Flashcards
Oxidation Skills
Cards in this deck (18)
- 1What has been oxidised? \(Cu^{2+} + Zn → Cu + Zn^{2+}\)Zn
- 2What is the oxidising agent? \(Cu^{2+} + Zn → Cu + Zn^{2+}\)\(Cu^{2+}\)
- 3What has been reduced? \(Cl_2 + 2Na → 2Na^+ + 2Cl^- \)\(Cl_2\)
- 4What is the reducing agent? \(Cl_2 + 2Na → 2Na^+ + 2Cl^- \)Na
- 5What has been oxidised? \(Cr_2O_7^{2-} + ClO_2^- → Cr^{3+} + ClO_4^-\)\(ClO_4^-\)
- 6What is the oxidising agent? \(Cr_2O_7^{2-} + ClO_2^- → Cr^{3+} + ClO_4^-\)\(Cr_2O_7^{2-}\)
- 7What has been reduced? \(Pb + Fe^{3+} → Fe^{2+} + Pb^{2+}\)\(Fe^{3+}\)
- 8What is the reducing agent? \(Pb + Fe^{3+} → Fe^{2+} + Pb^{2+}\)Pb
- 9Balance this half equation \(Zn → Zn^{2+} \)\(Zn → Zn^{2+} + 2e^- \)
- 10Balance this half equation \(Fe^{3+} → Fe^{2+} \)Balance this half equation
\(Fe^{3+} + e^- → Fe^{2+} \) - 11Balance this half equation \(Cl^- → Cl_2 \)Balance this half equation
\(2Cl^- → Cl_2 + 2e^- \) - 12Balance this half equation \(O_2 → O^{2-} \)Balance this half equation
\(O_2 + 4e^- → 2O^{2-} \) - 13Balance this half equation in acidic conditions \(MnO_4^- → Mn^{2+}\)\(MnO_4^- + 8H^+ + 5e^- → Mn^{2+}\ + 4H_2O \)
- 14Balance this half equation in acidic conditions \(S_2O_3^{2-} → SO_4^{2-}\)\(S_2O_3^{2-} + 5H_2O → 2SO_4^{2-} + 10H^+ + 8e^-\)
- 15Balance this half equation in acidic conditions \(Cr^{3+} → Cr_2O_7^{2-} \)\(2Cr^{3+} + 7H_2O → Cr_2O_7^{2-} + 14H^+ + 6e^-\)
- 16Combine these half equations \(Ag^+ + e^- → Ag\) \(Cu → Cu^{2+} + 2e^-\)\(2Ag^+ + Cu → 2Ag + Cu^{2+}\)
- 17Combine these half equations \(Co^{3+} + e^- → Co^{2+}\) \(2Cl^- → Cl_2 + 2e^-\)\(2Co^{3+} 2Cl^- → 2Co^{2+} + Cl_2 \)
- 18Combine these half equations \(Fe → Fe^{3+} + 3e^-\) \(MnO_4^- + 8H^+ + 5e^- → Mn^{2+} + 4H_2O\)\(5Fe + 3MnO_4^- + 24H^+ → 5Fe^{3+} + 3Mn^{2+} + 12H_2O\)
Information
- Description
- A Level Year 13 Chemistry Flashcards on Oxidation Skills, created by Robert Hebbs on 03/11/2014.
- Created by
- Robert Hebbs
- Views
- 214
- Rating