Balancing Redox Reactions Lecture 5 Balancing Redox Reactions

What about the fun stuff--Balancing Redox reactions Look at Lecture Problem: We are balancing this in an acidic solution. Fe2+ + Cr2O72- ==> Fe3+ + Cr3+ Where do we start?

Step 1: Write half reactions Fe2+ + Cr2O72- ==> Fe3+ + Cr3+

Step 1: Write half reactions Fe2+ + Cr2O72- ==> Fe3+ + Cr3+ 1) Fe2+ ==> Fe3+ 2) Cr2O72- ==> Cr3+

Step 2: Balance all elements except H & O 1) Fe2+ ==> Fe3+ 2) Cr2O72- ==> Cr3+

Step 2: Balance all elements except H & O 1) Fe2+ ==> Fe3+ no worries 2) Cr2O72- ==> 2 Cr3+

Step 3: Balance oxygen using H2O 1) Fe2+ ==> Fe3+ 2) Cr2O72- ==> 2 Cr3+

Step 3: Balance oxygen using H2O 1) Fe2+ ==> Fe3+ 2) Cr2O72- ==> 2 Cr3+ + 7H2O

Step 4: Balance hydrogen using H+ 1) Fe2+ ==> Fe3+ 2) Cr2O72- ==> 2Cr3+ + 7H2O

Step 4: Balance hydrogen using H+ 1) Fe2+ ==> Fe3+ still nothing…yet 2) 14 H+ + Cr2O72- ==> 2Cr3+ + 7H2O

Step 5: Balance charge using e-…attack the highest # 1) Fe2+ ==> Fe3+ 14 H+ + Cr2O72- ==> 2Cr3+ + 7H2O

Step 5: Balance charge using e-…attack the highest # 1) Fe2+ ==> Fe3+ + 1e- 6e- + 14 H+ + Cr2O72- ==> 2Cr3+ + 7H2O Look where electrons are--we are on the right track!

Almost there: Use common multiple for e- to cancel them out. 1) Fe2+ ==> Fe3+ + 1e- 2) 6e- + 14 H+ + Cr2O72- ==> 2Cr3+ + 7H2O

Almost there: Use common multiple for e- to cancel them out. 1) (Fe2+ ==> Fe3+ + 1e-) x 6 6 Fe2+ ==> 6Fe3+ + 6e- 2) (6e- + 14 H+ + Cr2O72- ==> 2Cr3+ + 7H2O) x 1 6e- + 14 H+ + Cr2O72- ==> 2Cr3+ + 7H2O

Add half reactions: cancel out identical species 6 Fe2+ ==> 6Fe3+ + 6e- 6e- + 14 H+ + Cr2O72- ==> 2Cr3+ + 7H2O 6 Fe2+ + 14 H+ + Cr2O72- ==> 6Fe3+ + 2Cr3+ + 7H2O

What needs to be balanced? 1) Elements Check 2) Charges 6 Fe2+ + 14 H+ + Cr2O72- ==> 6Fe3+ + 2Cr3+ + 7H2O

What would happen if we had a basic solution? You do everything the same except the very end. What do we have to get rid of? 6 Fe2+ + 14 H+ + Cr2O72- ==> 6Fe3+ + 2Cr3+ + 7H2O

6 Fe2+ + 14 H+ + Cr2O72- ==> 6Fe3+ + 2Cr3+ + 7H2O We need to neutralize the H+ with OH-. Little problem: Math tells if we add something to one side, we have to do it to the other side. When we add OH- to H+ we make H2O. We have to do some canceling.

Basic solution 6 Fe2+ + 14 H+ + Cr2O72- ==> 6Fe3+ + 2Cr3+ + 7H2O add 14 OH- to cancel out H+….but I have to do it to both sides

Basic solution combine the H+ and OH- 6 Fe2+ + 14 H+ + 14 OH- + Cr2O72- ==> 6Fe3+ + 2Cr3+ + 7H2O + 14OH- combine the H+ and OH-

Basic solution cancel out waters 6 Fe2+ + 14 H2O + Cr2O72- ==> 6Fe3+ + 2Cr3+ + 7H2O + 14OH- cancel out waters

Basic solution 6 Fe2+ + 14 H2O + Cr2O72- ==> 6Fe3+ + 2Cr3+ + 7H20 + 14OH- 7 H2O

Basic solution 6 Fe2+ + 7 H2O + Cr2O72- ==> 6Fe3+ + 2Cr3+ + 14OH- Final answer

Try one for us to check! Balance the following equation in acidic solution: IO3- + Mn2+  I2 + MnO4-