- An electroplating process uses a current of 3.5 A to deposit silver from AgNO₃ solution onto a steel substrate. After 25 minutes, the actual mass of silver deposited is 4.1 g. What is the current efficiency, and what is the most likely cause of the efficiency loss?
Answer: 70% efficiency; loss due to hydrogen gas evolution competing with silver reduction
Theoretical calculation: Q = 3.5 A × 25 × 60 s = 5250 C. Moles of electrons = 5250/96485 = 0.0544 mol. Since Ag⁺ + e⁻ → Ag, theoretical mass = 0.0544 mol × 107.9 g/mol = 5.87 g. Current efficiency = (4.1/5.87) × 100% = 70%. The main cause of efficiency loss in silver electroplating is competing reduction of H⁺ or H₂O to form H₂ gas at the cathode.
- A compound with the molecular formula C₃H₈O is oxidized by heating with acidified potassium dichromate(VI) solution to form a product, Y. Compound Y does not give a positive test with Tollens' reagent. What is the systematic name of the starting compound?
Answer: Propan-2-ol
The starting compound C₃H₈O is an alcohol. The product Y does not react with Tollens' reagent, which means it is not an aldehyde. Therefore, Y must be a ketone. Ketones are formed from the oxidation of secondary alcohols. The secondary alcohol with the formula C₃H₈O is propan-2-ol. Its oxidation produces propanone, a ketone. Propan-1-ol is a primary alcohol and would be oxidized to propanal (an aldehyde) and then to propanoic acid.
- In a fuel cell operating on the reaction H₂ + ½O₂ → H₂O, the theoretical cell voltage is 1.23 V under standard conditions. During operation at 0.8 V with a current density of 200 mA/cm², the cathode has an area of 50 cm². What is the power output per unit cathode area, and what is the primary cause of the voltage loss from theoretical?
Answer: Power density is 160 mW/cm²; voltage loss is primarily due to activation overpotential for the oxygen reduction reaction at the cathode surface
Power density = voltage × current density = 0.8 V × 200 mA/cm² = 160 mW/cm². The voltage loss is 1.23 - 0.8 = 0.43 V. In fuel cells, the largest voltage losses typically occur due to activation overpotential, especially for the oxygen reduction reaction (ORR) at the cathode, which is kinetically slow and requires significant overpotential to proceed at practical rates. At moderate current densities like 200 mA/cm², activation losses dominate over ohmic losses. The cathode area (50 cm²) affects total current but not current density or power density calculations. Choices C and D incorrectly calculate power density, while A and C incorrectly identify ohmic resistance as the primary loss mechanism.
- Consider the spontaneous reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). Which statement provides the most accurate description of this process?
Answer: Copper(II) ions are reduced by zinc metal.
In this reaction, zinc metal (Zn) loses electrons to become Zn²⁺ ions, so it is oxidized. It acts as the reducing agent. Copper(II) ions (Cu²⁺) gain electrons to become copper metal (Cu), so they are reduced. They act as the oxidizing agent. Therefore, the copper(II) ions are reduced by the zinc metal. A and C are incorrect because the species acting is the copper(II) ion, not copper metal. D is incorrect because the reducing agent is zinc metal, not zinc(II) ions.
- The reaction between ethanol and acidified dichromate(VI) ions produces ethanoic acid and chromium(III) ions. What is the total change in the oxidation state of chromium for one mole of dichromate(VI) ions, Cr₂O₇²⁻, that reacts?
Answer: A decrease of 6
First, determine the oxidation state of Cr in Cr₂O₇²⁻. Let the oxidation state be x. The oxidation state of O is -2. So, 2x + 7(-2) = -2, which gives 2x = +12, so x = +6. The final product is chromium(III) ion, Cr³⁺, where the oxidation state is +3. The change for one Cr atom is from +6 to +3, which is a change of -3. Since there are two Cr atoms in one Cr₂O₇²⁻ ion, the total change per ion is 2 × (-3) = -6. Therefore, the total change is a decrease of 6.
- What is the change in oxidation state of the carbonyl carbon atom when ethanal (CH₃CHO) is reduced to ethanol (CH₃CH₂OH)?
Answer: From +1 to -1
In ethanal (CH₃CHO), the carbonyl carbon is bonded to one H (+1), one C (0 contribution), and double-bonded to one O (-2). The oxidation state is +1. In ethanol (CH₃CH₂OH), this carbon is bonded to two H atoms (+1 each), one O (-2), and one C (0 contribution). The oxidation state is -1. The change is from +1 to -1, representing a reduction (gain of electrons).
- Propanal is treated with a reducing agent such as lithium aluminium hydride (LiAlH₄), followed by aqueous acid work-up. What is the major organic product?
Answer: Propan-1-ol
The reduction of an aldehyde yields a primary alcohol. Propanal is a three-carbon aldehyde. Its reduction will add two hydrogen atoms across the C=O double bond, converting the aldehyde functional group (-CHO) into a primary alcohol functional group (-CH₂OH). The product is therefore propan-1-ol. Propan-2-ol would be formed from the reduction of a ketone (propanone). Propanoic acid is an oxidation product of propanal.
- The following reactions are observed to occur spontaneously:
I. Zn(s) + Pb²⁺(aq) → Zn²⁺(aq) + Pb(s)
II. Fe(s) + Zn²⁺(aq) → No reaction
Based on this information, what is the correct order of increasing strength as reducing agents?
Answer: Pb < Fe < Zn
A stronger reducing agent is more easily oxidized (more reactive). From reaction I, Zn reduces Pb²⁺, so Zn is a stronger reducing agent than Pb. From reaction II, Fe does not reduce Zn²⁺, so Zn is a stronger reducing agent than Fe. To compare Fe and Pb, we can infer the reactivity series: Zn > Fe and Zn > Pb. A more reactive metal is a stronger reducing agent. The general reactivity series places Zn > Fe > Pb. Therefore, the order of increasing strength as reducing agents is Pb < Fe < Zn.
- A voltaic cell is set up with two unknown metals, P and Q, and their corresponding 1.0 mol dm⁻³ nitrate solutions. It is observed that the mass of electrode P increases and the concentration of Qⁿ⁺(aq) ions increases. Which conclusion must be correct?
Answer: Metal Q is more reactive than metal P, and electrons flow from Q to P.
An increase in the mass of electrode P means that reduction is occurring there (Pⁿ⁺(aq) + ne⁻ → P(s)), so P is the cathode. An increase in the concentration of Qⁿ⁺(aq) ions means that oxidation is occurring at electrode Q (Q(s) → Qⁿ⁺(aq) + ne⁻), so Q is the anode. In a voltaic cell, the more reactive metal is oxidized at the anode. Therefore, metal Q is more reactive than metal P. Electrons always flow from the anode (Q) to the cathode (P) through the external wire.
- Which of the following chemical equations represents a reaction that is not a redox reaction?
Answer: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)
A redox reaction involves a change in oxidation states. In reaction C, a precipitation reaction, the oxidation states of all elements remain constant: Ag is +1, N is +5, O is -2, Na is +1, and Cl is -1 on both sides of the equation. In A, Na changes from 0 to +1 and H from +1 to 0. In B, Cu changes from +2 to 0 and H from 0 to +1. In D, Cl changes from 0 to -1 and I from -1 to 0.
- But-2-ene reacts with hydrogen gas in the presence of a platinum catalyst. Which statement best describes this reaction?
Answer: It is an addition reaction where the carbon atoms are reduced.
The reaction is the hydrogenation of an alkene (but-2-ene + H₂ → butane), which is a type of addition reaction. During this reaction, the C=C double bond is broken and C-H bonds are formed. The oxidation state of the carbon atoms involved in the double bond decreases (from -1 to -2), so the carbon atoms are reduced. Therefore, it is both an addition and a reduction reaction. Option B correctly identifies both aspects.
- What are the expected products at the anode (positive electrode) and cathode (negative electrode) during the electrolysis of molten aluminium oxide, Al₂O₃?
Answer: Anode: oxygen; Cathode: aluminium
In the electrolysis of molten Al₂O₃, the ions present are Al³⁺ and O²⁻. At the cathode (negative electrode), reduction occurs, so positive Al³⁺ ions are attracted and reduced to aluminium metal: Al³⁺ + 3e⁻ → Al(l). At the anode (positive electrode), oxidation occurs, so negative O²⁻ ions are attracted and oxidized to oxygen gas: 2O²⁻ → O₂(g) + 4e⁻.
- In a redox titration, 25.0 mL of 0.150 M Fe²⁺ solution is titrated with 0.0200 M MnO₄⁻ in acidic conditions. The reaction is: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. What volume of MnO₄⁻ is required to reach the equivalence point?
Answer: 37.5 mL because 5 moles Fe²⁺ react per mole MnO₄⁻
Moles of Fe²⁺ = 0.025 L × 0.150 M = 0.00375 mol. From the balanced equation, 5 mol Fe²⁺ react with 1 mol MnO₄⁻, so moles of MnO₄⁻ needed = 0.00375 mol Fe²⁺ × (1 mol MnO₄⁻/5 mol Fe²⁺) = 0.000750 mol. Volume of MnO₄⁻ = 0.000750 mol ÷ 0.0200 M = 0.0375 L = 37.5 mL.
- A voltaic cell is constructed using a magnesium electrode in a 1.0 mol dm⁻³ Mg(NO₃)₂ solution and a silver electrode in a 1.0 mol dm⁻³ AgNO₃ solution. Magnesium is more reactive than silver. Which statement is correct as the cell operates?
Answer: The magnesium electrode is the negative electrode, and its mass decreases over time.
Since magnesium is more reactive, it will be oxidized at the anode: Mg(s) → Mg²⁺(aq) + 2e⁻. The anode is the negative electrode in a voltaic cell. As solid Mg is converted to Mg²⁺ ions, the mass of the magnesium electrode decreases. A: Ag⁺ is reduced to Ag, so its concentration decreases. B: Electrons flow from the anode (Mg) to the cathode (Ag). D: The magnesium half-cell is producing positive Mg²⁺ ions, so anions from the salt bridge will migrate there to balance the charge.
- Equal molar quantities of four different powdered metals are added to separate beakers containing 1.0 mol dm⁻³ hydrochloric acid. The time taken for the metal to be completely consumed is recorded. Which metal would be expected to be consumed in the shortest time?
Answer: Magnesium
The rate of reaction between a metal and an acid is determined by the metal's reactivity. A more reactive metal will react more vigorously and thus be consumed faster. The reactivity series for these metals is Magnesium > Zinc > Iron > Lead. Magnesium is the most reactive metal listed, so it will react the fastest with HCl and be consumed in the shortest time.
- In a titration, 25.00 cm³ of hydrogen peroxide, H₂O₂, reacts completely with 30.00 cm³ of 0.0500 mol dm⁻³ acidified potassium permanganate, KMnO₄. The unbalanced equation is:
MnO₄⁻(aq) + H₂O₂(aq) + H⁺(aq) → Mn²⁺(aq) + O₂(g) + H₂O(l)
The mole ratio of MnO₄⁻ to H₂O₂ in the balanced equation is 2:5. What is the concentration of the H₂O₂ solution?
Answer: 0.150 mol dm⁻³
1. Calculate moles of MnO₄⁻ used: n(MnO₄⁻) = C × V = 0.0500 mol dm⁻³ × (30.00/1000) dm³ = 0.00150 mol. 2. Use the mole ratio to find moles of H₂O₂: n(H₂O₂) = n(MnO₄⁻) × (5/2) = 0.00150 mol × 2.5 = 0.00375 mol. 3. Calculate the concentration of H₂O₂: C(H₂O₂) = n / V = 0.00375 mol / (25.00/1000) dm³ = 0.150 mol dm⁻³.
- Which statement best explains why a salt bridge is necessary for a voltaic cell to operate over a period of time?
Answer: It allows ions to migrate between the half-cells to maintain electrical neutrality.
The function of the salt bridge is to maintain charge neutrality in the half-cells. As the cell operates, positive ions (cations) are produced at the anode and consumed at the cathode. Without the salt bridge, a build-up of positive charge in the anode compartment and negative charge (from the anions of the electrolyte) in the cathode compartment would quickly stop the flow of electrons. The salt bridge allows anions to flow to the anode compartment and cations to flow to the cathode compartment, neutralizing this charge build-up and completing the electrical circuit.
- Which comparison between a primary (voltaic) cell and a secondary (rechargeable) cell is correct?
Answer: The reactions in a secondary cell can be reversed by applying an external current, but those in a primary cell cannot.
The key difference is reversibility. Both cell types use spontaneous redox reactions to generate electricity (discharge). However, the reactions in a secondary cell are designed to be reversible, allowing it to be recharged by an external power source. Primary cells are designed for single use. A is incorrect because both use redox. C is incorrect because both convert chemical to electrical energy during discharge. D is incorrect because oxidation always occurs at the anode, which is the negative electrode in any discharging voltaic/galvanic cell, primary or secondary.
- Consider the reaction: 2H₂S(g) + SO₂(g) → 3S(s) + 2H₂O(l). Which statement is correct?
Answer: Sulfur dioxide (SO₂) is the oxidizing agent because its sulfur atom is reduced.
In H₂S, the oxidation state of S is -2. In SO₂, the oxidation state of S is +4. In the product, elemental sulfur (S), the oxidation state is 0. The sulfur in H₂S is oxidized (-2 to 0), making H₂S the reducing agent. The sulfur in SO₂ is reduced (+4 to 0), making SO₂ the oxidizing agent. A is incorrect because the species that is oxidized is the reducing agent. C is incorrect as this is comproportionation, not disproportionation. D is incorrect as the oxidation state of oxygen remains -2.
- Based on these observations, what is the correct order of decreasing reactivity for these metals?
Answer: X > Z > Y
Experiment 1 shows that metal X can displace Y from its salt solution, so X is more reactive than Y (X > Y). Experiment 2 shows that metal Y cannot displace Z, so Z is more reactive than Y (Z > Y). Experiment 3 shows that metal X can displace Z, so X is more reactive than Z (X > Z). Combining these results gives the order of decreasing reactivity as X > Z > Y.
