Redox Equations Podcast

Slide 1: 

Podcast Created By: Silvia V. Oxidation Numbers & Oxidation-Reduction Equations

Oxidation Numbers : 

Oxidation Numbers Oxidation numbers are assigned to an element in a compound according to some rules This number enable us to describe oxidation- reduction states

Oxidation Number Rules : 

Oxidation Number Rules The oxidation state of any element such as Fe, H2, O2, P4, S8 is zero The oxidation # of oxygen in its compound is -2, except for peroxides like H2O2, and Na2O2, in which the # for O is -1 The oxidation # of hydrogen is +1 in its compound, except for metal hydrides, such as NaH, LiH, etc. in which the oxidation state for H is -1

Slide 4: 

4. The oxidation states of other elements are then assigned to make the algebraic sum of the oxidation states equal to the net charge on the molecule or ion. 5. The following elements usually have the same oxidation states in their compounds: +1 for alkali metals; Li, Na, K, Rb, Cs +2 for alkaline earth metals; Be, Mg, Ca, Sr, Ba -1 for halogens except when they form compounds with oxygen or one another

Examples : 

Examples a)Br2 b)SrO c) MoO4-2 d) As2S3 0 -2 -2 -2 +2 +6 +3

Redox Equations : 

Redox Equations They are equations that involve the transfer of electrons We use oxidation numbers to divide equations into half reactions Reduction Where the half reaction gains electrons Oxidation Where the half reaction loses electrons OIL RIG Cu(s) Cu+2 2 Ag+ 2 Ag s xidation s osing eduction aining

Balancing Redox Equations in Acidic Solution : 

Balancing Redox Equations in Acidic Solution Write the balanced reaction Write the net ionic equation Divide the equations into half reactions Balance atoms other than H and O Balance your oxygen by adding water to the side that needs oxygen Balance the hydrogen by adding H+ Make the # of e- gained = to the # of e- lost using coefficients Add the half reactions Cancel anything that is on both sides

Example 1 : 

2Ag+ + Cu Cu2+ + 2Ag 2Ag+ + 2NO3- + Cu Cu2+ + 2NO3- + 2Ag 2AgNO3(aq) + Cu(s) Cu(NO3)2(aq) + 2Ag(s) Example 1 Silver nitrate reacts with copper metal to form copper(II) nitrate and solid silver Step 1: write the balanced reaction Step 2: write the net ionic equation

Example 1 Continued2Ag+ + Cu Cu2+ + 2Ag : 

Example 1 Continued2Ag+ + Cu Cu2+ + 2Ag Oxidation Cu Cu2+ Reduction 2Ag+ 2Ag OIL RIG Step 3: Divide the equation into half reactions Step 4: Balance atoms other that H and O Step 5: Balance your oxygen by adding water to the side that needs oxygen Step 6: Balance the hydrogen by adding H+ Step 7: Make the # of e- gained = to # of e- lost Step 8: Add the half reactions Step 9: Cancel anything on both sides Cu + 2e- + 2Ag+ Cu2+ + 2e- + 2Ag Cu + 2Ag+ Cu2+ + 2Ag + 2e- 2e- +

Example 2 : 

H+(aq) + Cr2O72-(aq) + C2H5OH(l) Cr3+(aq) + CO2(g) + H20(l) Reduction Cr2O72-(aq) Cr3+(aq) Oxidation C2H5OH(l) CO2(g) Example 2 Potassium dichromate (K2Cr2O7) reacts with ethyl alcohol (C2H5OH) as follows: Step 3: Divide the equations into half reaction Step 4: Balance atoms other than H and O Step 5: Balance the oxygen by adding water Step 6: Balance the hydrogens by adding H+ Step 7: Balance the electrons by using coefficients Step 8: Add the half reactions Step 9: Cancel anything on either sides OIL RIG 2 2 3H20 + + 12 H+ + 12 e- + 7H20 14 H+ + 6e- + 2 3H2O + C2H5OH (l) + 12e- + 28H+ + 2Cr2O72- (aq) 2CO2(g) + 12H+ + 12e- + 4Cr3+(aq) + 7H2O 4 16 C2H5OH(l) + 2Cr2O72-(aq) + 16H+ 2CO2(g) + 4Cr3+ +4H2O

Balancing Redox Equations in Basic Solution : 

Balancing Redox Equations in Basic Solution Write the balanced reaction Write the net ionic equation Divide the equations into half reactions Balance atoms other than H and O Balance your oxygen by adding water to the side that needs oxygen Balance the hydrogen by adding H+ Make the # of e- gained = to the # of e- lost using coefficients Add the half reactions Cancel anything that is on both sides Eliminate H+ by adding OH- to form H2O Eliminate the water from both sides

Example 3 : 

Al(s) + OH- Al(OH)4- + H2(g) Example 3 Balance the following net ionic equation in basic solution 3: Divide the equations in half reactions 4: Balance the atoms other than H and O 5: Balance the O by adding water 6: Balance the H by adding H+ 7: Make the # of e- gained = the # of e- lost by using coefficients 8: Add the half reactions 9: Cancel anything on either sides OIL RIG + 4H+ Oxidation Al(s) Al(OH)4- 4H2O + + 3e- 2 Reduction OH- H2 + H2O 3H+ + 2e- + 3 8H2O + 2Al(s) + 6e- + 9H+ + 3OH- 2Al(OH)4- + 8H+ + 6e- + 3H2(g) + 3H2O + OH- 5 1 2 5H2O + 2Al(s)+ 1H+ + 2OH- 2Al(OH)4- + 3H2(g)

Example 3 Continued : 

Example 3 Continued Eliminate H+ by adding OH- to form H2O Eliminate the water from both sides 5H2O + 2Al(s)+ 1H+ + 2OH- 2Al(OH)4- + 3H2(g) 5H2O + 2Al(s)+ 1H+ + 2OH- 2Al(OH)4- + 3H2(g) + OH- 5H2O + 2Al(s)+ H2O + 2OH- 2Al(OH)4- + 3H2(g) 6H2O + 2Al(s)+ 2OH- 2Al(OH)4- + 3H2(g)