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Step 1 : Write the unbalanced formula equations

NaMnO 4 ( aq ) + Zn(s) + H 2 O(l) --------> MnO 2 (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.

meredith-anchoret
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Step 1 : Write the unbalanced formula equations

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  1. 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 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) 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 Mngoes 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)2is a solid it does NOT dissociate Step 1: Write the unbalanced formula equations 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–

  2. 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–

  3. 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

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