium in a high oxidation state
  2. H+H^+, because it is converted to water molecules
  3. Fe2+Fe^{2+}, because it loses electrons to become Fe3+Fe^{3+} (correct answer)
  4. Cr3+Cr^{3+}, because it is formed from dichromate reduction
  5. H2OH_2O, because it is the final product containing hydrogen
Explanation: When you encounter a redox equation like this, you need to identify which species are gaining electrons (reduction) and which are losing electrons (oxidation). The key is tracking oxidation state changes for each element. Let's examine what happens to each species. Iron changes from Fe2+Fe^{2+} to Fe3+Fe^{3+}, meaning it goes from a +2 oxidation state to a +3 oxidation state. This increase represents a loss of one electron per iron atom, which is the definition of oxidation. Chromium in Cr2O72−Cr_2O_7^{2-} has an oxidation state of +6, while Cr3+Cr^{3+} has an oxidation state of +3 – this decrease means chromium is gaining electrons and being reduced. Now let's address why the other choices are incorrect. Choice A is wrong because Cr2O72−Cr_2O_7^{2-} undergoes reduction, not oxidation – its chromium atoms decrease in oxidation state from +6 to +3. Choice B incorrectly identifies H+H^+ as being oxidized; hydrogen maintains its +1 oxidation state throughout and simply combines with oxygen to form water without changing oxidation state. Choice D confuses the product with the process – Cr3+Cr^{3+} is the product of reduction, not the species undergoing oxidation. Remember this key principle: oxidation involves an increase in oxidation state (loss of electrons), while reduction involves a decrease in oxidation state (gain of electrons). Always track the oxidation state changes of individual elements, not just the overall charge of ions.

Question 10

Which of the following represents the correct assignment of oxidation states in NH4NO3NH_4NO_3?

  1. N: -3 in NH4+NH_4^+, +5 in NO3−NO_3^-; H: +1; O: -2 (correct answer)
  2. N: +3 in NH4+NH_4^+, -5 in NO3−NO_3^-; H: +1; O: -2
  3. N: -3 in NH4+NH_4^+, +3 in NO3−NO_3^-; H: +1; O: -2
  4. N: +5 in NH4+NH_4^+, -3 in NO3−NO_3^-; H: -1; O: +2
  5. N: 0 in NH4+NH_4^+, 0 in NO3−NO_3^-; H: +1; O: -2
Explanation: When you encounter oxidation state problems with polyatomic compounds, start by identifying the individual ions and apply oxidation state rules systematically to each. In NH4NO3NH_4NO_3, you're dealing with ammonium nitrate, which contains NH4+NH_4^+ and NO3−NO_3^- ions. For NH4+NH_4^+: hydrogen is almost always +1, so with four hydrogens contributing +4, nitrogen must be -3 to give the ion a +1 charge overall. For NO3−NO_3^-: oxygen is typically -2, so three oxygens contribute -6. Since the ion has a -1 charge, nitrogen must be +5 to balance (-6 + 5 = -1). Option A correctly assigns N as -3 in NH4+NH_4^+ and +5 in NO3−NO_3^-, with H as +1 and O as -2. This follows standard oxidation state rules and produces the correct ionic charges. Option B incorrectly makes nitrogen +3 in NH4+NH_4^+ and -5 in NO3−NO_3^-. This would give NH4+NH_4^+ a charge of +7 and NO3−NO_3^- a charge of -11, which are impossible for these common ions. Option C assigns nitrogen as +3 in NO3−NO_3^-, which would make the ion's charge +3 instead of -1. Option D completely reverses the typical oxidation states, making hydrogen -1 and oxygen +2, which violates fundamental oxidation state rules for these elements in ionic compounds. Remember: hydrogen is +1 in ionic compounds, oxygen is -2 in most compounds, and nitrogen's oxidation state varies but must satisfy the overall charge requirement of each ion.

Question 11

In the reaction 3Br2+6OH−→5Br−+BrO3−+3H2O3 Br_2 + 6 OH^- \rightarrow 5 Br^- + BrO_3^- + 3 H_2O, what type of reaction is occurring?

  1. Simple oxidation, where bromine loses electrons uniformly
  2. Simple reduction, where bromine gains electrons uniformly
  3. Disproportionation, where bromine is both oxidized and reduced (correct answer)
  4. Acid-base neutralization with no electron transfer involved
  5. Precipitation reaction forming an insoluble bromine compound
Explanation: When you encounter a reaction where the same element appears in multiple oxidation states on the product side, you're likely looking at a disproportionation reaction. This occurs when one element simultaneously undergoes both oxidation and reduction. To identify this, examine the oxidation states of bromine throughout the reaction. In Br2Br_2, bromine has an oxidation state of 0. In the products, bromine appears as Br−Br^- (oxidation state -1) and BrO3−BrO_3^- (oxidation state +5). Starting from 0, some bromine atoms are reduced to -1 while others are oxidized to +5. This simultaneous oxidation and reduction of the same element defines disproportionation. Option A is incorrect because bromine doesn't lose electrons uniformly—some atoms gain electrons (reduction to Br−Br^-) while others lose electrons (oxidation to BrO3−BrO_3^-). Option B is wrong for similar reasoning; bromine doesn't uniformly gain electrons since some atoms are actually oxidized. Option D misses the point entirely—this reaction clearly involves electron transfer, as evidenced by the changes in oxidation states, and isn't an acid-base neutralization. The presence of hydroxide ions (OH−OH^-) and the basic conditions are typical for disproportionation reactions of halogens, where the halogen redistributes its electrons among different oxidation states. Remember: whenever you see the same element appearing in different oxidation states on the product side of a reaction, immediately consider disproportionation. Look for the telltale pattern of one element being both oxidized and reduced simultaneously.

Question 12

In the reaction 4NH3+5O2→4NO+6H2O4 NH_3 + 5 O_2 \rightarrow 4 NO + 6 H_2O, what is the change in oxidation state of nitrogen?

  1. From -3 to +2, an increase of +5 (correct answer)
  2. From -3 to +2, an increase of +1
  3. From +3 to -2, a decrease of -5
  4. From 0 to +2, an increase of +2
  5. From -3 to 0, an increase of +3
Explanation: When you encounter redox reactions, you need to track how electrons are transferred by calculating oxidation state changes for each element involved. To find nitrogen's oxidation state change, determine its oxidation state in both reactants and products. In NH3NH_3, hydrogen has an oxidation state of +1, so nitrogen must be -3 to make the compound neutral ((-3) + 3(+1) = 0). In NONO, oxygen has an oxidation state of -2, so nitrogen must be +2 to balance it ((+2) + (-2) = 0). The change in nitrogen's oxidation state is from -3 to +2, which represents an increase of 5 units: (+2) - (-3) = +5. This means nitrogen loses 5 electrons and is oxidized. Answer A correctly identifies this change as "from -3 to +2, an increase of +5." Answer B makes the right identification of the oxidation states (-3 to +2) but incorrectly calculates the magnitude of change as +1 instead of +5. Answer C gets both the direction and magnitude wrong, suggesting nitrogen goes from +3 to -2 (a decrease of -5), which reverses the actual process. Answer D incorrectly assigns nitrogen an oxidation state of 0 in NH3NH_3, when it's actually -3. Remember that oxidation state changes equal the difference between final and initial values. When the change is positive, the element is oxidized (loses electrons); when negative, it's reduced (gains electrons). Always double-check your arithmetic, as calculation errors are common traps in oxidation state problems.

Question 13

Which of the following half-reactions represents the oxidation of hydrogen peroxide in basic solution?

  1. H2O2+2H++2e−→2H2OH_2O_2 + 2 H^+ + 2 e^- \rightarrow 2 H_2O
  2. H2O2+2OH−→O2+2H2O+2e−H_2O_2 + 2 OH^- \rightarrow O_2 + 2 H_2O + 2 e^- (correct answer)
  3. H2O2→O2+2H++2e−H_2O_2 \rightarrow O_2 + 2 H^+ + 2 e^-
  4. H2O2+2e−→H2+O2−H_2O_2 + 2 e^- \rightarrow H_2 + O^{2-}
  5. 2OH−→H2O2+2e−2 OH^- \rightarrow H_2O_2 + 2 e^-
Explanation: When approaching half-reaction questions, you need to identify three key elements: whether it's oxidation or reduction, the correct species for the pH conditions, and proper electron balance. For hydrogen peroxide oxidation, you're converting H2O2H_2O_2 to O2O_2. The oxygen in H2O2H_2O_2 has an oxidation state of -1, while in O2O_2 it's 0. This increase means oxidation is occurring, so electrons must appear as products (being lost). Since we're in basic solution, any hydrogen ions must be balanced using OH−OH^- and H2OH_2O rather than H+H^+. Choice B correctly shows this process: H2O2+2OH−→O2+2H2O+2e−H_2O_2 + 2 OH^- \rightarrow O_2 + 2 H_2O + 2 e^-. The hydroxide ions react with the hydrogen peroxide, producing oxygen gas, water, and two electrons. Choice A represents reduction (electrons as reactants) rather than oxidation, and uses H+H^+ which is inappropriate for basic conditions. Choice C shows oxidation but incorrectly uses H+H^+ in basic solution - this would be the acidic form of the reaction. Choice D is completely incorrect, showing an impossible reduction to elemental hydrogen and oxide ion, with improper stoichiometry. Remember that in basic solution half-reactions, you'll always see OH−OH^- and H2OH_2O rather than H+H^+. For oxidation reactions, electrons appear as products; for reduction, they're reactants. Practice converting between acidic and basic forms of half-reactions to master this concept.

Question 14

What is the average oxidation state of sulfur in the thiosulfate ion S2O32−S_2O_3^{2-}?

  1. +1
  2. +2 (correct answer)
  3. +3
  4. +4
  5. +6
Explanation: When you encounter polyatomic ions with multiple atoms of the same element, you need to calculate the average oxidation state by considering the overall charge and the known oxidation states of other atoms. In the thiosulfate ion S2O32−S_2O_3^{2-}, oxygen typically has an oxidation state of -2. With three oxygen atoms contributing 3×(−2)=−63 \times (-2) = -6, and the overall ion charge being -2, the two sulfur atoms must collectively contribute +4+4 to balance the equation: 2S+(−6)=−22S + (-6) = -2, so 2S=+42S = +4. This means the average oxidation state of sulfur is +42=+2\frac{+4}{2} = +2, making answer B correct. However, it's worth noting that the two sulfur atoms in thiosulfate actually have different individual oxidation states (-2 and +6), but the question specifically asks for the average. Answer A (+1) would result if you incorrectly calculated the total charge contribution needed from sulfur. Answer C (+3) might come from assuming each sulfur needs to balance one oxygen atom individually. Answer D (+4) represents the total oxidation state contribution from both sulfur atoms combined, not the average per atom. When calculating average oxidation states, always remember to divide the total oxidation state contribution by the number of atoms of that element. This is particularly important in polyatomic ions where atoms of the same element can have different individual oxidation states but you're asked for their average.

Question 15

In the compound Fe3O4Fe_3O_4 (magnetite), what are the oxidation states of the iron atoms?

  1. All iron atoms have oxidation state +3
  2. All iron atoms have oxidation state +8/3
  3. Two iron atoms are +3, one iron atom is +2 (correct answer)
  4. One iron atom is +6, two iron atoms are +1
  5. Two iron atoms are +2, one iron atom is +4
Explanation: When you encounter compounds with transition metals like iron, remember that these metals can have multiple oxidation states within the same compound. The key is recognizing that the overall charge must equal zero in a neutral compound. To find the iron oxidation states in Fe3O4Fe_3O_4, start with what you know: oxygen typically has an oxidation state of -2. With four oxygen atoms, that's a total charge of -8. Since the compound is neutral, the three iron atoms must contribute a total charge of +8 to balance this. Here's the crucial insight: Fe3O4Fe_3O_4 is actually a mixed oxide that can be written as FeO⋅Fe2O3FeO \cdot Fe_2O_3. This reveals that it contains both Fe2+Fe^{2+} and Fe3+Fe^{3+} ions. Specifically, one iron atom has a +2 oxidation state and two iron atoms have +3 oxidation states. You can verify: (+2) + (+3) + (+3) = +8, which balances the -8 from oxygen. Answer A is incorrect because not all iron atoms are +3; this would give a total positive charge of +9, not +8. Answer B suggests all iron atoms are +8/3, but oxidation states must be whole numbers for individual atoms—fractional values only represent averages. Answer D proposes impossibly high and low oxidation states (+6 and +1) that don't reflect iron's typical chemistry and would give the wrong total charge anyway. Study tip: When dealing with mixed oxides of transition metals, look for common oxidation state combinations rather than assuming all atoms of the same element have identical charges.

Question 16

Which compound contains chlorine in the +7 oxidation state?

  1. HClOHClO
  2. HClO2HClO_2
  3. HClO3HClO_3
  4. HClO4HClO_4 (correct answer)
  5. NaClO3NaClO_3
Explanation: When you encounter oxidation state problems involving polyatomic ions, you need to systematically assign oxidation numbers using known rules. In oxoacids of chlorine, hydrogen is always +1 and oxygen is always -2, so you can solve for chlorine's oxidation state algebraically. For HClO4HClO_4 (perchloric acid), set up the equation: (+1) + (Cl) + 4(-2) = 0, since the overall charge is neutral. This gives you 1 + Cl - 8 = 0, so Cl = +7. This is chlorine's highest possible oxidation state, making HClO4HClO_4 the correct answer (D). Let's check why the other options are wrong. In HClOHClO (A), the calculation gives: 1 + Cl - 2 = 0, so Cl = +1. For HClO2HClO_2 (B): 1 + Cl - 4 = 0, so Cl = +3. In HClO3HClO_3 (C): 1 + Cl - 6 = 0, so Cl = +5. Each of these represents a progressively higher oxidation state, but none reaches +7. These compounds represent the complete series of chlorine oxoacids, where chlorine's oxidation state increases as more oxygen atoms are added: +1 in hypochlorous acid, +3 in chlorous acid, +5 in chloric acid, and +7 in perchloric acid. Remember this pattern: when solving oxidation state problems with polyatomic ions, always use the constraint that oxidation numbers must sum to the total charge. For neutral compounds, they sum to zero; for ions, they sum to the ion's charge. This systematic approach prevents calculation errors.

Question 17

In the reaction Zn+2AgNO3→Zn(NO3)2+2AgZn + 2 AgNO_3 \rightarrow Zn(NO_3)_2 + 2 Ag, which statement correctly describes the electron transfer?

  1. Silver gains 1 electron per atom, zinc loses 2 electrons per atom (correct answer)
  2. Silver loses 1 electron per atom, zinc gains 2 electrons per atom
  3. Silver gains 2 electrons per atom, zinc loses 1 electron per atom
  4. Each silver ion transfers 1 electron to each zinc atom
  5. Two zinc atoms each lose 1 electron to one silver atom
Explanation: When you encounter redox reactions, you need to track the oxidation states of each element to determine which atoms are gaining or losing electrons. In this reaction, zinc metal reacts with silver nitrate solution. Let's analyze the oxidation state changes. Zinc starts as a neutral metal (oxidation state = 0) and becomes Zn2+Zn^{2+} in Zn(NO3)2Zn(NO_3)_2, meaning it loses 2 electrons per atom. Silver starts as Ag+Ag^+ in AgNO3AgNO_3 and becomes neutral silver metal (oxidation state = 0), meaning each silver ion gains 1 electron to become a neutral atom. Looking at the answer choices: Choice A correctly identifies that silver gains 1 electron per atom while zinc loses 2 electrons per atom. Choice B reverses the electron flow entirely—this would describe the opposite reaction where silver is oxidized and zinc is reduced. Choice C gets the number of electrons wrong for each element, suggesting silver gains 2 electrons (but Ag+Ag^+ only needs 1 to become neutral) and zinc loses only 1 (but neutral zinc must lose 2 to become Zn2+Zn^{2+}). Choice D incorrectly describes the stoichiometry—it's not a 1:1 electron transfer between individual atoms, but rather 2 silver ions each accepting 1 electron from 1 zinc atom that loses 2 electrons total. Remember: in redox reactions, always determine oxidation states first, then track the changes. The substance being reduced (gaining electrons) is the oxidizing agent, while the substance being oxidized (losing electrons) is the reducing agent.

Question 18

Which of the following represents a disproportionation reaction?

  1. 2H2O2→2H2O+O22 H_2O_2 \rightarrow 2 H_2O + O_2 (correct answer)
  2. Zn+CuSO4→ZnSO4+CuZn + CuSO_4 \rightarrow ZnSO_4 + Cu
  3. 2KClO3→2KCl+3O22 KClO_3 \rightarrow 2 KCl + 3 O_2
  4. Fe2O3+3CO→2Fe+3CO2Fe_2O_3 + 3 CO \rightarrow 2 Fe + 3 CO_2
  5. AgNO3+HCl→AgCl+HNO3AgNO_3 + HCl \rightarrow AgCl + HNO_3
Explanation: A disproportionation reaction occurs when a single element simultaneously undergoes both oxidation and reduction, meaning one element changes to both higher and lower oxidation states in the products. To identify disproportionation, you need to track oxidation state changes for each element. In option A, hydrogen peroxide (H2O2H_2O_2) contains oxygen with an oxidation state of -1. In the products, oxygen appears as H2OH_2O (oxidation state -2) and O2O_2 (oxidation state 0). The same element, oxygen, has been both reduced (to -2) and oxidized (to 0) from its original -1 state. This is the defining characteristic of disproportionation. Option B shows a simple single displacement reaction where zinc is oxidized and copper is reduced, but these are different elements. Option C represents thermal decomposition of potassium chlorate - while oxygen changes oxidation states, it goes from -2 in KClO3KClO_3 to 0 in O2O_2, which is only oxidation, not simultaneous oxidation and reduction. Option D is a redox reaction where iron is reduced and carbon is oxidized, but again involves different elements changing oxidation states. When looking for disproportionation reactions, focus on finding a single element that appears in products with both higher and lower oxidation states than in the reactant. Hydrogen peroxide decomposition is a classic example you'll encounter frequently in chemistry courses.

Question 19

In the unbalanced equation Cr2O72−+Fe2++H+→Cr3++Fe3++H2OCr_2O_7^{2-} + Fe^{2+} + H^+ \rightarrow Cr^{3+} + Fe^{3+} + H_2O, how many electrons are transferred per mole of Cr2O72−Cr_2O_7^{2-} consumed?

  1. 3 electrons
  2. 6 electrons (correct answer)
  3. 9 electrons
  4. 12 electrons
  5. 14 electrons
Explanation: When you encounter redox equations, focus on identifying oxidation state changes to determine electron transfer. This question tests your ability to analyze half-reactions and balance electrons in redox processes. To find electrons transferred per mole of Cr2O72−Cr_2O_7^{2-}, examine the chromium oxidation state change. In Cr2O72−Cr_2O_7^{2-}, each chromium has a +6 oxidation state (since oxygen is -2, and the overall charge is -2: 2x+7(−2)=−22x + 7(-2) = -2, so x=+6x = +6). In the products, chromium becomes Cr3+Cr^{3+}. Each chromium atom changes from +6 to +3, gaining 3 electrons. Since Cr2O72−Cr_2O_7^{2-} contains two chromium atoms, the total electron gain is 2×3=62 \times 3 = 6 electrons per mole of Cr2O72−Cr_2O_7^{2-}. Looking at the wrong answers: (A) 3 electrons represents the electron change for only one chromium atom, missing that dichromate contains two chromium atoms. (C) 9 electrons might result from incorrectly calculating the oxidation state of chromium in Cr2O72−Cr_2O_7^{2-} or making arithmetic errors. (D) 12 electrons could come from doubling the correct answer or miscalculating the initial oxidation state. The correct answer is (B) 6 electrons. For redox problems, always write out half-reactions and carefully count atoms in polyatomic ions. Remember that Cr2O72−Cr_2O_7^{2-} contains two chromium atoms, so multiply individual atom changes by the number of atoms present. This systematic approach prevents common counting errors in electron transfer calculations.

Question 20

In the balanced equation 2MnO4−+5H2O2+6H+→2Mn2++5O2+8H2O2 MnO_4^- + 5 H_2O_2 + 6 H^+ \rightarrow 2 Mn^{2+} + 5 O_2 + 8 H_2O, what is the reducing agent?

  1. MnO4−MnO_4^-, because it contains manganese in a high oxidation state
  2. H2O2H_2O_2, because it is oxidized and provides electrons (correct answer)
  3. H+H^+, because it participates in the electron transfer process
  4. Mn2+Mn^{2+}, because it is the reduced form of manganese
  5. O2O_2, because it is produced from the oxidation reaction
Explanation: When you encounter redox reaction questions, focus on identifying which species loses electrons (gets oxidized) and which gains electrons (gets reduced). The reducing agent is always the species that gets oxidized—it "reduces" another species by giving up its own electrons. To find the reducing agent, track the oxidation state changes. In MnO4−MnO_4^-, manganese has an oxidation state of +7, and it becomes Mn2+Mn^{2+} with an oxidation state of +2. This is a reduction (gain of electrons). In H2O2H_2O_2, oxygen has an oxidation state of -1, and it becomes O2O_2 where oxygen has an oxidation state of 0. This is an oxidation (loss of electrons). Since H2O2H_2O_2 loses electrons and gets oxidized, it must be the reducing agent. Choice A incorrectly identifies MnO4−MnO_4^- as the reducing agent. While it does contain manganese in a high oxidation state, MnO4−MnO_4^- gains electrons and is reduced, making it the oxidizing agent, not the reducing agent. Choice C suggests H+H^+ is the reducing agent, but hydrogen maintains its +1 oxidation state throughout the reaction—it doesn't undergo oxidation or reduction. Choice D identifies Mn2+Mn^{2+} as the reducing agent, but this confuses products with reactants. Mn2+Mn^{2+} is the product formed when the oxidizing agent is reduced. Remember: the reducing agent gets oxidized (loses electrons) while the oxidizing agent gets reduced (gains electrons). Always check oxidation state changes in the reactants to identify which is which.