40daysofCHEM 9.3.2 A = Part 8

40daysofCHEM 9.4.2 C = Part 7

40daysofCHEM 9.4.2 B = Part 6

40daysofCHEM 9.4.2 A = Part 5

40daysofCHEM 9.2.5 B = Part 4

40 Days of Chem - 9.3.3 = Part 3

40 Days of Chem - 9.2.5 A = Part 2

40 Days of Chem - 9.4.1 = Part 1

9.6.7 HSC Feedback

2001 e)

EVALUATE means + / - and judgement (with criteria) – ie want to restore the artefact to as close to original condition as possible without further degradation. So for each step say the pros and cons. And when there is a decision to be made (ie a criteria/evaluation) say why eg mechanical vs acid concretion removal & stripping vs electrolysis of corrosion.
Remember to store in water first (why), leach (why – and NB leaching only removes SOLUBLE salts) remove concretions (here is a change to discuss pros and cons of mechanical vs chemical removal), electrolysis (pros and cons – cons mainly in terms of current and rate of electrolysis vs surface damage)

WHENEVER discussing electrolysis ALWAYS say a) It removes INSOLUBLE SALTS unlike leaching, b) It draws anions out of the artefact (eg Cl-) c) It restores the metal surface (ie metal ions reduced back to the surface)  AND d) WITHOUT FURTHER DAMAGE. ALWAYS use equations for the anode and cathode AND for the cathode reactions make sure you show the correct corrosion product being reduced (Fe(OH)Cl, Cu(OH)Cl, Ag₂S or PbS). NB in basic solutions the anode reaction will be 2OH^- => 1/2O₂ + H₂O + 2e^-

2002 cii)

Chemical procedures = leaching, concretion removal with acid, corrosion product stripping with citric acid/thiourea (Cu or EDTA (Pb), electrolysis and H-furnace (Fe) – make sure you show equations for the ones you chose.

If you have t speak about a specific artefact with chemical treatment choose the cannon over the anchor.

2003 e)

ASSESS = INTRO, +ve, -ve, judgement

INTRO = define electrolysis (perhaps brief history Davy/Faraday)

+ve = a) It removes INSOLUBLE SALTS unlike leaching, b) It draws anions out of the artefact (eg Cl-) c) It restores the metal surface (ie metal ions reduced back to the surface)  AND d) WITHOUT FURTHER DAMAGE.

-ve = trickier but   it requires energy, time, cannot remove concretions, cannot remove some oxides (eg why H-furnace is used for Fe artefacts sometimes)

Judgement = Has it developed artefact restoration? – ie before this could surface be restored? Could insoluble salts be removed?..

2004 aii)

Wood can be degraded and oxidised but not corroded (metals corrode)

2005 di)

The wax is a BARRIER to oxygen and electrolyte (water/humidity) preventing corrosion

2006 c)

The coins were encrusted (see diagram) thus you can’t just say that Cu artefacts wont have concretions

STEPS

1) Keep wet to avoid crystallisation of salt(and show what happens with equations)

2) Leach – Na2CO3 to prevent dissolution of Pb or Zn & Sn from Cu alloys

3) Concretion removal – with ACID for coins and don’t want to dameg surface engravings – perhaps only dilute/weak acid too else bubbling might damage surface

4) Leaching again?

5) For the Pb you might want to strip with EDTA as the surface is not important BUT for Cu the surface IS important (coins) thus you probably wont strip off the corrosion products as you want to RESTORE the Cu in the corrosion back to the surface

6) Electrolysis of coins – (and Pb?) make sure you use the correct equations – see notes for Q 2001e) above

VERY important – electrolysis removes corrosion NOT concretions

2010 a)

Don’t forget that the metal bands are part of the artefact and would be corroded!

2011 bi)

You need to know that the main corrosion product on silver is Ag2S and thus must show it being reduced back to silver (not just silver ion reduction)

2012 a)

Because the artefact material was not specified you should give SPECIFIC problems that can occur if salt crystallises in wood/leather, ceramics and metals. DON’T FORGET that acid can form on metals if salt crystallises and make sure you can write the equation for this.

9.6.3 HSC Question Feedback

2002 aii) – To reverse a galvanic cell and make it an electrolysis cell we have to reverse both the anode and cathode reactions. If we apply the exact opposite voltage produced by the galvanic cell all we do is stop the galvanic cell reaction. To reverse the cell we have to apply GREATER than the galvanic cell potential.

2003 bii) Any time you describe an experiment to measure the rate of electrolysis you MUST specify HOW you measured the rate – eg In our experiment we measured the mass change of the anode as an indication of the rate of reaction.

2006 bi) For the risk assessment and HOW it INFLUENCED our PROCEDURE– we used slightly toxic CuSO₄ – this may have limited our concentration to lower (safer) concentrations. We also used a power pack that limited the voltage to 12V to minimise the risk of electrocution but this limited our experiment to less than 12 V

2006 bii) When providing a conclusion from YOUR OBSERVATIONS you have to state the observations – eg the electrolysis cell that had the higher voltage had a greater mass loss at the anode in a fixed time. Thus it was concluded that higher voltages increase rates of electrolysis.

2007 ai) Graphite rods are used because they are INERT (meaning that they wont react in the redox reactions) CONDUCTORS (this allows them to transport re electrons required for electrolysis) – a lot of people forgot to mention that they are conductors. GRAPHITE IS NOT A METAL!

2008 ciii) It did say “Identify” the factors that affect rates of electrolysis but which of these two answers (that take basically the same time) is going to score more highly?:

·        Voltage, electrode spacing, electrode size and electrolyte concentration all affect electrolysis rates

OR

·        ­ voltage, ¯ electrode spacing, ­ electrode size and ­ electrolyte concentration all increase electrolysis rates.

2009 d) Usually w =e would use an Ag cathode so the Ag+ conc would stay constant in solution. This reaction gives a cell potential of 0V. However, i) if you think about you will see that a spontaneous metal displacement reaction would happen between the Ni and AgNO3. thus we do need to provide a -ve potential to the Ni to prevent it dissolving. ii) A potential of 0V actually means it is in equilibrium, the Ag will ppt and dissolve at both electrodes not plate the Ni. To actually get a silver plate we need to ensure that the reduction (deposition) of Ag only happens at the spoon (cathode) and oxidation (dissolution) of silver only happens at the silver anode - ie a voltage with the -ve terminal at the spoon is needed.

2010 bii) The cathode could be identified by the hydrogen gas or hydroxide ions  BUT HOW could you test for them – ie test with pop test or use a phenolphthalein indicator…

2012 di) Validity means measuring what you intended ie the effect of ONE variable (IV) on the dependent variable. As soon as you change multiple variables you can no longer validly say which changed variable affected the DV. Many people neglected to mention that in a valid test only one variable is changed.

2020 dii) Not the greatest Q for us. All through the Q student were drawing on the reactions at the electrodes until they get to the Q where it specifically states to draw the expected results on your diagrams and everyone stops drawing them on. You should have include 3 diagrams (control and two different variables changed – one in each of the other two pictures). You should have specified the constants and drawn on th results – ie bubbles and Cu deposition showing MORE (or less) depending on what variable you were testing.

Lastly – in their diagram they had a graphite and copper electrode. You should have used the sae chemicals in your example ie the Cu as the cathode th graphite as the anode.