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Section 20.3 lecture for Honors & Prep Chemistry

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- 1. Bellwork
- 2. Balancing redox reactions Read the 5 steps for balancing redox equations using the half-reaction method. You will get the steps on the test, but you must know what they mean in order to use them.
- 3. Balance the equation MnO 4 - + Fe 2+ Fe 3+ + Mn 2+ <ul><li>STEP 1 – split the complete equation into oxidation and reduction half-reactions </li></ul><ul><li>MnO 4 - Mn 2+ reduction ½ reaction </li></ul><ul><li>Fe 2+ Fe 3+ oxidation ½ reaction </li></ul>Each ½ reaction contains a reactant and a product that “look the same”
- 4. STEP 2- balance each ½ reaction using steps 2a-2d <ul><li>MnO 4 - Mn 2 + </li></ul><ul><li>2a- balance all elements except H and O Mn is already balanced with 1 on each side. </li></ul><ul><li>2b- balance oxygen by adding H 2 O There are 4 O on the left, add 4 H 2 O to the right. </li></ul>+ 4 H 2 O 2c- balance hydrogen using H + There are 8 H on the right, add 8H + on the left 2d- balance the charge using electrons The total charge is +7 on the left and +2 on the right, add 5e - to the left giving both sides a charge of +2. 5e - + 8H + +
- 5. <ul><li>Fe 2+ Fe 3+ </li></ul><ul><li>2a- Fe is already balanced </li></ul><ul><li>2b- there is no O </li></ul><ul><li>2c- there is no H </li></ul><ul><li>2d- the left side is +2 the right side is +3 </li></ul><ul><li>Adding one negative e - to the right side will give both sides an equal charge of +2 </li></ul>STEP 2- balance each ½ reaction using steps 2a-2d + e - +2 = +3 + (-1)
- 6. STEP 3- The 1/2 reactions must have the same number of e - If necessary, multiply one or both 1/2 reactions to make the number of electrons even. <ul><li>5 e - + 8 H + + MnO 4 - Mn 2 + + 4 H 2 O </li></ul><ul><li>Fe 2+ Fe 3+ + e - </li></ul><ul><li>The reduction ½ reaction shows a gain of 5e - </li></ul><ul><li>The oxidation ½ reaction shows a loss of 1 e - </li></ul><ul><li>Multiply the Fe reaction by 5 to equalize the number of e - </li></ul>5 5 5
- 7. STEP 4- add the ½ reactions together and cancel out anything that appears on both sides <ul><li>5 e - + 8 H + + MnO 4 - Mn 2 + + 4 H 2 O </li></ul><ul><li>5Fe 2 + 5Fe 3 + + 5e - </li></ul><ul><li>5e - + 8H + + MnO 4 - + 5Fe 2 + Mn 2 + + 4H 2 O + 5Fe 3 + + 5e - </li></ul>The electrons will always cancel out 8H + + MnO 4 - + 5Fe 2+ Mn 2+ + 4H 2 O + 5Fe 3+
- 8. STEP 5 – check that elements and charges balance <ul><li>8H + + MnO 4 - + 5Fe 2+ Mn 2+ + 4H 2 O + 5Fe 3+ </li></ul><ul><li>Left side right side </li></ul><ul><li>8H 8H 1Mn 1Mn 4 O 4 O 5 Fe 5 Fe (+1 x 8) + (-1) + (+2 x 5) = +17 (+2) + (+3x5) = +17 </li></ul>
- 9. Two Ways to Balance Redox Equations <ul><ul><li>Using Oxidation-Number Changes </li></ul></ul><ul><ul><ul><li>In the oxidation-number-change method , you balance a redox equation by comparing the increases and decreases in oxidation numbers. </li></ul></ul></ul>20.3
- 10. Two Ways to Balance Redox Equations <ul><ul><ul><li>Step 1 Assign oxidation numbers to all the atoms in the equation. </li></ul></ul></ul>20.3
- 11. Two Ways to Balance Redox Equations <ul><ul><ul><li>Step 2 Identify which atoms are oxidized and which are reduced. </li></ul></ul></ul><ul><ul><ul><li>Step 3 Use one bracketing line to connect the atoms that undergo oxidation and another such line to connect those that undergo reduction. </li></ul></ul></ul>20.3
- 12. Two Ways to Balance Redox Equations <ul><ul><li>In a balanced redox equation, the total increase in oxidation number of the species oxidized must be balanced by the total decrease in the oxidation number of the species reduced. </li></ul></ul>20.3
- 13. Two Ways to Balance Redox Equations <ul><ul><ul><li>Step 4 Make the total increase in oxidation number equal to the total decrease in oxidation number by using appropriate coefficients. </li></ul></ul></ul>20.3 2 <ul><ul><ul><li>Step 5 Finally, make sure that the equation is balanced for both atoms and charge. </li></ul></ul></ul>

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