1 question, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 1
When aluminum metal is placed in a solution of NiCl₂(aq), the half-reactions shown in the table occur. In the balanced overall equation with the smallest whole-number coefficients, what are the coefficients of Al(s) and Ni²⁺(aq)?
Answer and reasoning
AAl 3; Ni²⁺ 2 A student who multiplies each half-reaction by the number of electrons it contains picks this. Then 3 Al atoms lose 9 electrons while 2 Ni²⁺ ions gain only 4, so electrons and charge do not balance.
BAl 2; Ni²⁺ 3Correct Each Al atom loses 3 electrons and each Ni²⁺ ion gains 2. Multiplying the Al half-reaction by 2 and the Ni²⁺ half-reaction by 3 makes 6 electrons lost and 6 gained: 2 Al(s) + 3 Ni²⁺(aq) → 2 Al³⁺(aq) + 3 Ni(s), with +6 charge on each side.
CAl 1; Ni²⁺ 1 A student who balances only the atoms picks this: Al(s) + Ni²⁺(aq) → Al³⁺(aq) + Ni(s). The atoms balance, but the charge is +2 on the left and +3 on the right, because 3 electrons lost do not equal 2 gained.
DAl 6; Ni²⁺ 6 A student who multiplies both half-reactions by the total number of electrons transferred, 6, picks this. Then 18 electrons are lost and 12 gained; 6 is the number each half-reaction must reach, which needs 2 Al and 3 Ni²⁺.
Working Electrons lost per Al = 3; electrons gained per Ni²⁺ = 2. Least common multiple = 6: multiply the Al half-reaction by 2 and the Ni²⁺ half-reaction by 3. 2 Al(s) + 3 Ni²⁺(aq) → 2 Al³⁺(aq) + 3 Ni(s); charge +6 = +6; 6 e⁻ cancel.
In preparation: 0 of 1 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
4.9.A.1 Half-reaction Fix
Half-reaction
An equation that shows either the oxidation or the reduction part of a redox reaction on its own, with the electrons lost or gained written explicitly, for example Ni²⁺(aq) + 2 e⁻ → Ni(s).
Oxidation half-reaction
A half-reaction in which a species loses electrons, so its oxidation number increases; the electrons appear on the product side, as in Al(s) → Al³⁺(aq) + 3 e⁻.
Reduction half-reaction
A half-reaction in which a species gains electrons, so its oxidation number decreases; the electrons appear on the reactant side, as in Ni²⁺(aq) + 2 e⁻ → Ni(s).
Combining half-reactions
Each half-reaction is multiplied by the factor that makes the electrons lost in oxidation equal the electrons gained in reduction; the half-reactions are then added and the electrons cancel, so no electrons appear in the overall equation.
Balanced redox equation
An equation in which the atoms of every element and the total charge are the same on both sides. Equal charge follows from making the electrons lost equal the electrons gained.
Balancing a half-reaction in acidic solution
After balancing the atoms other than O and H, O atoms are balanced with H₂O, H atoms with H⁺, and then charge with electrons. Free O²⁻ ions are not used, and OH⁻ is not added in acidic solution.
Students often think A redox equation is balanced when the atoms of each element are equal on both sides; the total charge on each side need not be the same. In fact No. The total charge must also be equal on both sides, which happens only when the electrons lost equal the electrons gained. Fe³⁺ + Sn²⁺ → Fe²⁺ + Sn⁴⁺ balances the atoms but has a total charge of +5 on the left and +6 on the right.
Students often think When combining half-reactions, each half-reaction is multiplied by the number of electrons it contains. In fact No. Each half-reaction is multiplied by the factor that makes the electrons equal, which is the number of electrons in the OTHER half-reaction when the two numbers share no common factor: Al (3 e⁻) × 2 and Ni²⁺ (2 e⁻) × 3 give 6 e⁻ each.
5 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 5
In acidic solution, MnO₄⁻(aq) oxidizes Fe²⁺(aq) to Fe³⁺(aq) and is itself reduced to Mn²⁺(aq). Which equation is the balanced net ionic equation for the reaction? (States are omitted; all ions are aqueous and H₂O is liquid.)
Answer and reasoning
AMnO₄⁻ + 8 H⁺ + Fe²⁺ → Mn²⁺ + 4 H₂O + Fe³⁺ A student who balances only the atoms picks this. Every element balances, but the charge is +9 on the left and +5 on the right: the 5 electrons gained by Mn are not matched by the 1 electron lost by a single Fe²⁺.
BMnO₄⁻ + 5 Fe²⁺ → Mn²⁺ + 5 Fe³⁺ + 4 O²⁻ A student who treats the oxidation number of O (−2) as a real charge picks this, releasing O²⁻ ions. Free O²⁻ ions do not exist in aqueous solution; in acid the O atoms become H₂O, using H⁺.
CMnO₄⁻ + 8 H⁺ + 5 Fe²⁺ → Mn²⁺ + 4 H₂O + 5 Fe³⁺Correct The reduction half-reaction is MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O, and each Fe²⁺ loses one electron, so 5 Fe²⁺ are needed. Atoms balance and the total charge is +17 on each side.
DMnO₄⁻ + 8 H⁺ + Fe²⁺ + 4 e⁻ → Mn²⁺ + 4 H₂O + Fe³⁺ A student who adds the half-reactions as written, keeping the electrons that do not cancel, picks this: 5 e⁻ gained and 1 e⁻ lost leave 4 e⁻ in the equation. Electrons lost must equal electrons gained, so 5 Fe²⁺ are needed and no electrons remain.
Working Reduction: MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O (O balanced with H₂O, H with H⁺; charge +2 = +2). Oxidation: Fe²⁺ → Fe³⁺ + e⁻, multiplied by 5. Sum: MnO₄⁻ + 8 H⁺ + 5 Fe²⁺ → Mn²⁺ + 4 H₂O + 5 Fe³⁺; charge −1 + 8 + 10 = +17 on the left and +2 + 15 = +17 on the right. Adding the half-reactions as written (5 e⁻ gained, 1 e⁻ lost) leaves 4 e⁻ uncancelled.
Hydrogen peroxide, H₂O₂, is oxidized to O₂(g) in acidic solution. When the half-reaction for this oxidation is balanced with the smallest whole-number coefficients, what is the coefficient of e⁻?
Answer and reasoning
A1 A student who counts the oxidation-number change of one O atom only picks this. Both O atoms rise from −1 to 0, so 2 electrons are lost; with 1 e⁻ the charge would be 0 on the left and +1 on the right.
B4 A student who assigns O an oxidation number of −2 in H₂O₂ picks this, counting a rise of 2 for each O atom. In a peroxide each O is −1, so only 2 electrons are lost; with 4 e⁻ the charge would be 0 on the left and −2 on the right.
C0 A student who balances only the atoms picks this: H₂O₂ → O₂ + 2 H⁺ has every atom balanced. The charge is 0 on the left and +2 on the right, so 2 electrons must be added to the product side.
D2Correct In H₂O₂ each O atom has an oxidation number of −1 (each H is +1); in O₂ it is 0. Both O atoms rise by 1, so 2 electrons are lost: H₂O₂ → O₂ + 2 H⁺ + 2 e⁻, with a charge of 0 on each side.
Working Oxidation numbers: in H₂O₂, H is +1, so each O is −1; in O₂, O is 0. Two O atoms each rise by 1, so 2 electrons are lost: H₂O₂ → O₂ + 2 H⁺ + 2 e⁻ (charge 0 = +2 − 2). Distractors: one O atom only, 1; O taken as −2 in H₂O₂, 2 × 2 = 4; atoms balanced without electrons, H₂O₂ → O₂ + 2 H⁺, 0.
Iron(III) ions oxidize tin(II) ions in aqueous solution, forming Fe²⁺(aq) and Sn⁴⁺(aq). A student writes the equation Fe³⁺(aq) + Sn²⁺(aq) → Fe²⁺(aq) + Sn⁴⁺(aq) and claims that it is balanced. Which statement best evaluates the claim?
Answer and reasoning
AIt is not balanced: Sn²⁺ loses 2 electrons and each Fe³⁺ gains 1, so 2 Fe³⁺ are needed.Correct The atoms balance, but the charge is +5 on the left and +6 on the right. Each Sn²⁺ loses 2 electrons and each Fe³⁺ gains 1, so 2 Fe³⁺ must react with each Sn²⁺: 2 Fe³⁺ + Sn²⁺ → 2 Fe²⁺ + Sn⁴⁺, with +8 on each side.
BIt is balanced: each side has one Fe atom and one Sn atom, so every atom is conserved. A student who thinks an equation is balanced once the atoms balance picks this. The charges also have to balance: +5 on the left and +6 on the right shows that the electrons lost and gained are not equal.
CIt is not balanced: 2 Sn²⁺ are needed per Fe³⁺, since the Sn half-reaction has 2 electrons. A student who multiplies each half-reaction by its own number of electrons picks this. Then 2 Sn²⁺ lose 4 electrons while one Fe³⁺ gains 1; it is the iron half-reaction that must be doubled.
DIt is not balanced: the electrons that Sn²⁺ gives to Fe³⁺ must be written in the equation. A student who thinks the transferred electrons belong in the overall equation picks this. When the half-reactions are scaled so that electrons lost equal electrons gained, the electrons cancel and do not appear.
Working Half-reactions: Fe³⁺ + e⁻ → Fe²⁺ (1 e⁻ gained); Sn²⁺ → Sn⁴⁺ + 2 e⁻ (2 e⁻ lost). Electrons must match: 2 Fe³⁺ + Sn²⁺ → 2 Fe²⁺ + Sn⁴⁺. Student's equation: charge +5 on the left, +6 on the right.
A student places a coil of copper wire in an aqueous solution of AgNO₃. Silver crystals form on the wire, and copper ions enter the solution. When the reaction stops, the student removes, washes and dries the silver, then dries and weighs the wire. The table shows the data. Which conclusion is supported by the data?
Answer and reasoning
AOne Ag⁺ ion is reduced for each Cu atom oxidized, so each Cu atom loses one electron. A student who expects one metal to replace the other one for one picks this. The data give 0.0100 mol Ag for 0.00500 mol Cu, a 2 : 1 ratio, so each Cu atom reduces two Ag⁺ ions.
BOne Ag⁺ ion is reduced for each Cu atom oxidized, so each Cu atom loses two electrons. A student who balances only atoms picks this, writing Cu + Ag⁺ → Cu²⁺ + Ag. If each Cu lost two electrons but only one Ag⁺ gained one, charge would not be conserved; the data show 2 mol Ag per mol Cu.
CTwo Ag⁺ ions are reduced for each Cu atom oxidized, so each Cu atom loses two electrons.Correct Cu reacted: 1.524 g − 1.206 g = 0.318 g, or 0.318/63.55 = 0.00500 mol. Ag formed: 1.079/107.87 = 0.0100 mol. That is 2 Ag per Cu; each Ag⁺ gains one electron, so each Cu atom loses two and forms Cu²⁺.
DThree Ag⁺ ions are reduced for each Cu atom oxidized, so each Cu atom loses three electrons. A student who divides the masses instead of the moles picks this: 1.079 g / 0.318 g ≈ 3.4. Converted to moles, the amounts are 0.0100 mol Ag and 0.00500 mol Cu, a 2 : 1 ratio.
Working Mass of Cu that reacted = 1.524 g − 1.206 g = 0.318 g; n(Cu) = 0.318 g / 63.55 g/mol = 0.00500 mol. n(Ag) = 1.079 g / 107.87 g/mol = 0.0100 mol. Ratio Ag : Cu = 0.0100/0.00500 = 2.00, so each Cu atom gives 2 electrons (one to each of two Ag⁺): Cu(s) + 2 Ag⁺(aq) → Cu²⁺(aq) + 2 Ag(s). Mass ratio (wrong): 1.079/0.318 = 3.39 ≈ 3.
Aluminum metal reacts with an acidic solution. In the reaction, Al(s) is oxidized to Al³⁺(aq), and H⁺(aq) is reduced to H₂(g). When the equation for the reaction is balanced with the smallest whole-number coefficients, what are the coefficients of Al(s) and H⁺(aq)?
Answer and reasoning
AAl 2; H⁺ 6Correct Each Al atom loses 3 electrons, and forming one H₂ molecule from 2 H⁺ takes 2 electrons. Multiplying the oxidation by 2 and the reduction by 3 makes 6 electrons lost and 6 gained: 2 Al(s) + 6 H⁺(aq) → 2 Al³⁺(aq) + 3 H₂(g), with a total charge of +6 on each side.
BAl 1; H⁺ 2 A student who balances only the atoms picks this: Al(s) + 2 H⁺(aq) → Al³⁺(aq) + H₂(g). The atoms balance, but the charge is +2 on the left and +3 on the right, because 3 electrons lost do not equal 2 gained.
CAl 1; H⁺ 6 A student who counts the oxidation-number change of one H atom only writes 2 H⁺ + e⁻ → H₂ and picks this, using three of these for each Al. Both H atoms go from +1 to 0, so each H₂ takes 2 electrons; Al + 6 H⁺ → Al³⁺ + 3 H₂ has a charge of +6 on the left and +3 on the right.
DAl 3; H⁺ 4 A student who multiplies each half-reaction by the number of electrons it contains picks this: 3 Al and 2 × (2 H⁺). Then 9 electrons are lost and only 4 gained, and the charge is +4 on the left and +9 on the right.
Working Oxidation: Al(s) → Al³⁺(aq) + 3 e⁻. Reduction: 2 H⁺(aq) + 2 e⁻ → H₂(g) (two H atoms each go from +1 to 0, so 2 electrons are gained). Electrons lost must equal electrons gained; the least common multiple of 3 and 2 is 6, so the oxidation is multiplied by 2 and the reduction by 3: 2 Al(s) + 6 H⁺(aq) → 2 Al³⁺(aq) + 3 H₂(g). Atoms: 2 Al and 6 H on each side; charge: +6 on each side. Distractors: atoms only, Al + 2 H⁺ → Al³⁺ + H₂ (charge +2 and +3); reduction written with 1 e⁻, giving Al + 6 H⁺ → Al³⁺ + 3 H₂ (charge +6 and +3); each half-reaction multiplied by its own electrons, 3 Al + 4 H⁺ → 3 Al³⁺ + 2 H₂ (charge +4 and +9).
Compiled from the AP Chemistry Course and Exam Description (effective Fall 2024) and our question bank · Specialist review in progress. How these pages are made · Free, no account
