Electrochemistry Oxidation-reduction reactions

Review Ionic bonds : give and take of electrons (example NaCl) Ionic bonds : give and take of electrons (example NaCl) Covalent bonds: electrons shared (example CO 2 ) Covalent bonds: electrons shared (example CO 2 ) All elements in the molecule become STABLE (like a noble gas in electron configuration) and NEUTRAL (balanced in charge) All elements in the molecule become STABLE (like a noble gas in electron configuration) and NEUTRAL (balanced in charge)

Oxidation Numbers We can assign a specific charge to elements involved in bonding We can assign a specific charge to elements involved in bonding For simplicity, we assign full charges to elements even if they are sharing (example H 2 0…H is considered to be +1 and O as -2) For simplicity, we assign full charges to elements even if they are sharing (example H 2 0…H is considered to be +1 and O as -2) (H +1 ) x 2 + O - 2 = H 2 0 (H +1 ) x 2 + O - 2 = H 2 0 (Oxygen has a tighter hold on the electrons therefore is “negative”) (Oxygen has a tighter hold on the electrons therefore is “negative”)

Building molecules We can use oxidation numbers to predict the ratios of elements or elements with polyatomic negative ions in a stable and neutral molecule. We can use oxidation numbers to predict the ratios of elements or elements with polyatomic negative ions in a stable and neutral molecule. Example PbSO 4 Example PbSO 4 (SO 4 ) has a 2- charge,therefore Pb must be +2 (SO 4 ) has a 2- charge,therefore Pb must be +2 Name = Lead (II) Sulfate Name = Lead (II) Sulfate

Determining Oxidation Numbers What about SO 4 2- ? What about SO 4 2- ? Oxygen is given oxidation number -2 Oxygen is given oxidation number -2 4 oxygen atoms = -8 charge! 4 oxygen atoms = -8 charge! What must S’s charge be so that SO 4 has a total net charge of 2-? What must S’s charge be so that SO 4 has a total net charge of 2-? S must have a +6 Oxidation State S must have a +6 Oxidation State

Practice! Determine the oxidation numbers of each element in the following examples: Determine the oxidation numbers of each element in the following examples: HCl HCl PbO PbO Fe 2 O 3 Fe 2 O 3 CuNO 3 CuNO 3 NaOH NaOH PO 4 3- PO 4 3- HClO 4 HClO 4

OXIDATION Process whereby an element becomes MORE positive by losing electrons Process whereby an element becomes MORE positive by losing electrons REMEMBER that there are protons in the nucleus that are now unequal to the number of electrons REMEMBER that there are protons in the nucleus that are now unequal to the number of electrons

REDUCTION Process whereby elements become MORE negative by gaining electrons Process whereby elements become MORE negative by gaining electrons Element now has MORE electrons than it has positive protons in the nucleus and has a net charge that is negative. Element now has MORE electrons than it has positive protons in the nucleus and has a net charge that is negative.

Oxidizing agent This would be any element or molecule that TAKES electrons allowing another element to be oxidized (become more positive by losing electrons) This would be any element or molecule that TAKES electrons allowing another element to be oxidized (become more positive by losing electrons) The oxidizing agent is reduced in the process The oxidizing agent is reduced in the process

Reducing agent This would be any element or molecule that GIVES electrons to another element allowing it to be reduced- becoming more negative by gaining electrons This would be any element or molecule that GIVES electrons to another element allowing it to be reduced- becoming more negative by gaining electrons The reducing agent is oxidized in the process The reducing agent is oxidized in the process

Example Sodium reacts with carbon to produce sodium carbide Sodium reacts with carbon to produce sodium carbide Identify the oxidation number of each element Identify the oxidation number of each element Determine if any of the elements changed oxidation number Determine if any of the elements changed oxidation number Determine who is acting as the agents of change! Determine who is acting as the agents of change!

Writing half reactions Once we identify which elements are being oxidized and reduced, we can write out “half reactions” for the transfer of electrons involved Once we identify which elements are being oxidized and reduced, we can write out “half reactions” for the transfer of electrons involved Eg:Na + C--->Na 4 C Eg:Na + C--->Na 4 C Sodium is losing an electron= oxidized Sodium is losing an electron= oxidized Carbon is gaining electrons = reduced Carbon is gaining electrons = reduced

Balancing Na 0 Na e- (oxidized) Na 0 Na e- (oxidized) C e- C -4 (reduced) C e- C -4 (reduced) We need to make sure that the number of electrons given and taken is EQUAL! We need to make sure that the number of electrons given and taken is EQUAL! Therefore we will need 4 sodium atoms to donate all of the electrons needed for carbon Therefore we will need 4 sodium atoms to donate all of the electrons needed for carbon This is often a good way to help balance difficult reactions. This is often a good way to help balance difficult reactions.

Try this! Magnesium solid reacts with oxygen gas to produce magnesium oxide Magnesium solid reacts with oxygen gas to produce magnesium oxide STEP #1-write a chemical equation STEP #1-write a chemical equation STEP #2-assign oxidation numbers STEP #2-assign oxidation numbers STEP #3-write “half reactions” STEP #3-write “half reactions” STEP #4-label oxidation/reduction STEP #4-label oxidation/reduction STEP #5-balance electrons STEP #5-balance electrons STEP #6 -balance the equation! STEP #6 -balance the equation!

Try these for homework! Fe + CuS0 4 Cu + FeS0 4 Fe + CuS0 4 Cu + FeS0 4 H H 2 0 H H 2 0 Cu + AgN0 3 Cu(N0 3 ) 2 + Ag Cu + AgN0 3 Cu(N0 3 ) 2 + Ag HNO 3 + H 2 SS + NO + H 2 O HNO 3 + H 2 SS + NO + H 2 O

Redox Video