Welcome to Murph's World!

Hello Fellow Chemists,

Here is some information that you will find useful: 

 

Today's Mission:

• Read the information below, starting with "How to Access Lessons." 
• Look over the information in tab at the top of the blog that pertains to your class (for example, SCH3U Intro).  
• You and your parent/guardian must sign the Course Information Sheet, Syllabus and Safety Contract. 
  
• To save paper, write "I have read and understood the CIS, syllabus and safety contract.  Signed __________(you)    ________(parent)" on a piece of paper.  Send me this through an Edsby message.
• Do the Diagnostic Quiz at the bottom of that same tab.
  

How to Access Lessons:

All lessons (many with accompanying videos - mainly made by me), homework and answer keys can be found on this blog. 

A course timeline, with hyperlinks to the appropriate blog lesson, is posted on Edsby.

  

How to Access Help:

You can seek out help/clarification when the need arises by asking during class or contacting me on Edsby.  

I also strongly encourage you to start/become a part of a group chat.  The hive mind is typically stronger than the individual.  😃

SNC 2P/SCH 3U/SCH 4C/SCH 4U - Chemistry - Valences

Valences

A valence is the real or apparent charge on an atom in a compound.

Grab a fresh Periodic Table and several coloured pencils and let's get cracking.

 

This chart is also useful for determining valences.

 

 

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

SCH 4U - Corrosion

Read Section 10.6 in the textbook.


Homework:  #24-29

SCH 4U - Technology of Cells & Batteries

Read Section 10.3 in the textbook.

Homework: #20-23

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)    E^ored = -0.91 V

Au³⁺(aq) + 3e → Au(s)    E^ored = +1.50 V

SO₄^2-(aq) + H₂O(l) + 2e  →  SO₃^2-(aq) + 2OH^-(aq)    E^ored = -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°_red.  Note 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^-).*****

 

 

SCH 4U - Galvanic Cells

Galvanic Cells

The energy released in a spontaneous redox reaction can be used to perform electrical work.  This occurs in a galvanic cell (aka a voltaic cell) which is a device in which electron transfer is forced to take place through an external pathway, rather than directly between reactants.

Every galvanic cell consists of two beakers, filled with electrolyte solutions.  As well as, two electrodes (which are immersed in the solutions), an external circuit (wires) and usually a voltmeter (although a load, like a light bulb, could be placed here to harness the electrical energy produced by the cell).  Cells can be set up with many combinations of electrodes and electrolytes - this is just one possibility.          *****Let me take you through the cell*****  Starting on the left,  in the anode compartment, the solid zinc electrode undergoes oxidation to place zinc ions in solution (therby increasing the [zinc ion] in the electrolyte solution and decreasing the mass of the Zn electrode).  Two electrons are also produced in this process, which travel, via the external circuit, to the cathode compartment, where they are used in the reduction of the copper(II) ions in the electrolyte solution.  Solid copper is produced and it plates out onto the electrode (thereby decreasing the [copper(II) ion] in the electrolyte solution and increasing the mass of the Cu electrode).  The salt bridge closes the circuit and acts to  maintain electrical neutrality.  Notice that the oxidation reaction in the anode compartment produces positive zinc ions - so, negative nitrate ions flow from the salt bridge to balance out the charge.  Also, notice that the reduction reaction in the anode compartment uses up positive copper(II) ions - so, positive sodium ions flow from the salt bridge to balance out the charge.

    

The two solid metals (Zn & Cu) are called electrodes.  The electrode at which oxidation occurs is called the anode (Zn).  The electrode at which reduction occurs is called the cathode (Cu).

A voltaic cell consists of two half-reactions (reduction and oxidation).  Zn is oxidized at the anode to produce electrons, which flow through the external circuit to the cathode, where Cu²⁺ is reduced.  Oxidation of Zn introduced extra Zn²⁺ ions into the anode compartment – unless this positive charge is neutralized, no further oxidation can take place (also reduction of Cu²⁺ leaves excess negative charge in the cathode compartment).

Electrical neutrality is maintained by migration of ions through a salt bridge (a glass U-tube filled with cotton batting soaked electrolyte solution - usually sodium nitrate).  The Na⁺ ions migrate to the cathode and the NO₃^- ions migrate to the anode. The salt bridge also acts to close the circuit.

The resulting electrochemical cell can produce a voltage of 1.10 V

For the cell in which the reaction Zn°(s)   +   Cu²⁺(aq)   →   Zn²⁺(aq)   +   Cu°(s) takes place, it can be written in a short form, following the format below.  The │represents a phase boundary like that between the electrode and the electrolyte.  The ║represents a physical boundary like the salt bridge.

  anode(-) │electrolyte ║ electrolyte │cathode(+)

Zn(s)│Zn²⁺(aq)║ Cu²⁺(aq)│Cu(s)

TryIt! 

For the Cu(s)│Cu²⁺(aq)║ Ag⁺(aq)│Ag(s) cell:

1. Draw the cell, including beakers, specific electrodes, specific electrolytes, salt bridge , wires & voltmeter.
2. Label the anode & cathode.
3. Place the half cell reactions under the appropriate half cell.
4. Show the direction of electron flow.
5. Show the direction of ion flow.
6. Write out the net cell reaction.
   

Check out this video for the answer.

 

 

Cell Potential

So, why do electrons flow spontaneously through the external circuit?

Like all spontaneous processes, the answer involves energy.  It is energetically favourable for electrons to flow from the anode to the cathode of a voltaic cell, thus reducing their electrical potential energy.  This is similar to the way in which it is energetically favourable for a boulder to roll down a hill to reduce its gravitational potential energy.

In each half cell, the electrodes have different potential energies.  In the above cell, the E_P of the electrons is higher in the Zn electrode than in the Cu electrode.  If the electrodes are connected, the electrons lower their E_P by flowing from the Zn to the Cu electrode. 

The potential difference or voltage is measured in volts (V).  The potential difference between the electrodes of the cell is called the cell potential (E_cell or E°) or cell voltage.  The Zn/Cu cell is 1.10 V at standard conditions (1.0 M solns and 1 atm gases, usually 25°C).  So, 1.10 V is the standard cell potential (E°_cell).

Standard Electrode Potentials

E°_cell can be thought of as the sum of two half cell potentials (E°_ox and E°_red), the standard oxidation potential and the standard reduction potential:

E°_cell = E°_ox + E°_red

We can’t directly measure half cell potentials, so the standard reduction potential of the

2H⁺(aq)  +  2e^-   →   H₂(g)

half-cell has been assigned a value of 0 V and all other half cell potentials are measured relative to this.  This standard hydrogen electrode consists of platinum wire to serve as an inert surface for the cathode reaction.  The electrode is encased in a glass tube so H₂(g) can bubble over the platinum.

 

 

Still have questions?  Check out this video.

 Homework: #4-9 (found here)