Ion Electron Method Ch 20

Write an oxidation and a reduction half reaction. Sn 2+ → Sn 4+ Hg 2+ + Cl -1 → Hg 2 Cl 2

Balance each half reaction in terms of atoms. Sn 2+ → Sn 4+ 2Hg Cl -1 → Hg 2 Cl 2

Balance charges on opposite sides of each half-reaction equation by adding electrons to the appropriate side. Sn 2+ → Sn e - 2e - + 2Hg Cl -1 → Hg 2 Cl 2

The number of electrons lost in the oxidation half reaction must equal the number of electrons gained in the reduction half reaction. If necessary, multiply each half reaction equation by a stoichiometric coefficient to equalize the number of electrons transferred. Sn 2+ → Sn e- 2e- + 2Hg Cl -1 → Hg 2 Cl 2

Add the resulting half-reactions to obtain the balanced net ionic equation. Sn 2+ → Sn e- 2e- + 2Hg Cl- → Hg 2 Cl 2 2e- + Sn Hg Cl -1 → Sn 4+ + Hg 2 Cl 2 + 2e-

Cancel out any species that is the same on both sides of the reaction. Sn Hg Cl -1 → Sn 4+ + Hg 2 Cl 2 Note: Both atoms and charges are balanced.

*In many oxidation-reduction reactions that take place in aqueous solution, water plays an active role. *Any aqueous solution contains the species H 2 0, H +, and OH -. *In acidic solutions the predominant species are H 2 0 and H + ; *In basic solutions they are H 2 0 and OH -. *When balancing half reactions that occur in solution, we can use these species to achieve material balance.

If the reaction occurs in acidic solution … Cr 2 O H 2 S → Cr 3+ + S Write a half reaction for sulfur: H 2 S → S Balance the H + : H 2 S → S + 2H +

Balance the charge by adding electrons: H 2 S → S + 2H + + 2e- Now write the chromium half reaction: Cr 2 O 7 2- → Cr 3+ Balance the chromium atoms: Cr 2 O 7 2- → 2Cr 3+

Balance the oxygens on the left by adding water to the right side of the equation: Cr 2 O 7 2- → 2Cr 3+ + H 2 O Now add H +1 to the left: H +1 + Cr 2 O 7 2- → 2Cr 3+ + H 2 O Balance the H’s and O’s: 14H +1 + Cr 2 O 7 2- → 2Cr H 2 O

Now add electrons to balance the charge: 14H +1 + Cr 2 O 7 2- → 2Cr H 2 O There is 14+ and 2- on the left (12+) There is 6+ on the right Therefore, add 6e - to the left to balance the charge. 6e H +1 + Cr 2 O 7 2- → 2Cr H 2 O

Add the 2 half reactions together 3 (H 2 S → S + 2H + + 2e-) 6e- + 14H + + Cr 2 O 7 2- → 2Cr H 2 0 3H 2 S + 14H + + Cr 2 O e- → 3S + 6H + + 2Cr H e- Cancel out anything that is the same on both sides: 3H 2 S + 8H + + Cr 2 O 7 2- → 3S + 2Cr H 2 0 Note: notice how some of the H + ions cancel out.

In summary, when balancing half-reactions in acid solution: (a) To balance a hydrogen atom we add a hydrogen ion, H+, to the other side of the equation. (b) To balance an oxygen atom we add a water molecule to the side deficient in oxygen and then two H+ ions to the opposite side to remove the hydrogen imbalance.

If the reaction occurs in basic solution … Although you can use H 2 O and OH - directly, the simplest technique is to first balance the reaction as if it occurred in acidic solution, and then perform the "conversion" (described on the next slide) to adjust it to conform to conditions in basic solution.

Balance the Reaction in a Basic Solution Pb → PbO First we balance it as if it occurred in an acidic solution. H Pb → PbO + 2H + + 2e - Add water to balance the oxygens, add H + to balance the H’s then add e- to balance the charge.

The conversion to basic solution follows these three steps: Step 1 For each H+ that must be eliminated from the equation, add an OH - to both sides of the equation. In this example, we have to eliminate 2H +, so we add 2OH - to each side. H Pb + 2OH- → PbO + 2H + + 2OH - + 2e-

Step 2 Combine H + and OH - to form H 2 0. We have 2H + and 2OH - on the right, which give 2H 2 0. H Pb + 2OH- → PbO + 2H 2 O + 2e- 2H + + 2OH

Step 3 Cancel any H 2 0 that are the same on both sides. We can cancel one H 2 0 from each side. The final balanced half-reaction in basic solution is: Pb + 2OH- → PbO + H 2 O + 2e-

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