This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
ScanSolve AI Answer
You're on a roll — here are the solutions to the questions from the image. Question 11: Study the electrode potentials in the table below and answer the question that follow: (Letters are not the actual symbols of elements) | Half-reaction | E° / Volts | |---|---| | H²⁺(aq) + 2e⁻ → H(s) | +0.34 | | Z²⁺(aq) + 2e⁻ → Z(s) | -2.38 | | G⁺(aq) + e⁻ → G(s) | +0.80 | | T²⁺(aq) + 2e⁻ → T(s) | -2.87 | a) Which one is the strongest reducing agent? A strong reducing agent is easily oxidized, meaning it has the most negative (or least positive) standard reduction potential. Comparing the given E° values: • H²⁺/H: +0.34 V • Z²⁺/Z: -2.38 V • G⁺/G: +0.80 V • T²⁺/T: -2.87 V The most negative standard reduction potential is for T²⁺/T (-2.87 V). This indicates that T(s) is most easily oxidized. The strongest reducing agent is T(s). b) Write the ionic equation for the reaction that takes place when Z is dipped in a solution of G⁺ ions. Step 1: Identify the species that will be oxidized and reduced. Compare the standard reduction potentials for Z and G: • Z²⁺(aq) + 2e⁻ → Z(s) ; E° = -2.38 V • G⁺(aq) + e⁻ → G(s) ; E° = +0.80 V Since Z has a more negative reduction potential than G, Z will be oxidized (act as the reducing agent) and G⁺ will be reduced (act as the oxidizing agent). Step 2: Write the half-reactions. Oxidation (anode): Z(s) Z^2+(aq) + 2e^- Reduction (cathode): G^+(aq) + e^- G(s) Step 3: Balance the electrons and combine the half-reactions. Multiply the reduction half-reaction by 2 to balance the electrons: 2G^+(aq) + 2e^- 2G(s) Add the oxidation and balanced reduction half-reactions: Z(s) + 2G^+(aq) + 2e^- Z^2+(aq) + 2e^- + 2G(s) Cancel out the electrons: The ionic equation is Z(s) + 2G^+(aq) Z^2+(aq) + 2G(s). c) Calculate the E° cell value of the reaction in 11(b) above. The standard cell potential (E° cell) is calculated as: E^_cell = E^_reduction - E^_oxidation From part (b), G⁺ is reduced and Z is oxidized. E^_reduction = E^(G^+/G) = +0.80 V E^_oxidation = E^(Z^2+/Z) = -2.38 V E^_cell = (+0.80 V) - (-2.38 V) E^_cell = +0.80 V + 2.38 V E^_cell = +3.18 V Question 12: When a hydrocarbon was completely burnt in oxygen, 4.2g of Carbon (IV) oxide and 1.71g of water were formed. Determine the empirical formula of the hydrocarbon. (H=1.0, C=12.0, O=16.0) Step 1: Calculate the moles of carbon from CO₂. Molar mass of CO₂ = 12.0 + 2(16.0) = 44.0 g/mol Moles of C = Moles of CO_2 = Mass of CO_2Molar mass of CO_2 Moles of C = 4.2 g44.0 g/mol ≈ 0.09545 mol Step 2: Calculate the moles of hydrogen from H₂O. Molar mass of H₂O = 2(1.0) + 16.0 = 18.0 g/mol Moles of H_2O = Mass of H_2OMolar mass of H_2O Moles of H_2O = 1.71 g18.0 g/mol = 0.095 mol Since each mole of H₂O contains 2 moles of H atoms: Moles of H = 0.095 mol × 2 = 0.190 mol Step 3: Determine the simplest whole-number ratio of C to H. Divide