Balance the following equations and explain how you did it. ____ ClO 3 Solution

Question

Balance the following equations and explain how you did it.

____ClO3

Explanation / Answer

1) 1st half: ClO3^- + H^+ --> ClO2 + H2O balances for mass but not for charge. To balance for charge we must add electrons to the side of the reaction that is reduced. ClO3^- + H^+ + e^- --> ClO2 + H2O balanced for mass and charge. 2nd half: 2 Cl^- --> Cl2 + 2 e^- balanced for mass and charge. Now we must write each half reaction so that the number of electrons lost = number of electrons gained. 2 ClO3^- + 4 H^+ + 2 e^- --> 2 ClO2 + 2 H2O 2 Cl^- --> Cl2 + 2 e^- Now, add the two half-reactions and cancel out the quantities that are the same on opposite sides of the arrow. 2 ClO3^- + 4 H^+ + 2 e^- --> 2 ClO2 + 2 H2O 2 Cl^- --> Cl2 + 2 e^- --------------------------------------... 2 ClO3^- + 4 H^+ + 2 Cl^- --> 2 ClO2 + Cl2 + 2 H2O 2) 1st half: mno4^- + 8H^+ --> mn^2+ + 4H2O balances for mass but not for charge. To balance for charge we must add electrons to the side of the reaction that is reduced. mno4^-+8 H^+ + 5e^- --> mn^2++4 H2O balanced for mass and charge. 2nd half: so3^2-+H2o --> So4^2-+2H^+ balanced for mass and charge. So3^2-+H2o --> So4^2-+2H^+2e^-balanced for mass and charge. 2(mno4^-+8 H^+ + 5e^- --> mn^2++4 H2O 5(So3^2-+H2o --> So4^2-+2H^+2e^-) Now we must write each half reaction so that the number of electrons lost = number of electrons gained. Now, add the two half-reactions and cancel out the quantities that are the same on opposite sides of the arrow. (2mno4^-+16H^+ + 10e^- --> 2mn^2++8 H2O (5So3^2-+5H2o --> 5So4^2-+10H^+10e^-) ------------------------------------------------------------------------------------... 5SO3^2-+2MnO4^-+6H^+------------------->2Mn^2++5SO4^2-+3H2O 3) CN-(aq) + MnO4-(aq) --> CNO-(aq) + MnO2(s) balanced for mass and charge. 3(H2O(l) + CN-(aq) --> CNO-(aq) + 2H+(aq) + 2e-) we must write each half reaction so that the number of electrons lost = number of electrons gained. 2(3e- + 4H+(aq) + MnO4-(aq) --> MnO2(s) + 2H2O(l))we must write each half reaction so that the number of electrons lost = number of electrons gained. Now, add the two half-reactions and cancel out the quantities that are the same on opposite sides of the arrow (3H2O(l) + 3CN-(aq) --> 3CNO-(aq) + 6H+(aq) + 6e-) (6e- + 8H+(aq) +2 MnO4-(aq) --> 2MnO2(s) + 4H2O(l) --------------------------------------------------------------------------------------------- 2H^+ + 2MnO42-(aq) +3CN-(aq)--> 3CNO- + 2MnO2(s) + H2O(l)

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