Chemistry Final Exam (Answered)
Oxidation Rules
Rule 1: The oxidation number of an element in its free state = 0 (Also true for diatomic elements)
-Mg=0,H2=0, C=0
Rule 2: The oxidation number of a monatomic (one-atom) ion is the same as the charge on the ion
-Mg^2+=+2, Na^+=+1, O^2-=-2
...
Chemistry Final Exam (Answered)
Oxidation Rules
Rule 1: The oxidation number of an element in its free state = 0 (Also true for diatomic
elements)
-Mg=0,H2=0, C=0
Rule 2: The oxidation number of a monatomic (one-atom) ion is the same as the charge
on the ion
-Mg^2+=+2, Na^+=+1, O^2-=-2
Rule 3: The sum of all oxidation numbers in a neutral compound is zero. The sum of all
oxidation numbers in a polyatomic (many-atom) ion is equal to the charge on the ion.
Rule 4: The oxidation number of an alkali metal (IA family) in a compound is +1; the
oxidation number of an alkaline earth metal (IIA family) in a compound is +2.
Rule 5: The oxidation number of oxygen in a compound is usually -2. If, however, the
oxygen is in a class of compounds called peroxides (H2O2), then the oxygen has an
oxidation number of -1. If the oxygen is bonded to fluorine, the number is +1.
Rule 6: The oxidation state of hydrogen in a compound is usually +1.
Rule 7: The oxidation number of fluorine is always -1. Chlorine, bromine, and iodine
usually have an oxidation number of -1, unless they're in combination with an oxygen or
fluorine.
-In HF=-1, H=0 because F=-1
A) What is the oxidation state of an individual sulfur atom in MgSO4?
B) What is the oxidation state of an individual nitrogen atom in NH2OH?
C) What is the oxidation state of an individual phosphorus atom in PO33−?
D) What is the oxidation state of each individual carbon atom in C2O42−?
A) +6
B) -1
C) +3
D) +3
Which compound contains phosphorous in its highest oxidation state?
PO43-
PI3
P4
P2O2
PO4^3-
P+4(-2)=-3
,PI3 (P=3)
P4 (P=0)
2Na (s) + Cl2 (g) = 2NaCl (s)
What is the oxidation, reduction, oxidizing agent, and reducing agent
What are the half reactions?
Oxidation: Na= Na^+1 + 1e^-
Reduction: Cl + 2e^- = 2Cl
Na is oxidized by Cl2 (Cl2 is the oxidizing agent)
Cl is reduced by Na (Na is the reducing agent or reductant)
What is the oxidation and reduction half reaction for:
Zn (s) + Cu2+ (aq) = Zn2+ (aq) + Cu (s)
Oxidation 1⁄2 -reaction (Anode):
Zn (s) = Zn2+ (aq) + 2 e-
Reduction 1⁄2 -reaction (Cathode):
2 e- + Cu2+ (aq) = Cu (s)
Electrochemistry
Flow goes from anode to cathode
Anode=oxidation (Anode half cell = oxidation occurs)
Cathode=Reduction (Cathode half cell= reduction occurs)
Cations go from anode to cathode
Anions go from cathode to anode
The "salt bridge" is a reservoir for cations/anions to replenish electrons and close the
electrical circuit
Electrons/electricity always headed towards CATHODE
Eocell (Cell potential) = Eordn (cathode) - Eordn (anode)
*Half-reactions are always written as reductions
Substances with more NEGATIVE reduction potentials reduce substances with more
positive reduction potentials
Substances with very NEGATIVE reduction potentials are good reducing agents (More
negative= more likely to be a reduction)
Substances with very POSITIVE reduction potentials are good oxidizing agents
Nernst Equation
Ecell= E°cell - (RT/nF) (lnQ)
At 25 DegC, Eo - (0.0592/n) logQ
Among the following reactions, identify which ones are redox reactions (and the
oxidizing agent if they are redox reactions)
A) NO2- (aq) + H+ (aq) = HNO2 (aq)
,B) 4Al(s) + 3O2 (g) = 2Al2O3 (s)
C) 3Cl2 (g) + 2Fe(s) 2FeCl3 (s)
A) No redox
B) Oxidizing agent = O2
C) Oxidizing agent = Cl2
What is the emf for a galvanic cell that employs the following balanced overall
cell reaction:
2 Al (s) + 3 I2 (s) = 2 Al3+ (aq) + 6 I- (aq)
I2 (s) + 2e^- = 2I (aq) = 0.54
Al3+ (aq) + 3e^- = Al (s) = -1.66
What is the emf for a cell employing the reaction below if [Al3+] = 0.004M and [I- ]
= 0.010M at 298K?
2Al(s) + 3I2 (s) → 2Al3+ (aq) + 6I- (aq)
2.2 V
Eordn are independent of stoichiometric factors !
V = J/C (energy per charge transferred) = const.
In a basic solution, H+ need to be replaced by OH- in the final equation!
Take H2O on right, move to left
Take 4H+ on left, add 4OH- on right
Calculate Oxidation values for:
1. FeO3
2. O3
3. MgCl
4. LiOH
5. H2O2
6. NH4+
7. HSO4-
8. ClO2-
9. Cr2O7^2-
10. K2O2
1. Fe (+6), O (3*-2)
2. O (3*0=0)
3. Mg (+2), Cl (-2)
4. Li (+1), O (-2), H (+1) - Start with Li value
, 5. H2 (21), O (2-1)
6. N (-3), H (4*1) = 1
7. H (1), S (6), O (4*-2)= -1
8. Cl (3), O (2*-2)= -1
9. Cr (26), O (7-2) = -2
10. K (21), O (21) - Start with K (rule 4)
What is the value of ΔGo (in kJ/mol) for the following reaction 25.0 oC?
Al (s) + Ag+ (aq) = Ag (s) + Al3+ (aq)
Eo = + 2.46 V
- 712 kJ/mol
ΔGo = - 3 x 96,500 J/V/mol x (+2.46 V)
Chapter 20 Equations
ΔGo = - RT lnK
Nernst: Eo = (RT/nF) lnK
Spontaneity of redox reactions: ΔGo = - n F Eo
n = # electrons transferred
F = Faraday's constant
Spontaneity of Redox Reactions
DeltaGo=-nFEo
n = # electrons transferred
F = Faraday's constant
Oxidation-Reduction Reactions (Redox Reactions)
-Reactions that involve the transfer of electrons from one species to another.
-In general, one element will lose electrons (oxidation), with the result that it will
increase in oxidation number, and another element will gain electrons (reduction),
thereby decreasing in oxidation number.
An oxidizing agent is an element or compound in a redox reaction that oxidizes another
species and itself gets reduced and is therefore the electron acceptor in the reaction.
A reducing agent is an element or compound in a redox reaction that reduces another
species and itself gets oxidized and is therefore the electron donor in the reaction.
-Oxidation means an increase in oxidation state and a loss of electrons and involves a
reducing agent.
-Reduction means a decrease in oxidation state and a gain of electrons and involves an
oxidizing agent.
FeO+CO→Fe+CO2
A) Which element is oxidized in this reaction?
B) Which substance is the oxidizing agent in this reaction?
Cr2O72−+3HNO2+5H+→2Cr3++3NO3−+4H2O
C) Which element is reduced in this reaction?
D) Which substance is the reducing agent in this reaction?
A) Carbon is oxidized
B) FeO is the oxidizing agent
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller LectDan. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for £10.09. You're not tied to anything after your purchase.
