Redox Reactions: Ion-Electron Method and Stoichiometry Problems
100 mL of an aqueous solution of iron(II) chloride are reacted, in acidic medium, with a 0.35 M solution of K2Cr2O7, requiring 64.4 mL of the latter to complete the oxidation. In the reaction, iron(II) is oxidized to iron(III) and the Cr2O72- ion is reduced to chromium(III).
Balance the ionic equation of the reaction using the ion-electron method.
Calculate the molarity of the iron(II) chloride solution.
Oxidation half-reaction: (Fe2+ → Fe3+ + 1e-) × 6
Reduction half-reaction: (Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O) × 1
Ionic equation: 6Fe2+ + Cr2O72- + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O
Once the reaction is balanced and taking its stoichiometry into account: 0.0644 L × 0.35 mol Cr2O7/1 L × 6 mol Fe/1 mol Cr2O7 = 0.135 mol FeCl2 → M = 0.135/0.1 = 1.35 M
Balancing a redox equation and calculating the mass of KClO3
Using the ion-electron method, balance the redox equation corresponding to the following reaction: KClO3(s) + SbCl3(s) + HCl(aq) → SbCl5(s) + KCl(s) + H2O(l)
Calculate the grams of KClO3 needed to obtain 200 g of SbCl5, if the yield of the reaction is 50%.
Oxidation half-reaction: (Sb3+ → Sb5+ + 2e-) × 3
Reduction half-reaction: (ClO3- + 6H+ + 6e- → Cl- + 3H2O) × 1
Ionic equation: 3Sb3+ + ClO3- + 6H+ → 3Sb5+ + Cl- + 3H2O
Molecular equation: 3SbCl3 + KClO3 + 6HCl → 3SbCl5 + KCl + 3H2O
Once the reaction is balanced and taking its stoichiometry into account: 200 g SbCl5 × 1 mol SbCl5/299.3 g × 1 mol KClO3/3 mol SbCl5 × 122.6 g KClO3/1 mol KClO3 × 100/50 = 54.6 g KClO3
Oxidation of iron(II) by potassium dichromate
Iron(II) can be oxidized by an acidic solution of potassium dichromate according to the following ionic equation: Cr2O72- + Fe2+ → Cr3+ + Fe3+
Balance the ionic reaction that takes place using the ion-electron method.
If 26.0 mL of a 0.0250 M potassium dichromate solution are used to titrate 25.0 mL of a solution containing Fe2+, what is the concentration of the Fe2+ solution?
Oxidation half-reaction: (Fe2+ → Fe3+ + 1e-) × 6
Reduction half-reaction: (Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O) × 1
Ionic equation: 6Fe2+ + Cr2O72- + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O
Once the reaction is balanced and taking its stoichiometry into account: 0.026 L × 0.025 mol Cr2O72-/1 L × 6 mol Fe2+/1 mol Cr2O72- = 0.039 mol Fe2+ → M = 0.039/0.025 = 0.156 M
Reaction of copper with nitric acid
Given the following reaction: Cu(s) + HNO3(aq) → Cu(NO3)2(aq) + NO(g) + H2O(l)
Write and balance the molecular equation using the ion-electron method, indicating the corresponding half-reactions.
Calculate the volume of NO measured under normal conditions that will be released for every 100 g of copper that reacts, if the yield of the process is 80%.
Oxidation half-reaction: (Cu → Cu2+ + 2e-) × 3
Reduction half-reaction: (NO3- + 4H+ + 3e- → NO + 2H2O) × 2
Ionic equation: 3Cu + 2NO3- + 8H+ → 3Cu2+ + 2NO + 4H2O
Molecular equation: 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O
Once the reaction is balanced and taking its stoichiometry into account: 100 g Cu × 1 mol Cu/63.5 g Cu × 2 mol NO/3 mol Cu × 80/100 = 0.843 mol NO
To calculate the volume of NO, we know that 1 mol of gas under normal conditions occupies 22.4 L → 0.843 × 22.4 = 18.8 L NO
Preparation of chlorine gas
Chlorine gas is obtained by the oxidation of HCl with HNO3, producing NO2 and H2O as well.
Balance the molecular reaction using the ion-electron method.
Calculate the volume of chlorine obtained, at 25 °C and 1 atm, when 500 mL of a 2 M aqueous HCl solution react with excess HNO3, if the yield of the reaction is 80%.
Oxidation half-reaction: 2Cl- → Cl2 + 2e-
Reduction half-reaction: (NO3- + 2H+ + 1e- → NO2 + H2O) × 2
Ionic equation: 2Cl- + 2NO3- + 4H+ → Cl2 + 2NO2 + 2H2O
Molecular equation: 2HCl + 2HNO3 → Cl2 + 2NO2 + 2H2O
Once the reaction is balanced and taking its stoichiometry into account: 0.5 L × 2 mol HCl/1 L × 1 mol Cl2/2 mol HCl × 80/100 = 0.4 mol Cl2
PV = nRT → 0.4 × 0.082 × 298/1 = 9.8 L