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.
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Here are the answers to the questions: a) Determine the voltage of the electrochemical cell. Given half-reactions: Zn^2+(aq) + 2e^- Zn(s) E^ = -0.76 V Ni^2+(aq) + 2e^- Ni(s) E^ = -0.25 V Step 1: Identify the anode (oxidation) and cathode (reduction). The species with the more negative standard reduction potential will be oxidized. In this case, Zn has a more negative E^ (-0.76 V) compared to Ni (-0.25 V). Therefore, Zn will be oxidized (anode) and Ni^2+ will be reduced (cathode). Step 2: Write the half-reactions for the cell. Anode (oxidation): Zn(s) Zn^2+(aq) + 2e^- E^_ox = +0.76 V Cathode (reduction): Ni^2+(aq) + 2e^- Ni(s) E^_red = -0.25 V Step 3: Calculate the standard cell voltage (E^_cell). E^_cell = E^_cathode - E^_anode E^_cell = (-0.25 V) - (-0.76 V) E^_cell = -0.25 V + 0.76 V E^_cell = 0.51 V The voltage of the electrochemical cell is 0.51 V. b) State what would be observed when ethanoic acid reacts with CaCO_3. When ethanoic acid (an acid) reacts with calcium carbonate (a carbonate), carbon dioxide gas is produced. The observation would be the effervescence or fizzing of a colourless gas. c) Calculate the volume of oxygen required to burn completely 4 moles of C_3H_8 at s.t.p. (H = 1.0, C = 12.0, O = 16.0, 1 mole of gas occupies 22.4 dm^3 at s.t.p.) Step 1: Write the balanced chemical equation for the complete combustion of propane (C_3H_8). C_3H_8(g) + 5O_2(g) 3CO_2(g) + 4H_2O(l) Step 2: Determine the mole ratio between C_3H_8 and O_2. From the balanced equation, 1 mole of C_3H_8 reacts with 5 moles of O_2. Step 3: Calculate the moles of O_2 required for 4 moles of C_3H_8. Moles of O_2 = 4 mol C_3H_8 × 5 mol O_21 mol C_3H_8 = 20 mol O_2 Step 4: Calculate the volume of O_2 at s.t.p. At s.t.p., 1 mole of any gas occupies 22.4 dm^3. Volume of O_2 = 20 mol × 22.4 dm^3/mol = 448 dm^3 The volume of oxygen required is 448 dm^3. What's next?