chap-20-2011W-a

# Aq acidic the solution takes on the purple color of

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Unformatted text preview: olution. (aq) acidic The solution takes on the purple color of permanganate ion (MnO4-), permanganate and Bi3+ (aq) is present. Write a balanced equation of this reaction. (aq) 1. Write the unbalanced equation for the reaction in ionic form NaBiO3 (s) + Mn 2+ (aq) → Bi3+ (aq) + MnO-4 (aq) 2. Separate the equation into two half-reactions NaBiO3 (s) → Bi3+ (aq) Mn 2+ (aq) → MnO-4 (aq) 3. Balance the half-reactions halfa. Balance the atoms other than O and H in half-reactions atoms other halfNaBiO3 (s) → Bi3+ (aq) +Na + Mn 2+ (aq) → MnO-4 (aq) b. add H2O to balance O atoms and H+ to balance H atoms in atoms atoms in acidic acidic soln NaBiO3 (s) → Bi3+ (aq) +Na + NaBiO3 (s) + 6H + → Bi3+ (aq) + 3H 2 O(l) + Na + Mn 2+ (aq) → MnO-4 (aq) Mn 2+ (aq) + 4H 2 O(l) → 8H + + MnO-4 (aq) c. Add electrons to balance the charges on the half-reactions electrons balance halfNaBiO3 (s) + 6H + + 2e → Bi3+ (aq) + 3H 2 O(l) + Na + Mn 2+ (aq) + 4H 2 O(l) → 8H + + MnO-4 (aq) + 5e ×5 ×2 4. Equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients. 5NaBiO3 (s) + 30H + + 10e → 5Bi3+ (aq) + 15H 2 O(l) + 5Na + 2Mn 2+ (aq) + 8H 2 O(l) → 16H + + 2MnO-4 (aq) + 10e 5. Add the two half-reactions together and balance the final equation half5NaBiO3 (s) + 2Mn 2+ (aq)+14H + → 5Bi3+ (aq)+5Na + +2MnO-4 (aq)+7H 2 O(l) In a qualitative cation analysis lab, a student added 6 drops of 3% H2O2 to a strongly basic solution containing Mn2+ (aq), an...
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