balancing redox equations oxidation number method

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Question:Okay its really confusing me, can someone help? Acid rain is the result of atmospheric chemical reactions such as the following acid formation reaction: NO2(g) + H2O(l) --> NO(g) + H(+)(aq) + NO3(-)(aq) Balance this redox reaction equation. In the reaction, which element is reduced and which is oxidised? Identify changes in oxidation numbers. Can someone explain step by step with half reaction equations pleaseeee? :) Thanks!

Answers:Balanced redox reaction equation: 3NO2(g) + H2O(l) --> NO(g) + 2H(+)(aq) + 2NO3(-)(aq) That's all i can answer. Sorry =(

Question:The equation is: HBrO3 + Bi -> HBrO2 + Bi2O3 I figured out that the oxidation numbers are H+1, Br+5, O-6, Bi0, H+1, Br+3, O-4, Bi+3, O-2. I understand that Br is reduced, but from my numbers it seems that Bi and O is oxidized. Is that possible? and if so how would i put that into separate equations for the oxidation and reduction reactions?

Answers:I'm not sure about the oxygen but Bi is oxidized because on the reactant side it was an element , element state = 0 charge, and on the right it is part of a compound, and it has a charge

Question:1. ClO- + I - --> Cl- + I3 - 2. Br- + MnO4- --> Br2 + Mn +2 3. CH3OH + CrO7 (charge of 2-) --> CH2O + Cr +3 Please show all of the steps and the half reactions, balancing H and O, balancing charge using e- etc. Thanks and PLEASE HELP MEEE ASAP!

Answers:1. ClO- + I - --> Cl- + I3 - Break the reaction into half reactions (reduction and oxidation) ClO^- ==> Cl^- . . .and . . .I^- ==> I3^- ClO^- ==> Cl^- . . .balance O by adding 1 H2O to the right side. ClO^- ==> Cl^- + H2O . . .balance H by adding 2H^+ to the left side. ClO^- + 2H^+ ==> Cl^- + H2O . . . add 2e^- to the left to balance the charge ClO^- + 2H^+ + 2e^- ==> Cl^- + H2O (Equation 1) = reduction (gaining electrons) I^- ==> I3^- . . .Put a 3 in front of I^- to balance I. 3I^- ==> I3^- . . .add 2e^- to the right side to balance the charge. 3I^- ==> I3^- + 2e^- (Equation 2) = oxidation (loss of electrons) Add equations 1 and 2 together to cancel the 2e-. ClO^- + 2H^+ + 2e^- ==> Cl^- + H2O 3I^- ==> I3^- + 2e^- =============================== ClO^- + 3I^- + 2H^+ ==> Cl^- + I3^- + H2O ------------------------------------------------------------------------------------------- 2. Br- + MnO4- --> Br2 + Mn +2 Br^- ==> Br2 . . .and . . .MnO4^- ==> Mn^2+ Br^- ==> Br2 . . .put 2 in front of Br^- to balance Br 2Br^- ==> Br2 . . .add 2e- to the right side to balance charge. 2Br^- ==> Br2 + 2e^- (Equation 1) = oxidation = loss of electrons MnO4^- ==> Mn^2+ . . .Add 4 H2O to the right side to balance O. MnO4^- ==> Mn^2+ + 4H2O . . .add 8H^+ to the left side to balance H MnO4^- + 8H^+ ==> Mn^2+ + 4H2O . . .add 5e^- to the left side to balance the charge. MnO4^- + 8H^+ + 5e^- ==> Mn^2+ + 4H2O (Equation 2) = reduction = gaining electrons The least common multiple of 2e^- in Equation 1 and 5e^- in Equation 2 is 10^e-. So multiply Equation 1 by 5 to get 10e^-, multiply Equation 2 by 2 to get 1oe^-, then add the two equations together. 10Br^- ==> 5Br2 + 10e^- 2MnO4^- + 16H^+ + 10e^- ==> 2Mn^2+ + 8H2O ==================================== 2MnO4^- + 10Br^- + 16H+ ==> 2Mn^2+ + 5Br2 + 8H2O ---------------------------------------------------------------------------------------------- 3. Try to do this one yourself using the method I decribed above. The two half reactions are: (CrO7^2- SHOULD be Cr2O7^2-) CH3OH ==> CH2O . . .and . . .Cr2O7^2- ==> Cr^3+ Scroll down to check your answer. Cr2O7^2- + 8H^+ + 3CH3OH ==> 2Cr^3+ + 7H2O + 3CH2O

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Chemistry: Balancing redox reactions (2) :Chemistry: Oxidation and reduction. Oxidizing agent and reducing agent. Oxidation number, also known as oxidation state. Balancing oxidation-reduction reactions, also known as redox reactions, using the half-reaction method in acidic and basic solutions. This is a recording of a tutoring session, posted with the students' permission. These videos are offered on a "pay-what-you-like" basis. You can pay for the use of the videos at my website: www.freelance-teacher.com For a printable handout covering the material discussed in this video series, go to my website. For a list of all the available video series, arranged in suggested viewing order, go to my website. For a playlist containing all the videos in this series, click here: www.youtube.com (1) Oxidation and reduction. Oxidizing agent and reducing agent. Oxidation number (2) Formal charge and oxidation number (3) Oxidation numbers continued. Determining the oxidizing agent and reducing agent (4) Continued (5) Balancing redox reactions: a simple case (6) Balancing a redox reaction in acidic solution using the half-reaction method: an example (7) Continued (8) Example continued; balancing a redox reaction in basic solution using the half-reaction method (9) Another example of balancing a redox reaction using the half-reaction method (10) Continued (11) Another example (12) Continued (13) Continued (14) Continued (15) Another example (16) Continued (17) Another example (18) Another example (19) Continued

Chemistry: Balancing redox reactions (5) :Chemistry: Oxidation and reduction. Oxidizing agent and reducing agent. Oxidation number, also known as oxidation state. Balancing oxidation-reduction reactions, also known as redox reactions, using the half-reaction method in acidic and basic solutions. This is a recording of a tutoring session, posted with the students' permission. These videos are offered on a "pay-what-you-like" basis. You can pay for the use of the videos at my website: www.freelance-teacher.com For a printable handout covering the material discussed in this video series, go to my website. For a list of all the available video series, arranged in suggested viewing order, go to my website. For a playlist containing all the videos in this series, click here: www.youtube.com (1) Oxidation and reduction. Oxidizing agent and reducing agent. Oxidation number (2) Formal charge and oxidation number (3) Oxidation numbers continued. Determining the oxidizing agent and reducing agent (4) Continued (5) Balancing redox reactions: a simple case (6) Balancing a redox reaction in acidic solution using the half-reaction method: an example (7) Continued (8) Example continued; balancing a redox reaction in basic solution using the half-reaction method (9) Another example of balancing a redox reaction using the half-reaction method (10) Continued (11) Another example (12) Continued (13) Continued (14) Continued (15) Another example (16) Continued (17) Another example (18) Another example (19) Continued