Redox Processes (topic 9/19) Flashcards
Oxidation number
Roman numerals ex. v, vii
Oxidation state
+5, +7.
**Start with the plus or minus before the number otherwise there will be no point given.
Oxidation states of elements
0
The sum of the oxidation states
The sum of the oxidation states is the same as the charge on the ions. No charge = 0
Oxygen in a compound
Has an oxidation state of -2 **except in peroxides where it is -1
Hydrogen in a compound
Has an oxidation state of +1 **except if bonded to a metal, where it is -1
Flourine in a compound
Has an oxidation state of -1
Group 1, 2, and 13 in a compound
Have oxidation states of:
Group 1: +1
Group 2: +2
Group 13: +3
Halogens in a compound (group 17)
Probably -1 with exceptions. Except flourine which is -1
Tricky ones
- VNO3 = +1, +5, -2
- CuSO4 * 5H2O = +2, +6, -2, +1, -2
- Fe(ClO4)2 = +2, ClO4-, ClO4-
Upsetting ones
- H2CO (methanal) = +1, 0, -2
- C3H8 (propane) = -8/3, +1
Oxidation (definitions)
- Adding oxygen
- Removing hydrogen
- Loss of electrons
- Oxidation state increases
- OIL (oxidation is loss of electrons)
Reduction (definitions)
- Removing oxygen
- Adding hydrogen
- Gain of electrons
- RIG (reduction is gain of electrons)
Oxidizing agents
Force other chemicals to be oxidized (to lose electrons). Therefore the oxidizing agent itself is gaining electrons (reduction). An oxidizing agent is itself reduced. Where the oxidation arrow begins, that’s where the reducing agent is.
Reducing agents
Reducing agent is itself oxidized. Where the reduction arrow begins, that’s where the oxidizing agent is.
How to turn a formula into a name
In a compound, the second element’s subscript is multiplied to its oxidation number divided by the first element’s subscript.
How to turn a name into a formula
In a compound, the oxidation number of the elements are their own subscripts using the crossover method.
**Empirical formula is used so double check!!!
Deducing whether an element undergoes oxidation or reduction
- Writing out the oxidation number of elements
- Looking at which elements have changed the most
- Deduce whether the change is oxidation or reduction
Metal activity series (in the data booklet)
Top:
- most reactive
- “easy” to be oxidized
- best reducing agent
- low i.e.
Bottom:
- least reactive
Activity series tell you which metals react with which ions
“Most reactive ends up as an ion”
Zinc + Copper sulfate
Zn(s) + CuSO4(aq) -> Cu(s) + ZnSO4(aq)
Zinc metal(s) + copper sulfate(aq) = copper + zinc sulfate
Solid zinc metal dipped in aqueous copper sulfate solution makes copper on the zinc metal and zinc sulfate in the copper sulfate solution.
Zinc becomes an ion. Reactivity: zinc > copper
Silver nitrate + copper
2AgNO3(aq) + Cu(s) -> Ag(s) + Cu(NO3)2(aq)
Silver nitrate(aq) + copper(s) = silver + copper nitrate
Solid copper wire dipped into silver nitrate solution = silver nitrate turning into silver coating around the copper and the liquid solution making copper nitrate
Copper becomes an ion. Reactivity: copper > silver
Zinc nitrate + copper
Zn(NO3)2(aq) + Cu(s) = no reaction because zinc > copper in the reactivity series
Zinc nitrate + silver
Zn(NO3)2(aq) + Ag(s) = no reaction because zinc > silver in the reactivity series
Reactivity series for metals
- Top is most reactive metal
- Metals hate e- (electrophobic)
- Wants to lose electrons
- Wants to oxidize
- Is a good reducing agent
Ways of depleting oxygen dissolved in water
- Raw sewage when mixed with water become oxidized, thus removing oxygen from the water itself
- Nitrate fertilizers will encourage the growth of algae which will reduce oxygen levels
- Phosphates that are used in many detergents
Winkler experiment for dissolved oxygen
- Pouring water from a lake into a volumetric flask, fill it up completely and measure the amount of liquid contained after. Slow pouring, funnel touching glass to avoid aeration.
- Take the temperature of the liquid.
- Add manganous sulfate monohydrate (white powder) (not specified how much) into the volumetric flask. Supplies the manganese 2 ions (Mn2+)
- Add alkaline iodide azide pillow (small white powder packet) into the volumetric flask. Supplies iodide and increases the pH cause we need hydroxide ions.
- Cork up the volumetric flask. Make sure there are no air bubbles and shake.
**If it turns brown there is dissolved oxygen.
**The brown is the manganese 2 turning to manganese 4
- Set the volumetric flask down and let the precipitate settle down halfway.
- Once it is halfway, shake it up again and set the volumetric flask down and let the precipitate settle down halfway.
- Carefully remove the stopper.
- Add 2 ml of concentrated sulfuric acid (H2SO4). The new brown colour is the iodine molecules being formed. Manganese 4 turns into manganese 2.
- Carefully transfer solution into a beaker. DO NOT AERATE.
- Do a titration with sodium thiosulfate (Na2S2O3). Sodium thiosulfate is gonna turn the brown I2 iodine molecules into colorless iodide molecules.
- Getting closer to the end of the reaction (when the solution turns lighter in colour), add starch to sharpen the end point (two pinches approx.).
**Make sure to write down the titration readings
- Use a graduated cylinder (or another accurate glassware) to measure the liquid inside the volumetric flask.
Winkler method using equations
2Mn^2+(aq) + 4OH-(aq) + O2(aq) -> 2MnO2(s) + 2H2O(l)
MnO2(s) + 2I-(aq) + 4H+(aq) -> Mn^2+(aq) + I2(aq) + 2H2O(l)
I2(aq) + 2S2O3^2-(aq) -> S4O6^2-(aq) + 1I-(aq)
- If you know the volume and concentration of sodium thiosulfate (Na2S2O3 / 2S2O3 ^2-)(aq) you can get the moles. Then you can get the moles for I2(aq)
- Knowing the moles of iodine, you can get the moles for manganese oxide (MnO2)(s)
- Once you know that, you can work out the dissolved oxygen content (mol) and knowing the volume of water used, you can get the concentration
3 measurements of concentration
- conc (mol dm^-3) = mol / dm^3
- conc (g dm^-3) = mass of solute / volume
- conc (ppm) = (mass of constituent / mass of sample) x 10^6
Biological oxygen demand
- Get sample of water (which probably has pollution and bacteria)
- Saturate it with oxygen until it can’t take any more oxygen
- Close the lid
The organisms are going to digest their food and use the oxygen up. The pollution is food for the organisms so they use up more oxygen.
You need to wait 5 days until you can do the Winkler test / BOD
BOD
Is the measurement of nitrates, phosphates and sewage allowing organisms to remove oxygen as they grow and die. A high BOD is unhealthy for the ecosystem.
Salt bridge
- Completes the circuit
- Balances the charge in the half cells
Electrolysis / electrolytic cell
Put electricity IN (add electricity). Electrical energy to chemical energy.
Voltaic / electrochemical / galvanic cell
Get electricity OUT (make electricity). Chemical energy to electrical energy.
Wires
Conductor (allows the passage of electricity and is unchanged by it
Electrode
Normally graphite or platinum. Solids used in the electrolytic cell.
Electrolyte
Substance that conducts electricity as a liquid / aqueous and is chemically decomposed in the process. As a solid, it is an insulator.
Positive electrode
Anode
Negative electrode
Cathode
Diatomic elements
Hydrogen (H), nitrogen (N), oxygen (O), fluorine (F), chlorine (Cl), bromine (Br), iodine (I)
Convention on writing the cell (cell notation)
LHS || RHS
LHS = oxidation half-cell. Loss of electrons
RHS = reduction half-cell. Gain of electrons
|| = salt bridge
Standard electrode potential
- Make a standard half-cell
- Make a 1 mol solution of an ion you’re interested in
- Attach an electrode with a metal you’re interested in / platinum electrode
At 100 kPa (standard pressure)
At 298 K (standard temperature) - Attach a voltmeter on the half-cell
- Attach a standard hydrogen electrode on the voltmeter
The voltage that the voltmeter reads is the standard electrode potential = Eθ
Standard hydrogen electrode
Only HCl and HNO3 (any other acids won’t work)
1 mol dm^-3 H+(aq)
100 kPa hydrogen gas
Platinum electrode
25 degrees C or 298K
If the battery works
= spontaneous battery
= + voltage
If the battery will not work
= non-spontaneous battery
= - voltage
Electrolysis experiments
- Diluted NaCl(aq)
- Concentrated NaCl(aq)
- CuSO4(aq) with inert electrodes (such as graphite or platinum)
- CuSO4(aq) with copper electrodes
Dilute NaCl(aq)
- Label the positive electrode and negative electrodes
- Write down the ions that are present (Na+, Cl-, H+, OH-)
- Looking at the positive ions because they are attracted to the negative electrodes. **lowest on the list (of standard electrode potential) which is H+
- Looking at the negative ions because they’re attracted to the positive electrodes. **preferentially discharged which is OH-
You can observe colourless bubbles from each electrodes
Concentrated NaCl(aq)
- Label the positive electrode and negative electrodes
- Write down the ions that are present (Na+, Cl-, H+, OH-)
- Looking at the positive ions because they are attracted to the negative electrodes. **lowest on the list (of standard electrode potential) which is H+
- In a concentrated reaction, the more concentrated one is discharged (forms an element), Cl-
You can observe green bubbles
CuSO4(aq) with inert electrodes (such as graphite or platinum)
- Label the positive electrode and negative electrodes
- Write down the ions that are present (Cu2+, SO4^2-, H+, OH-)
- Looking at the positive ions because they are attracted to the negative electrodes. **lowest on the list (of standard electrode potential) which is Cu2+
- Looking at the negative ions because they’re attracted to the positive electrodes. **preferentially discharged which is OH-
Copper 2+ is gonna stick on the negative electrode (plated). Oxygen gas is gonna be released at the positive electrode
CuSO4(aq) with copper electrodes
- Label the positive electrode and negative electrodes
- Cu2+ will go to the negative electrode making the electrode bigger (reduction)
- Positive electrode will dissolve into Cu2+ (oxidation)
The colour of the electrolyte stays the same (blue)
