Redox Practice Problems Key For each of following skeletal redox reactions: a. Identify the oxidized species and state how many electrons are lost per atom b. Identify the reduced species and state how many electrons are gained per atom c. Identify the oxidizing agent d. Identify the reducing agent e. Write balanced half reactions (in acid) f. Write a balanced redox reaction (in acid) g. Write a balanced redox reaction (in base) 1. MnO4‐(aq) + H2O2(l) Æ Mn2+(aq) + O2(g) First write oxidation states for each element, per atom. +7 4(‐2) 2(+1) ‐2 +2 2(0) MnO4‐(aq) + H2O2(l) Æ Mn2+(aq) + O2(g) a. Identify the oxidized species and state how many electrons are lost per atom • Each O lost 1e‐ b. Identify the reduced species and state how many electrons are gained per atom • Each Mn gained 5e‐ c. Identify the oxidizing agent • MnO4‐(aq) d. Identify the reducing agent • H2O2(l) e. Write balanced half reactions (in acid) • Write Unbalanced half reactions including electron change per atom H2O2(l) Æ O2(g) + 1e‐ ‐
5e + MnO4‐(aq) Æ Mn2+(aq) • Balance half reactions for same # of oxidized/reduced species H2O2(l) Æ O2(g) + 2(1e‐) 5e‐ + MnO4‐(aq) Æ Mn2+(aq) • Balance half reactions for # of O, with H2O, and # of H, with H+ H2O2(l) Æ O2(g) + 2e‐ + 2H+(aq) 8H+(aq) + 5e‐ + MnO4‐(aq) Æ Mn2+(aq) + 4H2O(l) Now the half reactions are balanced within themselves, but not balanced for # of electrons gained and lost • Balance half reactions for # of electrons gained and lost 5(H2O2(l) Æ O2(g) + 2e‐ + 2H+(aq)) +
2(8H (aq) + 5e‐ + MnO4‐(aq) Æ Mn2+(aq) + 4H2O(l)) • Rewriting the half reactions with the multiplication done 5H2O2(l) Æ 5O2(g) + 10e‐ + 10H+(aq) +
16H (aq) + 10e‐ + 2MnO4‐(aq) Æ 2Mn2+(aq) + 8H2O(l) f. Write a balanced redox reaction (in acid) • Combine half reactions 16H+(aq) + 10e‐ + 5H2O2(l) + 2MnO4‐(aq) Æ 2Mn2+(aq) + 5O2(g) + 8H2O(l) + 10e‐ + 10H+(aq) • Cross out spectators 6H+(aq) + 5H2O2(l) + 2MnO4‐(aq) Æ 2Mn2+(aq) + 5O2(g) + 8H2O(l) g. Write a balanced redox reaction (in base) • Count the # of H+(aq). Add the same # of OH‐(aq) to both sides of the equation 6OH‐(aq) + 6H+(aq) + 5H2O2(l) + 2MnO4‐(aq) Æ 2Mn2+(aq) + 5O2(g) + 8H2O(l) + 6OH‐(aq) • On the side with both H+(aq) & OH‐(aq), those become water 6H2O(l)+ 5H2O2(l) + 2MnO4‐(aq) Æ 2Mn2+(aq) + 5O2(g) + 8H2O(l) + 6OH‐(aq) • Cross out spectators 5H2O2(l) + 2MnO4‐(aq) Æ 2Mn2+(aq) + 5O2(g) + 2H2O(l) + 6OH‐(aq) 2. CrO42‐(aq) + Cu(s) Æ Cr(OH)3(s) + Cu(OH)2(s) First write oxidation states for each element, per atom. +6 4(‐2) 0 +3 2(‐1) +2 2(‐1) CrO42‐(aq) + Cu(s) Æ Cr(OH)3(s) + Cu(OH)2(s) a. Identify the oxidized species and state how many electrons are lost per atom • Each Cu lost 2e‐ b. Identify the reduced species and state how many electrons are gained per atom • Each Cr gained 3e‐ c. Identify the oxidizing agent • CrO42‐(aq) d. Identify the reducing agent • Cu(s) e. Write balanced half reactions (in acid) • Write Unbalanced half reactions including electron change per atom Cu(s) Æ Cu(OH)2(s)+ 2e‐ 3e‐ + CrO42‐(aq) Æ Cr(OH)3(s) • Balance half reactions for same # of oxidized/reduced species Cu(s) Æ Cu(OH)2(s)+ 2e‐ 3e‐ + CrO42‐(aq) Æ Cr(OH)3(s) • Balance half reactions for # of O, with H2O, and # of H, with H+ 2H2O(l) + Cu(s) Æ Cu(OH)2(s)+ 2e‐+ 2H+(aq) 5H+(aq)+ 3e‐ + CrO42‐(aq) Æ Cr(OH)3(s) + H2O(l) Now the half reactions are balanced within themselves, but not balanced for # of electrons gained and lost • Balance half reactions for # of electrons gained and lost 3(2H2O(l) + Cu(s) Æ Cu(OH)2(s)+ 2e‐+ 2H+(aq)) 2(5H+(aq)+ 3e‐ + CrO42‐(aq) Æ Cr(OH)3(s) + H2O(l)) • Rewriting the half reactions with the multiplication done 6H2O(l) + 3Cu(s) Æ 3Cu(OH)2(s)+ 6e‐+ 6H+(aq) 10H+(aq)+ 6e‐ + 2CrO42‐(aq) Æ 2Cr(OH)3(s) + 2H2O(l) f. Write a balanced redox reaction (in acid) • Combine half reactions 6H2O(l) + 3Cu(s) + 10H+(aq)+ 6e‐ + 2CrO42‐(aq) Æ 3Cu(OH)2(s)+ 2Cr(OH)3(s) + 2H2O(l) + 6e‐+ 6H+(aq) • Cross out spectators 4H2O(l) + 4H+(aq)+ + 3Cu(s) + 2CrO42‐(aq) Æ 3Cu(OH)2(s)+ 2Cr(OH)3(s) g. Write a balanced redox reaction (in base) • Count the # of H+(aq). Add the same # of OH‐(aq) to both sides of the equation 4OH‐(aq) + 4H2O(l) + 4H+(aq)+ + 3Cu(s) + 2CrO42‐(aq) Æ 3Cu(OH)2(s)+ 2Cr(OH)3(s) + 4OH‐(aq) • On the side with both H+(aq) & OH‐(aq), those become water 4 H2O(l) + 4H2O(l) + 3Cu(s) + 2CrO42‐(aq) Æ 3Cu(OH)2(s)+ 2Cr(OH)3(s) + 4OH‐(aq) • Cross out spectators 8H2O(l) + 3Cu(s) + 2CrO42‐(aq) Æ 3Cu(OH)2(s)+ 2Cr(OH)3(s) + 4OH‐(aq) 3. AsO43‐(aq) + NO2‐(l) Æ AsO2‐(aq) + NO3‐(l) First write oxidation states for each element, per atom. +5 4(‐2) +3 2(‐2) +3 2(‐2) +5 3(‐2) AsO43‐(aq) + NO2‐(l) Æ AsO2‐(aq) + NO3‐(l) a. Identify the oxidized species and state how many electrons are lost per atom • Each N lost 2e‐ b. Identify the reduced species and state how many electrons are gained per atom • Each As gained 2e‐ c. Identify the oxidizing agent d. e. f. g. • AsO43‐(aq) Identify the reducing agent • NO2‐(l) Write balanced half reactions (in acid) • Write Unbalanced half reactions including electron change per atom NO2‐(l) Æ NO3‐(l)+ 2e‐ 2e‐ + AsO43‐(aq) Æ AsO2‐(aq) • Balance half reactions for same # of oxidized/reduced species NO2‐(l) Æ NO3‐(l)+ 2e‐ ‐
2e + AsO43‐(aq) Æ AsO2‐(aq) • Balance half reactions for # of O, with H2O, and # of H, with H+ H2O(l) + NO2‐(l) Æ NO3‐(l)+ 2e‐ + 2H+(aq) +
4H (aq) + 2e‐ + AsO43‐(aq) Æ AsO2‐(aq) + 2H2O(l) Now the half reactions are balanced within themselves, but not balanced for # of electrons gained and lost • Balance half reactions for # of electrons gained and lost H2O(l) + NO2‐(l) Æ NO3‐(l)+ 2e‐ + 2H+(aq) 4H+(aq) + 2e‐ + AsO43‐(aq) Æ AsO2‐(aq) + 2H2O(l) • Rewriting the half reactions with the multiplication done H2O(l) + NO2‐(l) Æ NO3‐(l) + 2e‐ + 2H+(aq) +
4H (aq) + 2e‐ + AsO43‐(aq) Æ AsO2‐(aq) + 2H2O(l) Write a balanced redox reaction (in acid) • Combine half reactions H2O(l) + NO2‐(l) + 4H+(aq) + 2e‐ + AsO43‐(aq) Æ AsO2‐(aq) + 2H2O(l) + NO3‐(l) + 2e‐ + 2H+(aq) • Cross out spectators NO2‐(l) + 2H+(aq) + AsO43‐(aq) Æ AsO2‐(aq) + H2O(l) + NO3‐(l) Write a balanced redox reaction (in base) • Count the # of H+(aq). Add the same # of OH‐(aq) to both sides of the equation 2OH‐(aq) + NO2‐(l) + 2H+(aq) + AsO43‐(aq) Æ AsO2‐(aq) + H2O(l) + NO3‐(l) + 2OH‐(aq) • On the side with both H+(aq) & OH‐(aq), those become water 2H2O(l) + NO2‐(l) + AsO43‐(aq) Æ AsO2‐(aq) + H2O(l) + NO3‐(l) + 2OH‐(aq) • Cross out spectators H2O(l) + NO2‐(l) + AsO43‐(aq) Æ AsO2‐(aq) + NO3‐(l) + 2OH‐(aq) 4. BH4‐(aq) + ClO3‐(aq) Æ H2BO3‐(aq) + Cl‐(aq) First write oxidation states for each element, per atom. ‐5 4(+1) +5 3(‐2) 2(+1) +3 3(‐2) ‐1 BH4‐(aq) + ClO3‐(aq) Æ H2BO3‐(aq) + Cl‐(aq) a. Identify the oxidized species and state how many electrons are lost per atom • Each B lost 8e‐ b. Identify the reduced species and state how many electrons are gained per atom • Each Cl gained 6e‐ c. Identify the oxidizing agent • ClO3‐(aq) d. Identify the reducing agent • BH4‐(aq) e. Write balanced half reactions (in acid) • Write Unbalanced half reactions including electron change per atom BH4‐(aq) Æ H2BO3‐(aq) + 8e‐ 6e‐ + ClO3‐(aq) Æ Cl‐(aq) • Balance half reactions for same # of oxidized or reduced species BH4‐(aq) Æ H2BO3‐(aq) + 8e‐ 6e‐ + ClO3‐(aq) Æ Cl‐(aq) •
Balance half reactions for # of O, with H2O, and # of H, with H+ 3H2O(l) + BH4‐(aq) Æ H2BO3‐(aq) + 8e‐ + 8H+(aq) 6H+(aq) + 6e‐ + ClO3‐(aq) Æ Cl‐(aq) + 3H2O(l) Now the half reactions are balanced within themselves, but not balanced for # of electrons gained and lost • Balance half reactions for # of electrons gained and lost 3(3H2O(l) + BH4‐(aq) Æ H2BO3‐(aq) + 8e‐ + 8H+(aq)) 4(6H+(aq) + 6e‐ + ClO3‐(aq) Æ Cl‐(aq) + 3H2O(l)) • Rewriting the half reactions with the multiplication done 9H2O(l) + 3BH4‐(aq) Æ 3H2BO3‐(aq) + 24e‐ + 24H+(aq) 24H+(aq) + 24e‐ + 4ClO3‐(aq) Æ 4Cl‐(aq) + 12H2O(l) f. Write a balanced redox reaction (in acid) • Combine half reactions 9H2O(l) + 3BH4‐(aq) + 24H+(aq) + 24e‐ + 4ClO3‐(aq) Æ 4Cl‐(aq) + 12H2O(l) + 3H2BO3‐(aq) + 24e‐ + 24H+(aq) • Cross out spectators 3BH4‐(aq) + 4ClO3‐(aq) Æ 4Cl‐(aq) + 3H2O(l) + 3H2BO3‐(aq) g. Write a balanced redox reaction (in base) • Count the # of H+(aq). Add the same # of OH‐(aq) to both sides of the equation 3BH4‐(aq) + 4ClO3‐(aq) Æ 4Cl‐(aq) + 3H2O(l) + 3H2BO3‐(aq) • No H+(aq) so it’s balanced in acid and in base 3BH4‐(aq) + 4ClO3‐(aq) Æ 4Cl‐(aq) + 3H2O(l) + 3H2BO3‐(aq) 5. CrO42‐(aq) + N2O(g) Æ Cr3+(aq) + NO(g) First write oxidation states for each element, per atom. +6 4(‐2) 2(+1) ‐2 +3 +2 ‐2 CrO42‐(aq) + N2O(g) Æ Cr3+(aq) + NO(g) a. Identify the oxidized species and state how many electrons are lost per atom • Each N lost 1e‐ b. Identify the reduced species and state how many electrons are gained per atom • Each Cr gained 3e‐ c. Identify the oxidizing agent • CrO42‐(aq) d. Identify the reducing agent • N2O(g) e. Write balanced half reactions (in acid) • Write Unbalanced half reactions including electron change per atom N2O(g) Æ NO(g) + 1e‐ 3e‐ + CrO42‐(aq) Æ Cr3+(aq) • Balance half reactions for same # of oxidized/reduced species N2O(g) Æ 2NO(g) + 2(1e‐) 3e‐ + CrO42‐(aq) Æ Cr3+(aq) • Balance half reactions for # of O, with H2O, and # of H, with H+ H2O(l) + N2O(g) Æ 2NO(g) + 2e‐ + 2H+(aq) 8H+(aq) + 3e‐ + CrO42‐(aq) Æ Cr3+(aq) + 4H2O(l) Now the half reactions are balanced within themselves, but not balanced for # of electrons gained and lost • Balance half reactions for # of electrons gained and lost 3(H2O(l) + N2O(g) Æ 2NO(g) + 2e‐ + 2H+(aq)) 2(8H+(aq) + 3e‐ + CrO42‐(aq) Æ Cr3+(aq) + 4H2O(l)) • Rewriting the half reactions with the multiplication done 3H2O(l) + 3N2O(g) Æ 6NO(g) + 6e‐ + 6H+(aq) 16H+(aq) + 6e‐ + 2CrO42‐(aq) Æ 2Cr3+(aq) + 8H2O(l) f. Write a balanced redox reaction (in acid) • Combine half reactions 3H2O(l) + 3N2O(g) + 16H+(aq) + 6e‐ + 2CrO42‐(aq) Æ 2Cr3+(aq) + 8H2O(l) + 6NO(g) + 6e‐ + 6H+(aq) • Cross out spectators 3N2O(g) + 10H+(aq) + 2CrO42‐(aq) Æ 2Cr3+(aq) + 5H2O(l) + 6NO(g) g. Write a balanced redox reaction (in base) • Count the # of H+(aq). Add the same # of OH‐(aq) to both sides of the equation 10OH‐(aq) + 3N2O(g) + 10H+(aq) + 2CrO42‐(aq) Æ 2Cr3+(aq) + 5H2O(l) + 6NO(g) + 10OH‐(aq) • On the side with both H+(aq) & OH‐(aq), those become water 10H2O(l) + 3N2O(g) + 2CrO42‐(aq) Æ 2Cr3+(aq) + 5H2O(l) + 6NO(g) + 10OH‐(aq) • Cross out spectators 5H2O(l) + 3N2O(g) + 2CrO42‐(aq) Æ 2Cr3+(aq) + 6NO(g) + 10OH‐(aq) 6. Br2(l) Æ BrO3‐(aq) + Br‐(aq) 7. NO2(g) Æ NO3‐(aq) + NO2(g) 8. Zn(s) + NO3‐(aq) Æ Zn(OH)42‐(s) + NH3(g) First write oxidation states for each element, per atom. 0 +5 3(‐2) +2 4(‐1) ‐3 3(+1) Zn(s) + NO3‐(aq) Æ Zn(OH)42‐(s) + NH3(g) a. Identify the oxidized species and state how many electrons are lost per atom • Each Zn lost 2e‐ b. Identify the reduced species and state how many electrons are gained per atom • Each N gained 8e‐ c. Identify the oxidizing agent • NO3‐(aq) d. Identify the reducing agent • Zn(s) e. Write balanced half reactions (in acid) • Write Unbalanced half reactions including electron change per atom Zn(s) + NO3‐(aq) Æ Zn(OH)42‐(s) + NH3(g) Zn(s) Æ Zn(OH)42‐(s) + 2e‐ 8e‐ + NO3‐(aq) Æ NH3(g) • Balance half reactions for same # of oxidized/reduced species Zn(s) Æ Zn(OH)42‐(s) + 2e‐ 8e‐ + NO3‐(aq) Æ NH3(g) • Balance half reactions for # of O, with H2O, and # of H, with H+ 4H2O(l) + Zn(s) Æ Zn(OH)42‐(s) + 2e‐ + 4H+(aq) 9H+(aq) + 8e‐ + NO3‐(aq) Æ NH3(g) + 3H2O(l) Now the half reactions are balanced within themselves, but not balanced for # of electrons gained and lost • Balance half reactions for # of electrons gained and lost 4(4H2O(l) + Zn(s) Æ Zn(OH)42‐(s) + 2e‐ + 4H+(aq)) 9H+(aq) + 8e‐ + NO3‐(aq) Æ NH3(g) + 3H2O(l) • Rewriting the half reactions with the multiplication done 16H2O(l) + 4Zn(s) Æ 4Zn(OH)42‐(s) + 8e‐ + 16H+(aq) 9H+(aq) + 8e‐ + NO3‐(aq) Æ NH3(g) + 3H2O(l) f. Write a balanced redox reaction (in acid) • Combine half reactions 16H2O(l) + 4Zn(s) + 9H+(aq) + 8e‐ + NO3‐(aq) Æ NH3(g) + 3H2O(l) + 4Zn(OH)42‐(s) + 8e‐ + 16H+(aq) • Cross out spectators 13H2O(l) + 4Zn(s) + NO3‐(aq) Æ NH3(g) + 4Zn(OH)42‐(s) + 7H+(aq) g. Write a balanced redox reaction (in base) •
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Count the # of H+(aq). Add the same # of OH‐(aq) to both sides of the equation 7OH‐(aq) + 13H2O(l) + 4Zn(s) + NO3‐(aq) Æ NH3(g) + 4Zn(OH)42‐(s) + 7H+(aq) + 7OH‐(aq) On the side with both H+(aq) & OH‐(aq), those become water 7OH‐(aq) + 13H2O(l) + 4Zn(s) + NO3‐(aq) Æ NH3(g) + 4Zn(OH)42‐(s) + 7H2O(l) Cross out spectators 7OH‐(aq) + 6H2O(l) + 4Zn(s) + NO3‐(aq) Æ NH3(g) + 4Zn(OH)42‐(s) 9. SO32‐(aq) + Cl2(g) Æ SO42‐(aq) + Cl‐(aq) First write oxidation states for each element, per atom. +4 3(‐2) 2(0) +6 4(‐2) ‐1 SO32‐(aq) + Cl2(g) Æ SO42‐(aq) + Cl‐(aq) a. Identify the oxidized species and state how many electrons are lost per atom • Each S lost 2e‐ b. Identify the reduced species and state how many electrons are gained per atom • Each Cl gained 1e‐ c. Identify the oxidizing agent • Cl2(g) d. Identify the reducing agent • SO32‐(aq) e. Write balanced half reactions (in acid) • Write Unbalanced half reactions including electron change per atom SO32‐(aq) Æ SO42‐(aq) + 2e‐ 1e‐ + Cl2(g) Æ Cl‐(aq) • Balance half reactions for same # of oxidized/reduced species SO32‐(aq) Æ SO42‐(aq) + 2e‐ 2(1e‐) + Cl2(g) Æ 2Cl‐(aq) • Balance half reactions for # of O, with H2O, and # of H, with H+ H2O(l) + SO32‐(aq) Æ SO42‐(aq) + 2e‐ + 2H+(aq) 2e‐ + Cl2(g) Æ 2Cl‐(aq) Now the half reactions are balanced within themselves, but not balanced for # of electrons gained and lost • Balance half reactions for # of electrons gained and lost H2O(l) + SO32‐(aq) Æ SO42‐(aq) + 2e‐ + 2H+(aq) 2e‐ + Cl2(g) Æ 2Cl‐(aq) • Rewriting the half reactions with the multiplication done H2O(l) + SO32‐(aq) Æ SO42‐(aq) + 2e‐ + 2H+(aq) 2e‐ + Cl2(g) Æ 2Cl‐(aq) f. Write a balanced redox reaction (in acid) • Combine half reactions H2O(l) + SO32‐(aq) + 2e‐ + Cl2(g) Æ 2Cl‐(aq) + SO42‐(aq) + 2e‐ + 2H+(aq) • Cross out spectators H2O(l) + SO32‐(aq) + Cl2(g) Æ 2Cl‐(aq) + SO42‐(aq) + 2H+(aq) g. Write a balanced redox reaction (in base) • Count the # of H+(aq). Add the same # of OH‐(aq) to both sides of the equation 2OH‐(aq) + H2O(l) + SO32‐(aq) + Cl2(g) Æ 2Cl‐(aq) + SO42‐(aq) + 2H+(aq) + 2OH‐(aq) • On the side with both H+(aq) & OH‐(aq), those become water 2OH‐(aq) + H2O(l) + SO32‐(aq) + Cl2(g) Æ 2Cl‐(aq) + SO42‐(aq) + 2H2O(l) • Cross out spectators 2OH‐(aq) + SO32‐(aq) + Cl2(g) Æ 2Cl‐(aq) + SO42‐(aq) + H2O(l) 10. Fe(CN)63‐(aq) + Re(s) Æ Fe(CN)64‐(aq) + ReO4‐(g) First write oxidation states for each element, per atom. +3 6(‐1) 0 +2 6(‐1) +7 4(‐1) Fe(CN)63‐(aq) + Re(s) Æ Fe(CN)64‐(aq) + ReO4‐(g) a. Identify the oxidized species and state how many electrons are lost per atom • Each Re lost 7e‐ b. Identify the reduced species and state how many electrons are gained per atom • Each Fe gained 1e‐ c. Identify the oxidizing agent • Fe(CN)63‐(aq) d. Identify the reducing agent • Re(s) e. Write balanced half reactions (in acid) • Write Unbalanced half reactions including electron change per atom Re(s) Æ ReO4‐(g) + 7e‐ 1e‐ + Fe(CN)63‐(aq) Æ Fe(CN)64‐(aq) • Balance half reactions for same # of oxidized/reduced species Re(s) Æ ReO4‐(g) + 7e‐ 1e‐ + Fe(CN)63‐(aq) Æ Fe(CN)64‐(aq) • Balance half reactions for # of O, with H2O, and # of H, with H+ 4H2O(l) + Re(s) Æ ReO4‐(g) + 7 e‐ + 8H+(aq) 1e‐ + Fe(CN)63‐(aq) Æ Fe(CN)64‐(aq) Now the half reactions are balanced within themselves, but not balanced for # of electrons gained and lost • Balance half reactions for # of electrons gained and lost 4H2O(l) + Re(s) Æ ReO4‐(g) + 7 e‐ + 8H+(aq) 7(1e‐ + Fe(CN)63‐(aq) Æ Fe(CN)64‐(aq)) • Rewriting the half reactions with the multiplication done 4H2O(l) + Re(s) Æ ReO4‐(g) + 7 e‐ + 8H+(aq) 7e‐ + 7Fe(CN)63‐(aq) Æ 7Fe(CN)64‐(aq) f. Write a balanced redox reaction (in acid) • Combine half reactions 4H2O(l) + Re(s) + 7e‐ + 7Fe(CN)63‐(aq) Æ 7Fe(CN)64‐(aq) + ReO4‐(g) + 7 e‐ + 8H+(aq) • Cross out spectators 4H2O(l) + Re(s) + 7Fe(CN)63‐(aq) Æ 7Fe(CN)64‐(aq) + ReO4‐(g) + 8H+(aq) g. Write a balanced redox reaction (in base) • Count the # of H+(aq). Add the same # of OH‐(aq) to both sides of the equation 8OH‐(aq) + 4H2O(l) + Re(s) + 7Fe(CN)63‐(aq) Æ 7Fe(CN)64‐(aq) + ReO4‐(g) + 8H+(aq) + 8OH‐(aq) • On the side with both H+(aq) & OH‐(aq), those become water 8OH‐(aq) + 4H2O(l) + Re(s) + 7Fe(CN)63‐(aq) Æ 7Fe(CN)64‐(aq) + ReO4‐(g) + 8H2O(l) • Cross out spectators 8OH‐(aq) + Re(s) + 7Fe(CN)63‐(aq) Æ 7Fe(CN)64‐(aq) + ReO4‐(g) + 4H2O(l) 11. MnO4‐(aq) + HCOOH(aq) Æ Mn2+(aq) + CO2(g) First write oxidation states for each element, per atom. +7 4(‐2) +1+2‐2‐2+1 +2 +4 2(‐2) MnO4‐(aq) + HCOOH(aq) Æ Mn2+(aq) + CO2(g) a. Identify the oxidized species and state how many electrons are lost per atom • Each C lost 2e‐ b. Identify the reduced species and state how many electrons are gained per atom • Each Mn gained 5e‐ c. Identify the oxidizing agent • MnO4‐(aq) d. Identify the reducing agent • HCOOH(aq) e. Write balanced half reactions (in acid) •
Write Unbalanced half reactions including electron change per atom HCOOH(aq) Æ CO2(g) + 2e‐ 5e‐ + MnO4‐(aq) Æ Mn2+(aq) • Balance half reactions for same # of oxidized/reduced species HCOOH(aq) Æ CO2(g) + 2e‐ 5e‐ + MnO4‐(aq) Æ Mn2+(aq) • Balance half reactions for # of O, with H2O, and # of H, with H+ HCOOH(aq) Æ CO2(g) + 2e‐ + 2H+(aq) +
8H (aq) + 5e‐ + MnO4‐(aq) Æ Mn2+(aq) + 4H2O(l) Now the half reactions are balanced within themselves, but not balanced for # of electrons gained and lost • Balance half reactions for # of electrons gained and lost 5(HCOOH(aq) Æ CO2(g) + 2e‐ + 2H+(aq)) +
2(8H (aq) + 5e‐ + MnO4‐(aq) Æ Mn2+(aq) + 4H2O(l)) • Rewriting the half reactions with the multiplication done 5HCOOH(aq) Æ 5CO2(g) + 10e‐ + 10H+(aq) +
16H (aq) + 10e‐ + 2MnO4‐(aq) Æ 2Mn2+(aq) + 8H2O(l) f. Write a balanced redox reaction (in acid) • Combine half reactions 5HCOOH(aq) + 16H+(aq) + 10e‐ + 2MnO4‐(aq) Æ 2Mn2+(aq) + 8H2O(l) + 5CO2(g) + 10e‐ + 10H+(aq) • Cross out spectators 5HCOOH(aq) + 6H+(aq) + 2MnO4‐(aq) Æ 2Mn2+(aq) + 8H2O(l) + 5CO2(g) g. Write a balanced redox reaction (in base) • Count the # of H+(aq). Add the same # of OH‐(aq) to both sides of the equation 6OH‐(aq) + 5HCOOH(aq) + 6H+(aq) + 2MnO4‐(aq) Æ 2Mn2+(aq) + 8H2O(l) + 5CO2(g) + 6OH‐(aq) • On the side with both H+(aq) & OH‐(aq), those become water 6H2O(l) + 5HCOOH(aq) + 2MnO4‐(aq) Æ 2Mn2+(aq) + 8H2O(l) + 5CO2(g) + 6OH‐(aq) • Cross out spectators 5HCOOH(aq) + 2MnO4‐(aq) Æ 2Mn2+(aq) + 2H2O(l) + 5CO2(g) + 6OH‐(aq) 12. Mn2+(aq) + Co3+(aq)Æ MnO2(s) + Co2+(aq) First write oxidation states for each element, per atom. +2 +3 +4 2(‐2) +2 Mn2+(aq) + Co3+(aq)Æ MnO2(s) + Co2+(aq) a. Identify the oxidized species and state how many electrons are lost per atom • Each Mn lost 2e‐ b. Identify the reduced species and state how many electrons are gained per atom • Each Co gained 1e‐ c. Identify the oxidizing agent • Co3+(aq) d. Identify the reducing agent • Mn2+(aq) e. Write balanced half reactions (in acid) • Write Unbalanced half reactions including electron change per atom Mn2+(aq) Æ MnO2(s) + 2e‐ 1e‐ + Co3+(aq) Æ Co2+(aq) • Balance half reactions for same # of oxidized/reduced species Mn2+(aq) Æ MnO2(s) + 2e‐ 1e‐ + Co3+(aq) Æ Co2+(aq) • Balance half reactions for # of O, with H2O, and # of H, with H+ 2H2O(l) + Mn2+(aq) Æ MnO2(s) + 2e‐ + 4H+(aq) 1e‐ + Co3+(aq) Æ Co2+(aq) Now the half reactions are balanced within themselves, but not balanced for # of electrons gained and lost •
Balance half reactions for # of electrons gained and lost 2H2O(l) + Mn2+(aq) Æ MnO2(s) + 2e‐ + 4H+(aq) 2(1e‐ + Co3+(aq) Æ Co2+(aq)) • Rewriting the half reactions with the multiplication done 2H2O(l) + Mn2+(aq) Æ MnO2(s) + 2e‐ + 4H+(aq) 2e‐ + 2Co3+(aq) Æ 2Co2+(aq) f. Write a balanced redox reaction (in acid) • Combine half reactions 2H2O(l) + Mn2+(aq) + 2e‐ + 2Co3+(aq) Æ 2Co2+(aq) + MnO2(s) + 2e‐ + 4H+(aq) • Cross out spectators 2H2O(l) + Mn2+(aq) + 2Co3+(aq) Æ 2Co2+(aq) + MnO2(s) + 4H+(aq) g. Write a balanced redox reaction (in base) • Count the # of H+(aq). Add the same # of OH‐(aq) to both sides of the equation 4OH‐(aq) + 2H2O(l) + Mn2+(aq) + 2Co3+(aq) Æ 2Co2+(aq) + MnO2(s) + 4H+(aq) + 4OH‐(aq) • On the side with both H+(aq) & OH‐(aq), those become water 4OH‐(aq) + 2H2O(l) + Mn2+(aq) + 2Co3+(aq) Æ 2Co2+(aq) + MnO2(s) + 4H2O(l) • Cross out spectators 4OH‐(aq) + Mn2+(aq) + 2Co3+(aq) Æ 2Co2+(aq) + MnO2(s) + 2H2O(l)