REDOX Sample Problem #4 in Base

advertisement
PROBLEM: Aqueous Sodium Permanganate plus solid Zinc plus liquid water yields
solid Manganese Dioxide plus solid Zinc Hydroxide. (Note: The reaction takes place in
basic solution)
Step 1: Write the unbalanced formula equations
NaMnO4(aq) + Zn(s) + H2O(l) --------> MnO2(s) + Zn(OH)2(s)
Step 2: Identify the species that are oxidized and reduced. Start by labeling the oxidation #s.
+1 +7 –2
0
+1 –2
+4 –2
+2 –2 +1
NaMnO4(aq) + Zn(s) + H2O(l) --------> MnO2(s) + Zn(OH)2(s)
Since Zn goes from 0 to +2 it lost e– and is oxidized.
Since Mn goes from +7 to +4 it gained e– and is reduced.
Now we will balance the equation as if it was in acid and then switch it to base after we have
the net ionic equation
Step 3: Write and balance the oxidation half reaction. Note – since Zn(OH)2 is a solid it does
NOT dissociate
Zn(s) --------> Zn(OH)2
Add 2 H2O to balance the O mass
2 H2O + Zn(s) --------> Zn(OH)2
Next balance the H mass by adding 2 H+
2 H2O + Zn(s) --------> Zn(OH)2 + 2 H+
Next add e– to balance the charge – since the left is 0, and the right is +2, add 2e– to the right
2 H2O + Zn(s) --------> Zn(OH)2 + 2 H+ + 2 e–
Step 4: Write and balance the reduction half reaction
MnO4– --------> MnO2
Balance the O by adding H2O
MnO4– --------> MnO2 + 2 H2O
Balance the H by adding H+
4 H+ + MnO4– --------> MnO2 + 2 H2O
Balance the charge by adding e–, since the left side is +3 and the right is 0, add 3 e– to the left
3 e– + 4 H+ + MnO4– --------> MnO2 + 2 H2O
Step 5: Next make the oxidation and reduction half-reactions have the same # of e–
2 H2O + Zn(s) --------> Zn(OH)2 + 2 H+ + 2 e– (X3)
6 H2O + 3 Zn(s) --------> 3 Zn(OH)2 + 6 H+ + 6 e–
3 e– + 4 H+ + MnO4– --------> MnO2 + 2 H2O (X2)
6 e– + 8 H+ + 2 MnO4– --------> 2 MnO2 + 4 H2O
Step 6: Combine the two half-reactions together, to make the net ionic equation. Remember to
cancel out the items on opposite sides of arrow (water, H+ ions, and e–). In this case all the e–,
6 H2O , and 6 H+.
2 H2O + 2 MnO4– + 3 Zn + 2 H+ --------> 2 Zn(OH)2 + MnO2
Step 7: Convert to base by adding an OH– ion for every H+ ion. But remember – to keep the
equation balanced you must add to both sides.
2 H2O + 2 MnO4– + 3 Zn + 2 H+ + 2 OH– --------> 3 Zn(OH)2 + 2 MnO2 + 2 OH–
Step 8: OH– plus H+ makes water – combine and cancel out waters from both sides
4 H2O + 2 MnO4– + 3 Zn --------> 3 Zn(OH)2 + 2 MnO2 + 2 OH–
Step 7: Add back spectator ions and combine with other ions to write complete compounds and the
balanced overall equation. Combine Na+ with the anions.
4 H2O + 2 NaMnO4 + 3 Zn --------> 3 Zn(OH)2 + 2 MnO2 + 2 NaOH
Download