Friday, June 26, 2020

SCH 4U - Electrolysis





Homework: #30-37


Success Criteria

- be able to determine the oxidation number of an element in a compound or a polyatomic ion.

- be able to define oxidation, reduction, half-reaction, redox reaction, spectator ion, inert electrode, cell, battery, primary/secondary cell, corrosion, sacrificial anode, cathodic protection, electrolysis, etc.

- be able to write out (i) total ionic equation, (ii) net ionic equation, (iii) balanced oxidation half-reaction (iv) balanced reduction half reaction for a redox reaction, (v) provide Eox & Ered, (vi) calculate Ecell and (vii) determine/explain if a cell reaction is spontaneous

- be able to balance a redox reaction equation (in acidic or basic solution) using the ion-electron &/or the oxidation number method

- be able to draw a fully labelled galvanic cell from a shorthand notation representing the cell and vice versa

- be able to choose the best/worst oxidizing/reducing agent from a list using E values 

- be able to answer multiple choice questions based on the "Cells and Batteries" and "Corrosion" reading

- be able to interpret an electrolytic cell (as in the lesson), with an emphasis on determining the minimum required voltage

Thursday, June 25, 2020

SCH 3U - SCH 3U Review

This is the exam review.  It is worth your while to be sure that you can do the questions in this booklet, as it will aid your success on the exam and in future chemistry courses (I'll see some of y'all next year 😉).

The most crucial concepts at which you MUST be proficient are:

  • nomenclature (surprise, surpise) ↝ #3-8
  • bonding ↝ #14, 18
  • predicting products ↝ #20,
  • using the Activity Series of Metals and the Solubility rules (as appropriate) ↝ #20
  • writing out balanced equations, with the correct state symbols ↝ #20
  • writing out total ionic and net ionic equations (as appropriate) ↝ #20
  • calculations involving moles, molar mass, mass ↝ #21, 22, 28, 32
  • calculations involving pressure, volume, temperature ↝ # 52-56
  • calculations involving moles, volume, concentration ↝ # 38, 41, 45, 46, 48
  • acid and base definitions and characteristics ↝ #47
  • pH scale & calculations ↝ # 39, 40, 49
  • stoichiometric calculations ↝ #33, 34, 44, 57-59
  • intermolecular forces and properties ↝ #35, 36

Obviously, there are a lot of concepts and skills that you must bring with you into the next step in your chemistry education. You should be able to do all of the above questions without looking at notes/textbook/internet or asking for help from friends/family/tutor.  It is only fair to warn you that the SCH 4U chemistry course is a large step up in terms of time, effort and ability compared to SCH 3U.  So, be prepared. 

Well folks, there's nothing left but the cheering.  We made it.  We survived and we prevailed. 

Homework:

 

 
  
Answers:

 

 

 

 

 

 

SCH 4U - Corrosion

Read Section 10.6 in the textbook.


Homework:  #24-29

Wednesday, June 24, 2020

SCH 3U/4C - Acid-Base Titration

Acid-Base Titration
A titration is a chemical analysis involving the progressive addition of a solution of known solute concentration, called the titrant, into a solution of unknown concentration, called the sample. 

The purpose is to determine the quantity of a specified chemical in the sample, from which the concentration and/or the molar mass of the chemical may be determined.  This is possible because the titrant and the sample contain substances that react according to known stoichiometry. 

Generally, the sample is placed in a flask and the titrant is placed in a burette. 

     
In an acid-base titration, we are interested in the point at which the number of moles of acid is equal to the number of moles of base; this is referred to as the equivalence point.  An indicator is chosen so that the endpoint occurs precisely at the equivalence point.  

na = nb
caVa = cbVb
                 ca(25.0 mL) = (0.10 mol/L)(34.6 mL)
              ca = 0.14 mol/L



Here is a video, showing how a titration is performed.
 


Homework:
Practice Problems p. 466 # 1-3 (hint: write out the balanced equation between the acid and the base first as stoichiometric ratios matter.)
 

 
Answers:



Success Criteria:

- know the Arrhenius and Bronsted-Lowry definitions of acid and base
- be able to define and explain the use of an indicator
- be familiar with the hydronium ion and what it represents
- know and apply the definition for ionization, dissociation, autoionization
- know the difference between a strong and weak electrolyte
- know the difference between a dilute and concentrated electrolyte solution
- be familiar with examples of strong, weak, mono-/di-/tri-/polyprotic acids
- be familiar with examples of strong, weak bases
- be able create an acid/base reaction and then label acid-conjugate base and base-conjugate acid pairs
- know the properties of acids and bases (taste, colour with litmus, etc)
- be able to draw the pH scale and label the regions
- be able to do acid-base calculations: pH, [H+], [OH-], titration (caVa = cbVb)
- know the info in the Acid Rain reading
- be able to create a balanced equation for an acid-base neutralization reaction 
- know the definitions and theory associated with titration


Acid-Base Review - do not do #6, 9, 13, 17


 
Answers:




SCH 4U - Technology of Cells & Batteries

Read Section 10.3 in the textbook.

Homework: #20-23

Monday, June 22, 2020

SCH 4U - Ecell

Standard Reduction Potentials, E°red
There are tables of standard reduction potentials, E°red, created by chemists.  All the reactions listed in the table are reversible.  As written, you have reduction half-reactions.  Flip the equation and you have oxidation half-reactions, which require the E° value to have a change of sign.  

Some extra half-reactions that aren't on the table, but might prove useful:

Cr2+(aq) + 2e →  Cr(s)    Eored = -0.91 V

Au3+(aq) + 3e → Au(s)    Eored = +1.50 V

SO42-(aq) + H2O(l) + 2e    SO32-(aq) + 2OH-(aq)    Eored = -0.93 V




Oxidizing & Reducing Agents
The more positive the E° for a ½-reaction, the more likely the reaction will occur as written.  A negative E° indicates that the species is more difficult to reduce than H+(aq).

From the table, 
F2(g)  +  2e-    2F-(aq)          E°red = 2.87 V
Li+(aq)  +  e-    Li(s)              E°red = -3.05 V

So F2(g) is the most easily reduced (strongest oxidizing agent) and Li+(aq) is the most difficult to reduce (poorest oxidizing agent).

Example  Which is the strongest oxidizing agent, MnO4- (in acid solution), I2(s) or Zn2+(aq)?

Answer  Using the table, find and write out the half reaction for each substance and its corresponding E° value.  I have dropped the state symbols here, so that the info would fit on the line - normally, we include the states for each entity.
MnO4-  +  8H+  +  5e-    Mn2+  +  4H2O        E°red = +1.51 V
I2  +  2e-    2I-                                                E°red = +0.536 V
Zn2+  +  2e-    Zn                                           E°red = -0.763 V

Thus, since the MnO4-  has the most positive E°red, it is the most easily reduced so it is the strongest oxidizing agent.



Spontaneity of Redox Reactions
We have seen that voltaic cells use redox reactions that proceed spontaneously.  Any reaction that occurs in a voltaic cell that produces a positive E°cell will be spontaneous.

Example  Are the following reactions spontaneous under standard conditions?
(a) Cu(s)  +  2K+(aq)    Cu2+(aq)  +  2K(s)
(b) Cl2(g)  +  2I-(aq)    2Cl-(aq)  +  I2(s)              

Answer
(a)
  • Split the reaction into two half-reactions – one for oxidation and one for reduction.  Do this by grouping like substances together, maintaining their positions as reactants or products.
Cu(s)    Cu2+(aq) 
K+(aq)    K(s)

  • Add electrons to each half-reaction to balance out the charges.
Cu(s)    Cu2+(aq)  +  2e-
K+(aq)  +  e-    K(s)

  • If the number of electrons in each half reaction does not match, multiply each half reaction by the lowest factor needed to balance.
Cu(s)    Cu2+(aq)  +  2e-
2K+(aq)  +  2e-    2K(s)

  • Look up the half reactions in the table.  If you find the half reaction as written, it is the reduction.  Write down the E°redNote that we DO NOT double the E°red even though the half reaction is doubled.
2K+(aq)  +  2e-    2K(s)        E°red = -2.93 V

  • If the reaction in the table but it is backward, flip it.  Write down the E°ox.  
Cu(s)    Cu2+(aq)  +  2e-       E°ox = -0.34 V

  • Now add together the two half reactions and their corresponding values.  The electrons should cancel out completely.
                  Cu(s)    Cu2+(aq)  +  2e-       E°ox = -0.337 V
    2K+(aq)  +  2e-    2K(s)                       E°red = -2.925 V
______________________________________________
Cu(s)  +  2K+(aq)    Cu2+(aq)  +  2K(s)   E°cell = -3.262 V

If the E°cell is positive, the reaction is spontaneous; if it is negative, the reaction is not spontaneous.
∴ not spontaneous

(b)
     Cl2(g)  +  2e-    2Cl-(aq)                E°red = +1.359 V
               2I-(aq)    I2(s)  +  2e-           E°ox = -0.536 V
__________________________________________
Cl2(g)  +  2I-(aq)    2Cl-(aq)  +  I2(s)   E°cell = +0.823 V
spontaneous


Homework: #10, 11a, 18a

*****TIP: When faced with multiple half-reactions containing the same substance (for instance copper(II)), use the half-reaction that has the solid metal in combination with an ion (for instance, Cu(s)    Cu2+(aq)  +  2e-).*****