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Chapter 10: Electrochemistry

Re: Chapter 10: Electrochemistry

Postby Lavi » Tue Jun 07, 2016 12:08 am

Bit confused about Q4 from the chapter review-
4) What should happen when a piece of copper is placed in 1M HCl?

A) The copper is completely dissolved by the acid.
B) The copper is dissolved by the acid with the release of hydrogen gas.
C) The copper bursts into greenish flames.
D) Nothing happens.

How do you know to compare the reduction potential of copper to H, rather than copper to Cl?

Thank you
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Re: Chapter 10: Electrochemistry

Postby goldstanda3269 » Wed Jun 08, 2016 2:34 am

Lavi wrote:How do you know to compare the reduction potential of copper to H, rather than copper to Cl?



To compare, you look at the table provided, and they you arrange to have the most positive Eo value and that always will give you the reaction which is most likely to be spontaneous.


Cl2(g) + 2e- ↔ 2Cl-

1.36 V

2H+ + 2e- ↔ H2(g)

0.00 V

Cu2+ + 2e- ↔ Cu(s)

0.34 V


Notice that if you put Cl- on the left side of the equation, the reaction would have had a -1.36 which is not likely to occur spontaneously even if coupled with a + or -0.34 reaction.
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Re: Chapter 10: Electrochemistry

Postby ElleRT » Sat Jun 25, 2016 11:46 pm

Hi I'm a bit confused with this question:

3) A student sets up a galvanic cell with sodium and chlorine. What is the expected potential difference required to separate sodium and chlorine into their respective ions?

A) -4.07 V
B) -1.36 V
C) 1.36 V
D) 4.07 V


Explanation: Voltage = 2.71 + 1.36 = 4.07 V (the reverse of electrolysis is a galvanic cell which is a spontaneous reaction).Why is the -2.71V become positive? And am I right that the chlorine is being reduced and is where the cathode will be? Or has the -2.71 become postive as it would be Eo = 1.36 - (-2.71).

thanks
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Re: Chapter 10: Electrochemistry

Postby ElleRT » Sat Jun 25, 2016 11:51 pm

Hi, another question about this one:

4) What should happen when a piece of copper is placed in 1M HCl?

A) The copper is completely dissolved by the acid.
B) The copper is dissolved by the acid with the release of hydrogen gas.
C) The copper bursts into greenish flames.
D) Nothing happens.

Explanation: The relevant reaction is Cu + 2H+ -> Cu2+ + H2
Using the data in Table 1, E = -0.34 + 0.00 = -0.34 V
Since E < 0, the reaction is not spontaneous.

Isn't it Cu that is oxidized so the sum would be E= 0.00 - (-0.34) = 0.34??
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Re: Chapter 10: Electrochemistry

Postby goldstanda3269 » Mon Jun 27, 2016 3:10 am

ElleRT wrote:3) A student sets up a galvanic cell with sodium and chlorine. What is the expected potential difference required to separate sodium and chlorine into their respective ions?
...

Why is the -2.71V become positive? And am I right that the chlorine is being reduced and is where the cathode will be? Or has the -2.71 become postive as it would be Eo = 1.36 - (-2.71).


From the table:

Na+ + e- ↔ Na(s) Eo = -2.71 V
Cl2(g) + 2e- ↔ 2Cl- Eo = 1.36 V

We want ions created so ions must be to the right of each equation. So we must switch the equation for sodium which means that the sign of Eo switches also. Now here are the 2 half equations:

Na(s)↔ Na+ + e- Eo = 2.71 V
Cl2(g) + 2e- ↔ 2Cl- Eo = 1.36 V

Now multiply the first equation by 2 so that electrons cancel when the 2 equations are added. Do not multiply Eo because the "o" means that it is at standard conditions with standard pressure and concentration (unlike enthalpy/Hess' Law).

2Na(s)↔ 2Na+ + 2e- Eo = 2.71 V
Cl2(g) + 2e- ↔ 2Cl- Eo = 1.36 V

Add the 2 equations and add the 2 Eo values:

2Na(s) + Cl2(g) ↔ 2Na+ + 2Cl- Eo = 4.07

LEO is A GERC so sodium (s) lost electrons, reduction. Don't worry about memorising the cathode or anode's position, those details will be provided in the passage, if asked.
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Re: Chapter 10: Electrochemistry

Postby goldstanda3269 » Mon Jun 27, 2016 3:31 am

ElleRT wrote:4) What should happen when a piece of copper is placed in 1M HCl?

Isn't it Cu that is oxidized so the sum would be E= 0.00 - (-0.34) = 0.34??


From the table:

Cu2+ + 2e- ↔ Cu(s) Eo = 0.34 V
2H+ + 2e- ↔ H2(g) Eo = 0.0 V

Acid (H+) is combined with a piece of copper which means solid copper: Cu(s). So we need to put Cu(s) on the left and change the sign of Eo.

Cu(s) ↔ Cu2+ + 2e- Eo = -0.34 V
2H+ + 2e- ↔ H2(g) Eo = 0.0 V

Add the 2 equations while cancelling the electrons, and add the Eo values:

Cu(s) + 2H+ ↔ Cu2+ + H2(g) Eo = -0.34 V

You may have wondered: If HCl was used with a piece of copper, and HCl is a powerful acid which completely becomes H+ + Cl-, why don't we consider Cl-? It's because solid copper is an electron donor (loss electrons oxidation), and Cl- also losses electrons so the 2 cannot react (you need a donor + an acceptor for a Redox reaction).

After you finish the questions on our website, you may want to review GAMSAT Practice Questions (ACER's Red Booklet), all questions in Unit 12.

If you want to continue using the equation to sum potentials, you may want to review CHM 10.1 so you go through the examples step by step.
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Re: Chapter 10: Electrochemistry

Postby Neda » Tue Jul 12, 2016 9:22 pm

Query with regards to question 8.

for answering to this question, how should I figure out that 2 electrons are not shared between two different cytochromes? I think, the information given through the passage is not sufficient.
Regards
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Re: Chapter 10: Electrochemistry

Postby goldstanda3269 » Wed Jul 13, 2016 3:29 am

Neda wrote:Query with regards to question 8.

for answering to this question, how should I figure out that 2 electrons are not shared between two different cytochromes? I think, the information given through the passage is not sufficient.
Regards


Actually, the table is listed as "Reduction: Half Reactions" and "Table 1 Standard half-reaction reduction potentials". So if you have never seen redox reactions before, then the implication is that each reaction represents 1/2 of the overall chemical reaction. Thus any conclusion must imply that 2 chemical species are involved in the transfer (i.e. as opposed to 3 separate chemical species as required by answer choice B). If it were not for the preceding, the next 3 questions (and basically all questions on Eo) become impossible since they can only be considered if you are thinking about 2 chemical species at a time.

If you have done many redox reactions, then it would be clear from experience that redox reactions occur in pairs.
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Re: Chapter 10: Electrochemistry

Postby AnitaK » Sun Jan 21, 2018 3:18 am

Can someone please explain question

4) What should happen when a piece of copper is placed in 1M HCl?

I'm still trying to understand why we're not using the reduction potential for chlorine?
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Re: Chapter 10: Electrochemistry

Postby goldstanda3269 » Sun Jan 21, 2018 9:06 pm

AnitaK wrote:Can someone please explain question

4) What should happen when a piece of copper is placed in 1M HCl?

I'm still trying to understand why we're not using the reduction potential for chlorine?


No problem, let's try!

So we have HCl and that makes H+ and Cl- so let's see what the chloride ions can do, from Table 1:

Cl2(g) + 2e- ↔ 2Cl-

1.36 V

So then for chloride:

2Cl- ↔ Cl2(g) + 2e-

-1.36 V

Since Cu(s) is -0.34 then combining the reactions (adding the E values) we get - 1.70 and since it is minus that means it cannot occur spontaneously, and it is more minus that the reaction with H+ meaning it's even more unlikely as a possibility.
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Re: Chapter 10: Electrochemistry

Postby zimengy17106 » Tue Mar 02, 2021 1:48 pm

Hi,

I have a question on Q19. The question asked the rate of rusting in grams per hour.

In the solution:
"Current = (Quantity of electricity)/(Time)
Q = 0.2 A x (80 min x 60 s min-1) = 960 C
Number of faradays required to add 2 moles of electrons to one mole of Fe2+ (to obtain one mole of Fe, see Equation I) = 2 x F = 2 x 96000 C
Number of moles Fe = 960 C/(2 x 96000 C) = 1/(100 x 2) = 1/200 mol
Number of grams Fe = 1/200 mol x 56 g mol-1
Number of grams Fe = (1/200 x 56) g = 7/25 g = 28/100 g = 0.28 g"

So the Q you use in the calculation are for 80min instead of one hour, so the number of faradays is also for 80min, thus 0.28g is the rusting in grams in the 80min reaction. The questions asked for the rate per hour. Don't we need to divide by 1.3 to get the hourly rate? Thanks!
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Re: Chapter 10: Electrochemistry

Postby goldstanda3269 » Tue Mar 02, 2021 5:12 pm

Good question!

The short answer is, no.

Of course, this is one of the more challenging questions. Getting one of these in every 20 questions might be normal.

Time is used in this question for 2 separate processes. Let's take a step back to assess what is going on overall: Solid iron is in water and there is clearly a conversion of solid to iron ions (because Fe2+ has appeared in the water, which we assume is an indication of rust). The release of Fe2+ into the water occurred over a period of one hour. This means that once we calculate how much Fe2+ there is at the end of the hour, that is equal to the rate of rusting in grams per hour.

In the second process, which is to determine the quantity of Fe2+, it does not matter if it takes 80 minutes or 80 hours, we are only trying to determine how much Fe2+ is present. The timing of the second process is completely unrelated to the timing of the first process.
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Re: Chapter 10: Electrochemistry

Postby zimengy17106 » Tue Mar 02, 2021 5:32 pm

Thanks a lot! I just realized I didn't pay enough attention to the first sentence (which said the iron is placed in water and exposed to air for one hour).
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Re: Chapter 10: Electrochemistry

Postby horbowiec.7527 » Sun Aug 01, 2021 7:23 pm

Regarding your YouTube "How to solve GAMSAT Section 3 problems (ACER Practice Test 1 - Green Booklet): Unit 4, Q 14 and 15"

The example you provide there as A: H2 + CO --> CH3OH but in my green booklet (purchased this year, 2021) Unit 4, Question 14, answer A has 2H2 + CH3CN --> CH3CH2NH2. My reaction C in new green booklet is also different then in older booklets. Do you mind going through explanation for establishing oxidation numbers in those reactions if you have access to updated version of the green booklet?

Also question 15, you say answer is C "any odd integer from +3 to -3", and your explanation makes sense of course, but my answers in new green booklet state correct answer is A: "any integer from +3 to -3", and your correct answer is C: "any odd integer from +3 to -3"

Would you assume ACER made a typo mistake when **** the green booklet?

Regards,
Ania
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Re: Chapter 10: Electrochemistry

Postby goldstanda3269 » Thu Aug 05, 2021 2:39 pm

Because of the different versions, perhaps it would better to email your question?
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