1575454
Description
Flashcards by Robert Hebbs, updated more than 1 year ago
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Created by Robert Hebbs
about 10 years ago
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| Question | Answer |
| What has been oxidised? \(Cu^{2+} + Zn → Cu + Zn^{2+}\) | Zn |
| What is the oxidising agent? \(Cu^{2+} + Zn → Cu + Zn^{2+}\) | \(Cu^{2+}\) |
| What has been reduced? \(Cl_2 + 2Na → 2Na^+ + 2Cl^- \) | \(Cl_2\) |
| What is the reducing agent? \(Cl_2 + 2Na → 2Na^+ + 2Cl^- \) | Na |
| What has been oxidised? \(Cr_2O_7^{2-} + ClO_2^- → Cr^{3+} + ClO_4^-\) | \(ClO_4^-\) |
| What is the oxidising agent? \(Cr_2O_7^{2-} + ClO_2^- → Cr^{3+} + ClO_4^-\) | \(Cr_2O_7^{2-}\) |
| What has been reduced? \(Pb + Fe^{3+} → Fe^{2+} + Pb^{2+}\) | \(Fe^{3+}\) |
| What is the reducing agent? \(Pb + Fe^{3+} → Fe^{2+} + Pb^{2+}\) | Pb |
| Balance this half equation \(Zn → Zn^{2+} \) | \(Zn → Zn^{2+} + 2e^- \) |
| Balance this half equation \(Fe^{3+} → Fe^{2+} \) | Balance this half equation \(Fe^{3+} + e^- → Fe^{2+} \) |
| Balance this half equation \(Cl^- → Cl_2 \) | Balance this half equation \(2Cl^- → Cl_2 + 2e^- \) |
| Balance this half equation \(O_2 → O^{2-} \) | Balance this half equation \(O_2 + 4e^- → 2O^{2-} \) |
| Balance this half equation in acidic conditions \(MnO_4^- → Mn^{2+}\) | \(MnO_4^- + 8H^+ + 5e^- → Mn^{2+}\ + 4H_2O \) |
| Balance this half equation in acidic conditions \(S_2O_3^{2-} → SO_4^{2-}\) | \(S_2O_3^{2-} + 5H_2O → 2SO_4^{2-} + 10H^+ + 8e^-\) |
| Balance this half equation in acidic conditions \(Cr^{3+} → Cr_2O_7^{2-} \) | \(2Cr^{3+} + 7H_2O → Cr_2O_7^{2-} + 14H^+ + 6e^-\) |
| Combine these half equations \(Ag^+ + e^- → Ag\) \(Cu → Cu^{2+} + 2e^-\) | \(2Ag^+ + Cu → 2Ag + Cu^{2+}\) |
| Combine these half equations \(Co^{3+} + e^- → Co^{2+}\) \(2Cl^- → Cl_2 + 2e^-\) | \(2Co^{3+} 2Cl^- → 2Co^{2+} + Cl_2 \) |
| Combine these half equations \(Fe → Fe^{3+} + 3e^-\) \(MnO_4^- + 8H^+ + 5e^- → Mn^{2+} + 4H_2O\) | \(5Fe + 3MnO_4^- + 24H^+ → 5Fe^{3+} + 3Mn^{2+} + 12H_2O\) |
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