Redox Reactions

Oxidation numbers identify and indicate which element is oxidized and which is reduced. Here's an example - the reaction between sodium metal and chlorine gas: It is often useful to write the oxidation number for every element above the element in the equation. Thus for our reaction we have: balancing coefficients in the equation  do not affect the value of the oxidation numbers. 2 Na + Cl2 → 2 NaCl      +1   -1 2 Na + Cl2 → 2 NaCl

example Add in the oxidation numbers 2 Mg + O2 → 2 MgO Add in the oxidation numbers which are oxidized and which are reduced? An increase in oxidation number indicates oxidation A decrease in oxidation number indicates reduction

Reducing agent the substance that is oxidized. It allows another element to be reduced. Oxidizing agent the substance that is reduced. It allows another element to be oxidized.

example N2 + 2H2 → 2 NH3 element initial ox no final ox no e-   final ox no e- oxidized or reduced Agent

Assignments Practice 6.1.3 Assignment 6.1.3

Balancing Redox reactions Many redox reactions cannot easily be balanced just by counting atoms. Consider the following net ionic equation Cu(s) + Ag+(aq) → Cu2+(aq) + Ag(s) If you simply count atoms, the equation is balanced but the charges aren’t balanced! Charges represent gain or loss of electrons, and, like atoms, electrons are conserved during chemical reactions.

1. Balancing Equations using Oxidation Numbers Cu(s) + Ag+(aq) → Cu2+(aq) + Ag(s element initial ox no   final ox no change in e- balance for electrons

example MnO41- + Fe2+ + H1+ → Mn2+ + Fe3+ + H2O element initial ox no   final ox no change in e- balance for electrons

example NH3 + O2 → NO2 + H2O Before using a multiplier to get the electrons to match, notice the subscript with oxygen - O2. In our summary chart we base our oxidation number changes on a single atom, but our formula tells us that we must have at least two oxygen.   element initial ox no   final ox no change in e- balance for electrons

element initial ox no   final ox no change in e- No. atoms No. e- balance for electrons 28

Since we were counting oxygen atoms in the O2 molecule on the reactant side of the equation, that's where we'll use the "7". Since nitrogen's oxidation number also changed we will use nitrogen's balancing coefficient there 4 NH3 + 7 O2 → 4 NO2 + H2O The last step is to balance for hydrogen atoms (and finishing oxygen), which will mean placing a 6 in front of H2O: 4 NH3 + 7 O2 → 4 NO2 + 6 H2O

K2Cr2O7 + NaI + H2SO4 → Cr2(SO4)3 + I2 + H2O + Na2SO4 + K2SO4

K2Cr2O7 + NaI + H2SO4 → Cr2(SO4)3 + I2 + H2O + Na2SO4 + K2SO4 Next, check for any subscripts associated with either of these two elements - we see that Cr always has a subscript of "2" (in both K2Cr2O7 and Cr2(SO4)3), and I has a subscript in I2. So we'll add that to our summary chart to get a total number of electrons transferred, and then balance element initial ox no   final ox no change in e- No. atoms No. e- balance for electrons

element initial ox no   final ox no change in e- No. atoms No. e- balance for electrons

One more example One more tricky one. Balance Zn + HNO3 → Zn(NO3)2 + NO2 + H2O Determine oxidation numbers and create your summary chart: The main thing to notice is that N appears in two separate products - Zn(NO3)2 and NO2. Should we consider the subscript for nitrogen from Zn(NO3)2? In this case no, because this compound also contains Zn, the oxidized element. Also, the oxidation number for nitrogen does not change from HNO3 to Zn(NO3)2 . element initial ox no   final ox no change in e- No. atoms No. e- balance for electrons

We now get our balancing coefficients from our summary table We now get our balancing coefficients from our summary table. A "1" will be placed in front of Zn, but which N should we use for the "2"? If you put it in front of both HNO3 and NO2 you'll find you cannot balance for nitrogen atoms. Since the oxidation number for nitrogen changed in becoming NO2, we will try it there first. Some trial-and-error may be required: