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Chapter 22 Notes II

Chapter 22 Notes II. Balancing Redox Reactions. Steps to balancing a Redox reaction:. Lets balance the following equation to show the steps: HNO 3 + I 2 a HIO 3 + NO 2 + H 2 O. 1) Write the oxidation number above each element in the equation. HNO 3 + I 2 a HIO 3 + NO 2 + H 2 O.

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Chapter 22 Notes II

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  1. Chapter 22 Notes II Balancing Redox Reactions

  2. Steps to balancing a Redox reaction: Lets balance the following equation to show the steps: HNO3 + I2a HIO3 + NO2 + H2O

  3. 1) Write the oxidation number above each element in the equation. HNO3 + I2a HIO3 + NO2 + H2O +1 +5 -2 0 +1 +5 -2+4 -2 +1 -2

  4. 2) Draw a line connecting elements whose oxidation numbers change. HNO3 + I2a HIO3 + NO2 + H2O +1 +5 -2 0 +1 +5 -2+4 -2 +1 -2

  5. +1e- -5e- 3) Determine how much the oxidation number changed for each and write this number on the connecting line. HNO3 + I2a HIO3 + NO2 + H2O +1 +5 -2 0 +1 +5 -2+4 -2 +1 -2 The positive in front of the 1 means it is gaining 1 electron and the negative in front of the 5 means it is losing electrons.

  6. x 1 x 2 4) Multiply the number on the connecting line by the highest subscript on the element changing. HNO3 + I2a HIO3 + NO2 + H2O +1e- +1e- +1 +5 -2 0 +1 +5 -2+4 -2 +1 -2 -5e-

  7. x 1 x 2 5) Multiply the top and bottom line by numbers so that the electrons gained equal electrons lost. HNO3 + I2a HIO3 + NO2 + H2O +1e- +1e- +1 +5 -2 0 +1 +5 -2+4 -2 +1 -2 -5e- Right now, the top line multiplies to equal 1 and the bottom line multiplies to equal 10. What is the least common multiple of 1 and 10?

  8. x 1 x 2 5) Multiply the top and bottom line by numbers so that the electrons gained equal electrons lost. HNO3 + I2a HIO3 + NO2 + H2O x 10 +1e- +1e- +1 +5 -2 0 +1 +5 -2+4 -2 +1 -2 10! -5e- x 1 So we’ll need to multiply each line by a number that makes it equal to ten (the two numbers don’t have to be the same).

  9. x 1 x 2 6) Circle the last two numbers on each line. The product represents how many atoms of that element you need on each side. HNO3 + I2a HIO3 + NO2 + H2O x 10 +1e- +1e- =10 +1 +5 -2 0 +1 +5 -2+4 -2 +1 -2 -5e- x 1 =2 So we’ll need 10 nitrogens and 2 iodines on each side of the equation.

  10. x 1 x 2 7) Place the coefficients in the equation which will give you the determined number of atoms. HNO3 + I2a HIO3 + NO2 + H2O x 10 +1e- +1e- =10 +1 +5 -2 0 +1 +5 -2+4 -2 +1 -2 10 2 10 -5e- x 1 =2 Since there is already a two subscript on I2, it gets a one coefficient to bring the total number to two.

  11. 8) Check all atoms to see if everything is balanced. If hydrogen and oxygen are both unbalanced, try hydrogen first. HNO3 + I2a HIO3 + NO2 + H2O 10 2 10 4 Make sure not to disturb the ratio of elements oxidized to elements reduced that you just set up!

  12. Balancing the following using the steps: I2 + HClO + H2O a HIO3 + HCl 5 2 5

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