Electrochemistry : Electrochemistry The study of the interchange
between chemical and electrical energy
Oxidation-Reduction Reaction Review : Oxidation-Reduction Reaction Review “Redox” reactions involve the transfer of electrons (e-)
We use oxidation states to keep track of electron transfer
Some rules for assigning oxidation numbers (or states):
Oxidation numbers are roughly equivalent to hypothetical charge in compounds (there are some exceptions)
H is almost always +1 in compounds (Group 1 element)
O is almost always -2 in compounds (Group 6 element)
F is always -1 in compounds (Group 7 element)
For elements (H2, O2, F2, Ca, K, Mn, etc.) the oxidation state is always 0
Oxidation-Reduction Review Con’t : Oxidation-Reduction Review Con’t The overall charge of a compound = the sum of the oxidation states of all atoms in it
Neutral (i.e. H2O, CO2, CH4)
H2O : H = +1, O = -2
The overall charge is 2(1) + -2 = 0
CO2: What is the oxidation state of C?
Since C + 2 (O) = 0
C + 2(-2) = 0, thus C = +4
CH4: Is C still +4?
H is always +1
Therefore to remain neutral, 4(1) + C = 0
C must = -4
Oxidation-Reduction Review Con’t : Oxidation-Reduction Review Con’t Charged compounds (NO3-, CO32-, NH4+)
NO3- or (NO3)- : What is the oxidation # of N?
O is -2, and the overall charge is -1
Thus, N + 3(O) = -1 or N + 3(-2) = -1
N = +5
(CO3)2-: What is the oxidation # of C?
O is -2, and the overall charge is -2
Thus, C + 3(O) = -2 or C + 3(-2) = -2
C = +4
The oxidation # of charged atoms = charge
Mn3+ has an oxidation # of +3
S2- has an oxidation # of -2
Oxidation-Reduction Review Con’t : Oxidation-Reduction Review Con’t Try these…MnO4-, Cr2O72-, C2O42-
(MnO4)-
O = -2, so [4(-2) + Mn = -1]
Mn = +7
(Cr2O7)2-
O = -2, so [7(-2) + 2Cr = -2]
2Cr = 12, therefore Cr = +6
(C2O4)2-
O = -2, so [2C + 4(-2) = -2]
2C = 6, therefore C = +3
Assigning Oxidation # Practice : Assigning Oxidation # Practice Assign oxidation numbers to each atom
Cl2
Fe2+
ClO3-
ClO4-
IO2-
CrO42-
Fe3(PO4)2
CoSO4
H2CO3
Assigning Oxidation # Practice : Assigning Oxidation # Practice Assign oxidation numbers to each atom
Cl2: element ? oxidation # = 0
Fe2+ ion ? oxidation # same as charge… +2
ClO3- : O = -2, Cl = +5 (Cl is usually -1)
ClO4- : O = -2, Cl = +7
IO2- : O = -2, I = +3
CrO42- : O = -2, Cr = +6
Fe3(PO4)2 : O = -2, P = +5, Fe = +2
CoSO4 : O = -2, S = +6, Co = +2
H2CO3 : O = -2, C = +4, H = +1
Oxidation-Reduction Reactions : Oxidation-Reduction Reactions Two separate reactions occurring simultaneously
Oxidation: oxidation # of an atom increases
i.e. Fe(s) ? Fe3+(aq) oxidation state goes from 0 ? +3
Reduction: oxidation # of an atom is “reduced”
i.e. O2 ? O2- (oxidation state goes from 0 ? -2)
When occurring together…
Fe + O2 ? Fe3+ + O2-
This common redox reaction is the formation of rust
But, how do we balance this?
Balancing by Half-Reactions*in acidic solution : Balancing by Half-Reactions*in acidic solution CH3OH (aq) + Cr2O72-(aq) ? CH2O(aq) + Cr3+(aq)
1. Assign oxidation states to all atoms within the reaction
C-2H4+O2- + (Cr26+O72-)2- ? C0H2+O2- + Cr3+
2. Determine which atom is being reduced and which is oxidized and write separate reactions
Ox: C-2H4+O2- ? C0H2+O2- (C is going from -2 to 0)
Red: (Cr26+O72-)2- ? Cr3+
(Cr is being reduced from +6 to +3)
Balancing by Half-Reactions*in acidic solution : Balancing by Half-Reactions*in acidic solution 3. For each half-reaction:
Disregard all oxidation states except those that are changing (all others will cancel each other out)
C-2H4+O2- ? C0H2+O2- into C-2H4O ? C0H2O
Balance elements (except H and O) by using coefficients (like normal)
C-2H4O ? C0H2O Carbon is already balanced
Balance O’s by adding H2O to either side
C-2H4O ? C0H2O Oxygen is already balanced
Balancing by Half-Reactions*in acidic solution : Balancing by Half-Reactions*in acidic solution 3. For each half-reaction:
Balance H’s by adding H+ to either side (in acidic solution)
Where do we add H+?
C-2H4O ? C0H2O
Double check that all atoms are balanced
Balance charge by adding electrons to either side
Remember that electrons are negative charges
Use changing oxidation state as a guide + 2H+
Slide 12 : C-2H4O ? C0H2O + 2H+
Now that all elements are balanced, we have to balance the charges
Balance charges by adding e- to either side
All charges stay the same, except C
Therefore add e- to make the charge equal on each side
C-2H4O ? C0H2O + 2H+
-2 charge ? 0 charge
Now we have to use the same procedure on the reduction reaction!!!!!!!
I’ll bet you just can’t wait!!!!! + 2e- -2 charge = -2 charge
Time to balance the reduction half : Time to balance the reduction half Reduction half reaction: (Cr26+O72-)2- ? Cr3+
Disregard all other oxidation states (the ones that aren’t changing): (Cr26+O7)2- ? Cr3+
Balance all elements except H and O
(Cr26+O7)2- ? Cr3+
Balance O by adding H2O if necessary
(Cr26+O7)2- ? 2Cr3+
Balance H by adding H+ if necessary
(Cr26+O7)2- ? 2Cr3+ + 7H2O
Balance charge by adding e-
14H+ + (Cr26+O7)2- ? 2Cr3+ + 7H2O 2 + 7H2O 14H+ + 6e- +
Adding Half-Reactions*in acidic solution : Adding Half-Reactions*in acidic solution Now that we have balanced each half rxn, we need to add them together to get the full rxn
Ox: C-2H4O ? C0H2O + 2H+ + 2e-
Red: 6e- + 14H+ + (Cr26+O7)2- ? 2Cr3+ + 7H2O
4. In order to add the rxn’s, e- must be equal
So, multiply rxn’s by integers to equalize electrons
C-2H4O ? C0H2O + 2H+ + 2e-
6e- + 14H+ + (Cr26+O7)2- ? 2Cr3+ + 7H2O
3C-2H4O ? 3C0H2O + 6H+ + 6e- 3 ( )
Let the adding begin… : Let the adding begin… 6e- + 14H+ + (Cr26+O7)2- ? 2Cr3+ + 7H2O
3C-2H4O ? 3C0H2O + 6H+ + 6e-
3CH4O + + Cr2O72- ? 3CH2O + 2Cr3+ + 7H2O
…and the reaction is now balanced!
5. Double check that all atoms and charges are balanced 8H+
Balancing by Half-Reactions(in basic solution) : Balancing by Half-Reactions(in basic solution) Write oxidation and reduction half-reactions
Balance each half-reaction
Balance elements except for O, H
Balance O by adding H2O
Balance H by adding H+
Add enough OH- to both sides to cancel out any H+ and form H2O
Recall that H+ + OH- ? H2O
Balance charge by adding electrons
Equalize electrons and add half-reactions
Check elements and charges for balance
Balancing by Half-Reactions(in basic solution) : Balancing by Half-Reactions(in basic solution) Let’s balance a previous example in basic solution
Remember, it is all the same steps up to this point
14H+ + (Cr26+O7)2- ? 2Cr3+ + 7H2O
+ (Cr26+O7)2- ? 2Cr3+ + 7H2O + 14OH-
7H2O + (Cr26+O7)2- ? 2Cr3+ + 14OH-
Now, continue as you normally would… + 14OH- + 14OH- 14H2O
Practice Balancing Redox Reactions : Practice Balancing Redox Reactions Unbalanced reaction (in acid):
MnO4? + Fe2+ ? Mn2+ + Fe3+
Balanced Reduction half-reaction:
8H+ + MnO4? + 5e? ? Mn2+ + 4H2O
Balanced Oxidation half-reaction:
Fe2+ ? Fe3+ + e?
Balanced overall reaction:
8H+ + MnO4? + 5Fe2+ ? Mn2+ + 5Fe3+ + 4H2O 5( )
Redox Vocabulary : Redox Vocabulary MnO4? + Fe2+ ? Mn2+ + Fe3+
Oxidized species (atom, ion, molecule, or compound)
whichever species is being oxidized (? oxidation #)
Reduced species
whichever species is being reduced (? oxidation #)
Oxidizing agent
Whichever species CAUSES oxidation to occur—in other words, the oxidizing agent IS the reduced species
Reducing agent
Whichever species CAUSES reduction to occur—in other words, the reducing agent IS the oxidized species
Slide 20 : MnO4- + Fe2+ ? Mn2+ + Fe3+ Oxidized species Reduced species Reducing Agent Oxidizing Agent