I'll add it back in at the end.ġ6H + + 2MnO 4¯ + 5S 2¯ -> 2Mn 2+ + 5S + 8H 2Oĥ) Add two sulfides on each side to make MnS:ġ6H + + 2MnO 4¯ + 7S 2¯ -> 2MnS + 5S + 8H 2OĦ) This document balances the equation in basic solution. Note that I eliminated the sulfide from the MnS. Here it is, in all its glory:Ĭr 2O 7 2¯ + Cl¯ -> Cr 3+ + Cl 2 + O 2¯Ĭr 2O 7 2¯ + 6Cl¯ -> 2Cr 3+ + 3Cl 2 + 7O 2¯īut oxide ions would immediately react with waterĬr 2O 7 2¯ + 6Cl¯ + 7H 2O -> 2Cr 3+ + 3Cl 2 + 14OH¯īalancing with oxide ions!! You don't see that one every day.ħ) And then, since are in acidic solution, we use 14H + to react with the hydroxide:Ĭr 2O 7 2¯ + 6Cl¯ + 7H 2O + 14H + -> 2Cr 3+ + 3Cl 2 + 14H 2OĨ) And then remove seven waters from each side to arrive at the answer given in step 4. It winds up with the equation balanced in basic solution. What happened?Īnswer: the writer of that page represented hydrogen ion as H 3O + rather than H +, thus adding six H 2O to each side.Įxample #6: VO 2+ + MnO 4¯ -> V(OH) 4 + + Mn 2+ġ5H 2O + 5VO 2+ -> 5V(OH) 4 + + 10H + + 5e¯ġ1H 2O + 5VO 2+ + MnO 4¯ -> 5V(OH) 4 + + Mn 2+ + 2H +Įxample #7: Cr 2O 7 2¯ + Cl¯ -> Cr 3+ + Cl 2Ħe¯ + 14H + + Cr 2O 7 2¯ -> 2Cr 3+ + 7H 2Oġ4H + + Cr 2O 7 2¯ + 6Cl¯ -> 2Cr 3+ + 3Cl 2 + 7H 2Oĥ) A more detailed discussion about balancing this equation can be found here.Ħ) I once saw an unusual method to balancing this particular example equation. In the hits from that search, there will be a number of websites that will help you figure out the answer. Notice that I completely ignore capitals in the formulas. The search suggestion is to type the reactants plus an arrow, like this: Note how easy it was to balance the copper half-reaction. No need to equalize electrons since it turns out that, in the course of balancing the half-reactions, the electrons are equal in amount. The answer to 2b is the exact same as 2a in terms of the stoichiometry of the reaction.Įxample #3: MnO 4¯ + H 2S -> Mn 2+ + S 8ģ) Make the number of electrons equal (note that there are no common factors between 5 and 16 except 1):Ĥ0H 2S -> 5S 8 + 80H + + 80e¯ 16Mn 2+ + 64H 2O 5S 8 + 16Mn 2+ + 64H 2OĪnother possibility of removing a factor of 8 destroyed by an odd number, in this case, the 5 in front of the S 8. Notice that, in the answer, the S coefficient stays the same (but the subscript of 8 goes away) and the other coefficients are all reduced by a factor of 8. Also, note that duplicates of 48 electrons and 48 hydrogen ions were removed.ĥ) Sometimes, you will see the nitric acid in molecular form:Ģ4H 2S + 16HNO 3 -> 3S 8 + 16NO + 32H 2OĮxample #2b: H 2S + HNO 3 -> NO + S + H 2Oĭiscussion: Many times, teachers and textbooks will use S rather than S 8. The duplicates are 6e¯, 3H 2O, and 6H +Ģ4H 2S -> 3S 8 + 48H + + 48e¯ 16NO + 32H 2O 3S 8 + 16NO + 32H 2OĬomment: removing a factor of 8 does look tempting, doesn't it? However, the three in front of the S 8 (or the five in the next example) makes it impossible. Note that items duplicated on each side were cancelled out. These items are usually the electrons, water and hydrogen ion.Īfter Example #5c, I have a suggestion for searching for help on balancing a specific equation.Įxample #1: ClO 3¯ + SO 2 -> SO 4 2¯ + Cl¯ĦH 2O + 3SO 2 -> 3SO 4 2¯ + 12H + + 6e¯ 3SO 4 2¯ + Cl¯ + 6H + In order to get the electrons in each half-reaction equal, one or both of the balanced half-reactions will be multiplied by a factor.Ģ) Duplicate items are always removed. \) as necessary to either side of the equation to balance the charge.Balancing redox reactions in acidic solution Balancing redox reactions in acidic solutionįifteen Examples Problems 1-10 Problems 26-50 Balancing in basic solution Problems 11-25 Only the examples and problems Return to Redox menuġ) Electrons NEVER appear in a correct, final answer.
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