H2 Chem - Electrochemistry
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Hi UltimaOnline,
I have some questions on electrochem:
1: VJC 2012 Prelim P1/Q8
Remarks: Depending on the answer scheme provided by different parties, the answer is either B or C. I would like to confirm if the answer is C. Here are my thoughts: comparing Eox potentials, water is oxidised preferentially to Cl- at the anode. Hence, H+ ions move towards Z, thus turning the colour of the litmus red. However, I would think after a few minutes, when sufficient oxygen has been evolved, owing to LCP, the oxidation potential of water would have been lower than -1.36V, thus chloride ions in the KCl starts to oxidise instead, thus turning the litmus white. Am I right?
2. Which of the following compounds do not exist under aqueous conditions?
1 Cobalt(II) chloride
2 Chromium(III) carbonate
3 Iron(III) iodide
Answer: 2 and 3.
Remarks: For 3 is it because the Ecell for the redox reaction between Fe3+ and that of I- is positive (+0.23V)? For 2 chromium (III) cannot undergo a redox reaction with water as the Ecell is negative - so why can’t it exist? What about 1?Thank you! :)
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Hi Gohby,
Q1. Yes you're right. But in addition, even before LCP has a chance to participate, the fact is the question didn't specify the molarity of the KCl solution. In theory, if significantly higher than 1M (ie. concentrated), Cl- would be oxidized to Cl2 at the anode immediately. If significantly lower than 1M (ie. dilute), then H2O would be oxidized instead. Data Booklet redox potentials are only for standard molarities, ie. all species 1M on both LHS and RHS. In practice (for almost all aqueous solutions, unless super dilute or super concentrated), you'll almost always get a mixture of both O2 and Cl2 generated at the anode.
Q2. You're right for option 3. For option 2, Cr3+ has a high cationic charge density, and thus the hexaaquated complex ion will, just like Al3+(aq), lose protons to the solution, making it acidic, thus protonating carbonate ion, generating carbonic acid, which exists in equilibrium with, and hence can decompose into CO2(g) and H2O(l) with thermodynamically favourable positive entropy change, and also as predicted by Le Chatelier's principle.
You're welcome :)Originally posted by gohby:Hi UltimaOnline,
I have some questions on electrochem:
1: VJC 2012 Prelim P1/Q8
Remarks: Depending on the answer scheme provided by different parties, the answer is either B or C. I would like to confirm if the answer is C. Here are my thoughts: comparing Eox potentials, water is oxidised preferentially to Cl- at the anode. Hence, H+ ions move towards Z, thus turning the colour of the litmus red. However, I would think after a few minutes, when sufficient oxygen has been evolved, owing to LCP, the oxidation potential of water would have been lower than -1.36V, thus chloride ions in the KCl starts to oxidise instead, thus turning the litmus white. Am I right?
2. Which of the following compounds do not exist under aqueous conditions?
1 Cobalt(II) chloride
2 Chromium(III) carbonate
3 Iron(III) iodide
Answer: 2 and 3.
Remarks: For 3 is it because the Ecell for the redox reaction between Fe3+ and that of I- is positive (+0.23V)? For 2 chromium (III) cannot undergo a redox reaction with water as the Ecell is negative - so why can’t it exist? What about 1?Thank you! :)
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Hi UltimaOnline,
I have further questions on electrochem:
I. The diagram below shows the sodium-nickel(II) chloride battery, which is a high power, high capacity cell suitable for electric traction applications.
The electrolyte used is molten sodium aluminium chloride, which has a melting point of 157°C. In the reaction, nickel(II) chloride is reduced to nickel. Which of the following statements is incorrect?
A: Terminal E is the negative terminal.
Question: How do I know which is the positive and negative terminal?
II: In the construction of pacemakers for the heart, a tiny magnesium electrode can be used to create an electrical cell with the inhaled oxygen. The relevant half�equations are as shown:
Mg2+ + 2e− ⇌ Mg (Equilibrium 1)
½ O2 + 2H+ + 2e− ⇌ H2O (Equilibrium 2)
In the body, a potential of 3.20V is usually obtained. What is the best explanation for this e.m.f.?
A The small size of the magnesium electrode
B The low concentration of Mg2+ surrounding the magnesium electrode
C The high resistance of the body fluids surrounding the electrodes
D The physiological pH of between 7 and 8 of the body fluid surrounding the electrodes
Answer: D
Remarks: Since the Ecell according to the Data Booklet for this reaction +3.61V, and in the body the usual emf is 3.20V, this would mean either the [Mg2+] is greater in reality, or that there or lower [H+] or [Oxygen] in reality. I can infer that D is correct because of the low [H+] in a pH environment between 7 to 8, but how do I know the other choices are wrong/not good explanations?
III: http://img.photobucket.com/albums/v700/gohby/Chemistry/sludge_zpsimhrdrbj.jpg
Answer: D
Remarks: Comparing the Ered values, I can gather that Cu2+ will be reduced to Cu at the cathode.
(i) What is meant by “Ag and Fe impurities”? If they refer to metallic compounds, wouldn’t it be unfavourable for the metallic ions to be oxidised (since it is at the anode) even further?
(ii) If “Ag and Fe impurities” refers to the metals itself, how do I know that Cu2+ and Ag+ will be formed, since the Eox (Cu/Cu2+) and Eox (Ag/Ag+) are negative? And how does Ag “fall off” the electrode and form the sludge?
IV: http://img.photobucket.com/albums/v700/gohby/Chemistry/doublelec_zpshzwpp2ja.jpg
Answer: A
Remarks:
(i) If it is an open-ended question, will I know if the product at S would be Fe2+ (instead of Fe)? I think it can be both (so long as the Ecell is positive) but I can’t be sure.
(ii) At electrode Q, comparing the Ered potentials between (Cu2+/Cu) and (H+/H2), wouldn’t Cu2+ be preferentially reduced compared to the latter - so why would hydrogen gas be the products formed at Q?
Thank you! :)
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Hi Gohby,
QI. Oxidation potential of Na to Na+ is more positive than oxidation potential of Ni to Ni2+. Hence Na is oxidized, so Terminal E is the anode.
QII. For electrochem electrodes, size doesn't matter. [Mg2+] should be larger, not smaller. If resistance is higher, voltage is higher, not lower.
QIII. Impurities could refer to either (reduced or unoxidized) metallic or (oxidized) ionic state. At the anode, both Fe and Cu are oxidized, while Ag remains unoxidized and hence metallic and hence falls off into the anode sludge. At the cathode, Cu2+ is reduced, while Fe2+ remains in aqueous state. Regarding the sludge, this is what I wrote in another thread :
"Anode sludge" is the term given for the gooey disguisting stuff (ie. sludge) that is deposited below the anode. Why does it fall off and drop down? Because metals are insoluble in water *and* denser than water. Imagine you have a cake with many small stones embedded in it. As the cake is being eaten away by ants, won't the stones fall? As the Cu atoms in the block of impure Cu ore rock (which is the anode) is being oxidized to *aqueous* Cu2+ (ie. cations are soluble (unlike solid metals in atomic state) thanks to ion - permanent dipole interactions can dissolve and become aqueous, making the block of impure Cu ore rock which is the anode appear to dissolve away), the less reactive (ie. less electropositive and hence harder to oxidize) metals such as silver, remaining as the unoxidized solid metal, and solid metals being insoluble in water and denser than water, naturally drop down from the impure copper block of ore rock (ie. the anode), which together with any unreactive (organic or inorganic) material left over from the ore rock (eg. sand, soil, etc), becomes the gooey disgusting mess below the anode, which we call "anode sludge".
QIV. Under aqueous conditions, you cannot reduce Fe2+ to Fe. You'll need molten iron for that. But since electrolysis is expensive, the blast furnace method is utilized instead. You cannot use standard redox potentials, because in aqueous soluions, molarity of water is much larger than 1M (molarity of pure water is 55.555M, and Cambridge may ask A level students to show this via calculations), hence position of equilibrium shifts to RHS, hence reduction potential of H2O becomes more positive than stated in Data Booklet.
At cathode Q, H+ is reduced to H2 gas instead of Cu2+ being reduced to Cu, because the electrolyte doesn't have any Cu2+ to begin with. Sure, Cu2+ is being generated at anode P, but at the start of the electrolysis (for which the qn is based on), the molarity of H+ far outweighs the molarity of Cu2+ (which is technically zero M at initial).
No prob :) -
Hi UltimaOnline,
(i) When oxygen is reduced at the cathode, how do I know which of the following half-equations should I use, and what is the basis for it? I understand that the latter is used for the cathode reaction for fuel cells..
(ii) When a compound is left to oxidise in air, which half-equation should I use? How would I know whether oxygen reacts with H+ or a water molecule?
Thank you!
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Originally posted by gohby:
Hi UltimaOnline,
(i) When oxygen is reduced at the cathode, how do I know which of the following half-equations should I use, and what is the basis for it? I understand that the latter is used for the cathode reaction for fuel cells..
(ii) When a compound is left to oxidise in air, which half-equation should I use? How would I know whether oxygen reacts with H+ or a water molecule?
Thank you!
Hi Gohby, which equation to use, depends on the pH of the electrolyte solution. In fuel cells, because the electrolyte has to be alkaline (in turn because H2 can only be oxidized to H2O in alkaline conditions), hence the 2nd equation is the correct one. -
Originally posted by UltimaOnline:
Hi Gohby, which equation to use, depends on the pH of the electrolyte solution. In fuel cells, because the electrolyte has to be alkaline (in turn because H2 can only be oxidized to H2O in alkaline conditions), hence the 2nd equation is the correct one.To confirm: you meant O2 being reduced to H2O in alkaline conditions right?
If the question asks if a compound would be oxidised in air, we would use the first equation (with E0 value being +1.23V) for the reduction of water. But where does the H+ come from?
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Originally posted by gohby:
To confirm: you meant O2 being reduced to H2O in alkaline conditions right?
If the question asks if a compound would be oxidised in air, we would use the first equation (with E0 value being +1.23V) for the reduction of water. But where does the H+ come from?
Edited : Typo, my bad.
I meant H2 can only be oxidized to H2O at the anode under alkaline pH, hence we need the cathode reaction to generate the required alkaline pH, therefore we reduce O2 under non-acidic conditions to OH-. If we had used acidic pH, O2 would be reduced to H2O, which although has a more positive reduction potential and hence would generate more electricity (ie. higher cell potential) in theory, but would not work because we need alkaline pH at the anode, and the cathode and anode need to be connected with the same electrolyte for current to flow.
The H+ comes from an inorganic acid which you used to set up the acidic pH in the first place. That is, if you're aware the pH is acidic (because of an inorganic acid used earlier to make it so), then use the half-equation which has H+ on the LHS. If you're aware the pH is neutral or alkaline, then use the half-equation which does not have H+ on the LHS. -
Originally posted by UltimaOnline:
Edited : Typo, my bad.
I meant H2 can only be oxidized to H2O at the anode under alkaline pH, hence we need the cathode reaction to generate the required alkaline pH, therefore we reduce O2 under non-acidic conditions to OH-. If we had used acidic pH, O2 would be reduced to H2O, which although has a more positive reduction potential and hence would generate more electricity (ie. higher cell potential) in theory, but would not work because we need alkaline pH at the anode, and the cathode and anode need to be connected with the same electrolyte for current to flow.
The H+ comes from an inorganic acid which you used to set up the acidic pH in the first place. That is, if you're aware the pH is acidic (because of an inorganic acid used earlier to make it so), then use the half-equation which has H+ on the LHS. If you're aware the pH is neutral or alkaline, then use the half-equation which does not have H+ on the LHS.What about those cases where oxygen in the air is the oxidising agent? Why do we choose the half equation with E value +1.23V? Where does the H+ comes from in that case?
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Originally posted by gohby:
What about those cases where oxygen in the air is the oxidising agent? Why do we choose the half equation with E value +1.23V? Where does the H+ comes from in that case?
Case-by-case. See whether the solution is acidic or alkali, then choose the half-equation accordingly. The H+ or OH- must have originally already been present (because you or someone else acidified or alkalinized it). -
Originally posted by UltimaOnline:
Case-by-case. See whether the solution is acidic or alkali, then choose the half-equation accordingly. The H+ or OH- must have originally already been present (because you or someone else acidified or alkalinized it).Hi UltimaOnline,
Ok, so further to this, with reference to A Level 2007 P1 Q36,
CS Toh’s A Level suggested solution are as follows:
For choices 2 and 3 - the compounds scrutinised are chromium (II) chloride and iron (II) hydroxide. Hence, shouldn’t we be using the half-equation which does not have H+ on the LHS (i.e +0.40V)? Why is the solution suggesting that we use the half equation with H+ on the LHS for all 3 compounds?
Secondly (as per the solution), why are we using the reduction potential of iron (III) hexacyanoferrate instead of potassium hexacyanoferrate (III) itself?
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Originally posted by gohby:
Hi UltimaOnline,
Ok, so further to this, with reference to A Level 2007 P1 Q36,
CS Toh’s A Level suggested solution are as follows:
For choices 2 and 3 - the compounds scrutinised are chromium (II) chloride and iron (II) hydroxide. Hence, shouldn’t we be using the half-equation which does not have H+ on the LHS (i.e +0.40V)? Why is the solution suggesting that we use the half equation with H+ on the LHS for all 3 compounds?
Secondly (as per the solution), why are we using the reduction potential of iron (III) hexacyanoferrate instead of potassium hexacyanoferrate (III) itself?
With regards to Fe(OH)2 being oxidized to Fe(OH)3 when left exposed to the air, this is a common experimental observation that both O level and A level students are expected to recognize and/or memorize. Your reasoning (regarding H+ vs OH-) is fair, but because you're also supposed to know (ie. memorize) that this reaction does occur, you can come up with a couple of plausible explanations for this, eg. unlimited O2 available in the atmosphere, hence non-standard molarities, Le Chatelier's principle, etc. Cambridge didn't ask the student to explain, just to be able to state the reaction does occur.
With regards to hexacyanoferrate(III), CS Toh is correctly using the coordination complex formula (the Fe3+ is not the counter ion, but the complex ion), and his explanation that Fe3+ cannot be readily oxidized further is also correct. To oxidize Fe3+ to Fe6+ in the form of FeO4 2-, you'll need a far stronger oxidizing agent than atmospheric O2. -
Hi UltimaOnline,
I have some further questions on electrochem:
1.
Remarks: The answer states that 2 is wrong. However, given the half-equation for reduction, i.e.
wouldn’t the presence of water shift the equilibrium to the left, thereby reducing the magnitude of the Ered, which would decrease the Ecell? From the answer I presume that the correct theory is that the addition of water will reduce both [VO2+] and [VO2+] so the Ecell will remain unchanged. However, why is the first theory wrong?
This does not reconcile with the suggestion in the answer that choice 3 is accurate in this question:
2. For QI posted on 21 Oct 4pm in the same thread, how is the battery able to work if the electrolyte and the sodium electrode are kept separate by a separator?
3. When nitric acid is added to iron filings, a brown gas that turns moist blue litmus red is observed. What is the standard cell potential of the reaction?
Remarks: The relevant half equations are as follows:
[R] NO3- + 2H+ + e <-> NO2 + H2O (+0.81V)
[O] Fe2+/3+ +2/3e <-> Fe (-0.44V/-0.04V)
How do I know which Fe half equation to use? Do I take the one which will lead to a more positive as cell?
4:
Remarks: In an electrolytic cell, is the Ecell of the reaction the voltage provided by the battery or do we not even talk about the Ecell at all? In addition, why could 1 be a possible reason for the failure?
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Originally posted by gohby:
Hi UltimaOnline,
I have some further questions on electrochem:
1.
Remarks: The answer states that 2 is wrong. However, given the half-equation for reduction, i.e.
wouldn’t the presence of water shift the equilibrium to the left, thereby reducing the magnitude of the Ered, which would decrease the Ecell? From the answer I presume that the correct theory is that the addition of water will reduce both [VO2+] and [VO2+] so the Ecell will remain unchanged. However, why is the first theory wrong?
This does not reconcile with the suggestion in the answer that choice 3 is accurate in this question:
2. For QI posted on 21 Oct 4pm in the same thread, how is the battery able to work if the electrolyte and the sodium electrode are kept separate by a separator?
3. When nitric acid is added to iron filings, a brown gas that turns moist blue litmus red is observed. What is the standard cell potential of the reaction?
Remarks: The relevant half equations are as follows:
[R] NO3- + 2H+ + e <-> NO2 + H2O (+0.81V)
[O] Fe2+/3+ +2/3e <-> Fe (-0.44V/-0.04V)
How do I know which Fe half equation to use? Do I take the one which will lead to a more positive as cell?
4:
Remarks: In an electrolytic cell, is the Ecell of the reaction the voltage provided by the battery or do we not even talk about the Ecell at all? In addition, why could 1 be a possible reason for the failure?
Q1. Stoichiometry is different for both cases. Only in the Cr case, does Le Chatelier's principle predict the position of equilibrium shifts to the side of 2 Cr3+ to generate more ions to oppose the dilution.
Q2. The separator allows for ions to pass through.
Q3. Yes, choose the half-equation that gives the more thermodynamically feasible overall reaction. Fe prefers to be oxidized to Fe2+.
Q4. Correct ; the Ecell of the reaction the voltage provided by the battery *HENCE* we not even talk about the Ecell at all. The electrolyte needs to contain Cu2+(aq), otherwise either Cr3+ or H2O will be reduced at the cathode (depending on molarities), which either contaminates the cathodic copper (ie. no longer pure), and/or wastes electricity (and thus not an acceptable setup) before the [Cu2+] reaches a sufficiently high molarity to be reduced at the cathode. -
Originally posted by UltimaOnline:
Q4. The electrolyte needs to contain Cu2+(aq), otherwise either Cr3+ or H2O will be reduced at the cathode (depending on molarities), which either contaminates the cathodic copper (ie. no longer pure), and/or wastes electricity (and thus not an acceptable setup) before the [Cu2+] reaches a sufficiently high molarity to be reduced at the cathode.Could you elaborate a bit on this? Actually the question said "The Data Booklet is relevant to the question" - so I looked at the E values previously - (Cr3+/Cr2+) = -0.41V, (Water/H2) = -0.83V and (Cu2+/Cu) = +0.34V. I suppose the molarities could potentially affect Cr3+ or water being reduced (even in spite of the significant disparity of the E values) instead, but wouldn't this be (i) insignificant; and (ii) goes against the line "The Data Booklet is relevant.."?
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Originally posted by gohby:
Could you elaborate a bit on this? Actually the question said "The Data Booklet is relevant to the question" - so I looked at the E values previously - (Cr3+/Cr2+) = -0.41V, (Water/H2) = -0.83V and (Cu2+/Cu) = +0.34V. I suppose the molarities could potentially affect Cr3+ or water being reduced (even in spite of the significant disparity of the E values) instead, but wouldn't this be (i) insignificant; and (ii) goes against the line "The Data Booklet is relevant.."?
(i) Molarities certainly and strongly affect the redox and cell potentials (see the Nernst equation), because redox and cell potentials are an expression of equilibrium, and equilibrium is about molarities ; the Data Booklet values are relevant only for standard molarities (ie. 1 mol/dm3 for all species with 1 mol as the stoichiometric coefficient). You certainly don't need the Data Booklet for this question (as well as many other MCQs that claim the Data Booklet is relevant). At the most fundamental level, in the first place as part of both the O and A level syllabuses, the student must memorize the compulsory use of a Cu2+(aq) electrolyte, most commonly CuSO4(aq) for the purification of copper setup, so just based on this fundamental point, statement 1 is already (and obviously) wrong.
