Exam 4

Wednesday 4/19
7:00-8:30pm

BUR 106


avg = 86.4

number of submissions: 181

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🔑 Here are all the KEYS to Exam 4


Exam protocol All you need to know about HOW to take our online exams.


ChemBook Chapter:
Chapter 12: Electrochemistry

Learning Outcomes for Electrochemistry

Students will be able to...

  1. Identify an oxidation – reduction (redox) reaction based on changes in oxidation numbers across the chemical change.
  2. Identify oxidizing/reducing agents in chemical reaction.
  3. Completely balance a redox reaction in acidic or basic solution.
  4. Recognize degrees of reactivity based on an activity series or a standard potential table.
  5. Apply standard potential data to determine the relative strength of oxidizing and reducing agents.
  6. Construct an electrochemical cell diagram or drawing which includes the identification of the anode, cathode, and salt bridge from either a redox reaction or from the short hand cell notation.
  7. Show the direction of electron flow, the sign of the electrodes, the direction of ion flow in the salt bridge, on a given cell diagram or drawing.
  8. Describe the standard hydrogen electrode (SHE) and state its function.
  9. Apply standard potential data (\(E^\circ\)) to calculate the standard cell potential (\(E^\circ_{\rm cell}\)) for an electrochemical cell and from the sign of the potential predict if the cell is voltaic or electrolytic.
  10. Calculate the cell potential for a nonstandard cell.
  11. Describe fully the relationship between the standard free energy (\(\Delta G^\circ\)) and the standard cell potential (\(E^\circ\)).
  12. Describe fully the relationship between standard cell potential (\(E^\circ\)) and the equilibrium constant (\(K\)).
  13. Explain thermodynamically the operation of a concentration cell, and be able to predict the concentration in the cell based on the measured cell potential.
  14. Understand the relationship between the quantity of charge delivered or produced (coulombs) and the amount of reactant used or product formed (moles) for both voltaic and electrolytic cells.
  15. Describe the basic principles of battery design and function.
  16. Identify the differences and similarities of the three fundamental types of batteries: primary cells, secondary cells, and fuel cells.
  17. Know the details of the chemical reactions used in a lead-acid battery.
  18. Know the reaction chemistry (reactants and products) of an alkaline battery (cell)


Electrochemistry

electrochem definitions

reduction is the gain of electrons and occurs at the cathode.
this means that electrons always flow into the cathode from the external circuitry

oxidation is the loss of electrons and occurs at the anode.
this means that electrons always flow out of the anode into the external circuitry



standard cell potential, E°cell
The BEST way to think about getting E°cell

If you really "think like a chemist" then you should be thinking about the push/pull nature of the electrochemical cell. The anode is pushing electrons out while the cathode is pulling electrons in. The standard potential is really the addition of two standard potentials, one is the driving force of the reduction (\(E^\circ_{\rm red}\), the pull) and the other is the driving force of the oxidation (\(E^\circ_{\rm ox}\), the push). Put these together and you get:

\[{\cal E}^\circ_{\rm cell} = {\cal E}^\circ_{\rm red} + {\cal E}^\circ_{\rm ox} \]

You must "flip" a reaction from the table of standard potentials in order to get an oxidation potential. Also remember, when you flip a reaction, you change the sign on the state function, which in this case is \(E^\circ\). Realize that you have to flip the oxidation reaction anyway when you write out the balanced equation. So this is the best way to do it.

Shortcut method Unfortunately, we tend to teach you the shortcut first and it leads to misconceptions. None-the-less, the fact is that the standard potential for any electrochemical cell is simply the difference in the standard potentials of the two half cells.

\[{\cal E}^\circ_{\rm cell} = {\cal E}^\circ_{\rm cathode} - {\cal E}^\circ_{\rm anode} \]

Realize that the standard cell potential (or reaction potential) can be positive (+) or negative (–). Also realize that the \(-E^\circ_{\rm anode}\) term IS the oxidation potential which is the push of \(E^\circ_{\rm ox}\) shown in the first equation.



shorthand cell notation

We "tell" you which 1/2 cell is the anode by writing it on the left side in the shorthand notation:

anode \(\bigl|\) anodic solution \(\bigl| \bigr|\) cathodic solution \(\bigr|\) cathode

Each of the vertical lines, |, marks a distinct phase-change in the cell as you traverse from the anode to the cathode through the cell. The double line, ||, is the salt bridge. If the 1/2 reaction has no conductor directly in the reaction (like say the H+/H2 reaction for S.H.E.), you use an inert electrode such as Pt, Au, or C(s,graphite). If a solid or gas is an active component, you have to show it directly next to the electrode. Solution species will be those listed adjacent to the salt bridge and the order of listed species does NOT matter.

To illustrate cases with solids and gases, consider a cell made using the Ag/AgCl (E°= +0.22V) and the S.H.E. (E° = 0.00 V) as the two half reactions. The correct way to show this with the Ag/AgCl as the anode is:

Ag(s) \(\bigl|\) AgCl(s) \(\bigl|\) KCl(aq) \(\bigl| \bigr|\) HCl(aq) \(\bigr|\) H2(g) \(\bigr|\) Pt(s)

Note how in this case the needed Cl- ions in the silver cell are shown as a full ionic formula of KCl(aq) - which would certainly provide the needed chloride ions. In a similar manner, the needed H+ ions are provided via hydrochloric acid, HCl(aq). Sometimes the ions are shown by themselves, and other times they are shown as a complete formula with counter ions. So you see, there is some flexibility in the details in writing out a shorthand cell notation.



cell potential and free energy

standard conditions:    \(\Delta G^\circ = -n F {\cal E}^\circ \)

non-standard conditions:    \(\Delta G = -n F {\cal E} \)



the Triforce of CH302

\(-n F {\cal E}^\circ = \Delta G^\circ = -RT\ln K\)



non-standand cell potentials

Non-standard cell potentials are calculated via the Nernst Equation:

\[{\cal E}_{\rm cell} = {\cal E}^\circ_{\rm cell} - {RT\over nF}\ln(Q) \]

The temperature (T) is usually 298.15 K for most of our problems. The faraday is always 96485 C/mol of e-, and R is always 8.314 J/mol K. Therefore you can use either of the following versions of the Nernst equation to save yourself from punching in all those long numbers into your calculator.

\[{\cal E}_{\rm cell} = {\cal E}^\circ_{\rm cell} - {0.02570\over n}\ln(Q) \]

\[{\cal E}_{\rm cell} = {\cal E}^\circ_{\rm cell} - {0.05916\over n}\log(Q) \]

FYI... calculate \(RT/F\) and you'll use (8.314)(298.15)/96485 to get 0.02569 which is where that 0.02570 came from up above. You have to use natural log for that number though. If you want to use log base 10 (log), then you'll need to multiply the log 10 version of Q by ln(e) which is 2.303 and that is why the 0.05916 number is there.



algebra, logarithms, and the Nernst equation

Know your basic logarithm algebra laws

\(x = 10^{\log(x)}\hskip36pt x = e^{\ln(x)}\)

\(\log(xy) = \log(x) + \log(y)\)

\(\log(x/y) = \log(x) - \log(y)\)

\(\log(x)^y = y\log(x)\)


Now lets combine two of the above relations to get our "formula" for natural log using base10 log.

\(x = 10^{\log(x)}\)     Now take the natural log (ln) of both sides.

\(\ln(x) = \ln(10)\cdot\log(x)\)     which will solve to

\(\ln(x) = 2.302585 \cdot \log(x) \)

So, natural log of any number is the same as the log base10 number times 2.303 (4 sig figs). So that is why the Nernst equation can be rewritten as

\({\cal E}_{\rm cell} = {\cal E}^\circ_{\rm cell} - {2.303RT\over nF}\log(Q) \)

\({\cal E}_{\rm cell} = {\cal E}^\circ_{\rm cell} - {0.05916\over n}\log(Q) \) when \(T=298.15\) K



batteries

3 Types of Batteries: Primary Cells (irreversible and non-rechargeable), Secondary Cells (reversible and rechargeable), and Fuel Cells (non-rechargeable, but refillable).

Pros and Cons: Each battery type and chemistry has their own specific set of attributes that can be advantanges (assets) or disadvantages. Here are many of the considerations of a battery type and chemistry: cost, potential or voltage available, energy density, weight, environmental impact, safety, shelf life, operating temperature range, heat dissipation rate, internal resistance, maximum current, power, energy (J). In general, the application will help with which attributes are the most important and help one to decide which type of battery and chemistry to use. There is no single "perfect" battery.

Solid State is the way to go: In ALL cases, a battery should be able to perform at the specification voltage for most of its working life. In order to "hold" this voltage, most all batteries depend on the use of solids as reactants and products. This helps Q for the reaction to stay fairly constant because any amount of solid will have an activity of one.

Sizing: Batteries cover a wide range of physical sizes as well. Small batteries do not have to deliver large amounts of current (TV remote, simple LCD watch, smoke detector) so you can have small devices which are convenient. Larger batteries deliver larger amounts of current and therefore power. A larger battery will generally have a much larger surface area on the electrodes and therefore able to pass more current. Innovation has lead to the maximum amount of surface area for the minimum amount of space.


The TWO Battery Chemistries You Should Know

Alkaline Cell:    Zn(s) + MnO2(s)  ⇌  Zn(OH)2(s) + Mn2O3(s)
there is also water (H2O) and hydroxide (OH–) on each side as well although they cancel out in the net equation/reaction. Also, Zn(OH)2 is often shown as ZnO + H2O (like on gchem)


Lead-Acid Battery:
DIScharge:   Pb(s) + PbO2(s) + 2H2SO4(aq)  ⇌  2PbSO4(s) + 2H2O(ℓ)
REcharge:   2PbSO4(s) + 2H2O(ℓ)  ⇌  Pb(s) + PbO2(s) + 2H2SO4(aq)



"electric world"

Potential is given the symbol \({\cal E}\) in equations and is measured in volts.

Volts (or voltage) is a unit of measure and is abbreviated with just V.

Electric current is given the symbol \(I\) in equations and is measured in amps.

Electric current flow is always in the opposite direction of electron flow.

Amps is a unit of measure and is abbreviated with an A.

An amp (A) is also just a rate of charge transfer which is coulombs/second or C/s.

Time is given the symbol t (lowercase!) and is measured on many different scales.

Those scales of time are typically seconds, minutes, hours, days, or years. The SI unit of time is the second (s).

Reminder, temperature is given the symbol T (uppercase!) and has 3 different scales that can be used (°F, °C, or K).

Power is given the symbol P in equations and is measured in watts.

Watts (or wattage) is a unit of measure of power and is abbreviated with just W.

Power is defined as a rate of energy use or energy/time. A watt is really just a joule/second or J/s. One common equation for calculating power is: \(P = I \cdot {\cal E}\). Yes, it spells "pie" but is really power = current × potential.

Charge is given the symbol q in equations and is measured in coulombs.

Charge is calculated via current × time or: \(q = I\cdot t\)

Energy (joules, J) is calculated via charge × potential or: Energy = \(q\cdot {\cal E}\)

Don't confuse energy and potential, they are typically both symbolized with an E. Sometimes (not always) a script looking \({\cal E}\) is used for potential to be different from the E used for energy. The unicode script E looks like this, â„°. Just note the context in which it is used - it is usually obvious which one we mean (blatantly obvious once you LEARN the concepts).

The main conversion factor to know in order to convert from "electricity world" to "chemistry world" is the faraday constant, F. It tells us the number of coulombs in a mole of ionic unary charge - to be even more specific, for our needs, the faraday is 96485 coulombs which is the total absolute charge on 1 mole of electrons. The faraday doesn't care about being positive or negative, it only cares about the magnitude of the charge. YOU need to remember the signs for charge... but no matter where the charge is coming from... 1 F = 96485 C/mol of electron charge.



electricity world ⇄ chemistry world (big picture)

So the flow of conversions is the following:

current → charge → moles of e– → moles of reaction

So taking all the conversions needed and putting them all into one big equation we get

\[{I\cdot t\over n\cdot F} = {\rm moles\;of \;reaction}\]

(that is moles of reaction that uses n moles of electrons for the way you balanced it)

It would be worth your while to READ THIS SECTION of my CHEMBOOK about the two worlds. The Stranger Things tie-in is a bonus. And, while you are there... keep reading other sections as well.