![]() ![]() 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. 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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