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- admin
**Posts:**1065**Joined:**Tue Jun 29, 2010 7:47 pm

Hello

I am struggling to figure out how the equation in 10.5.1 Electrolysis Problem is broken down.

I have it like this at the minute, but my answer seems too small...

0.025F x (1mol Cu/2F) x (63.5g Cu/mol Cu) =

0.025F x (63.5g/130000F) x (63.5/1) =

0.025F x 0.0005 x 63.5 = 0.00079

I'm not sure where I have gone wrong, thank you!

I am struggling to figure out how the equation in 10.5.1 Electrolysis Problem is broken down.

I have it like this at the minute, but my answer seems too small...

0.025F x (1mol Cu/2F) x (63.5g Cu/mol Cu) =

0.025F x (63.5g/130000F) x (63.5/1) =

0.025F x 0.0005 x 63.5 = 0.00079

I'm not sure where I have gone wrong, thank you!

- olirushworth
**Posts:**32**Joined:**Sun Jan 12, 2014 3:07 am

0.025F x (1mol Cu/2F) x (63.5g Cu/mol Cu) =

Notice in the above equation, F cancels, mol Cu cancels so we get:

0.025 x (1/2) x 63.5 = .025 x 31.75 = 1/4 x 10^-1 x 31.75 = 1/4 x 3.2 = 0.8 g Cu

Notes:

You should be able to do (1/2) x 63.5 in your head because it is similar to 1/2 x 60 which is 30 and then you just need to do 1/2 of the number 3.5.

Recognize that .025 = .25 x 10^-1 = 1/4 x 10^-1 (note that working with fractions is usually faster that multiplying/dividing using decimals).

Note that 3.175 was estimated at 3.2 which is almost always possible as long as the answer choices are sufficiently far apart.

When you see 1/4 x 3.2, you should think of 1/4 of 32 which is easy = 8 and then adjust for the decimal point.

www.gamsat-prep.com/gamsat-maths-tips-formulas/

Notice in the above equation, F cancels, mol Cu cancels so we get:

0.025 x (1/2) x 63.5 = .025 x 31.75 = 1/4 x 10^-1 x 31.75 = 1/4 x 3.2 = 0.8 g Cu

Notes:

You should be able to do (1/2) x 63.5 in your head because it is similar to 1/2 x 60 which is 30 and then you just need to do 1/2 of the number 3.5.

Recognize that .025 = .25 x 10^-1 = 1/4 x 10^-1 (note that working with fractions is usually faster that multiplying/dividing using decimals).

Note that 3.175 was estimated at 3.2 which is almost always possible as long as the answer choices are sufficiently far apart.

When you see 1/4 x 3.2, you should think of 1/4 of 32 which is easy = 8 and then adjust for the decimal point.

www.gamsat-prep.com/gamsat-maths-tips-formulas/

- goldstanda3269
**Posts:**1673**Joined:**Wed Aug 25, 2010 10:59 pm

Ok thank you, I see I went wrong by not cancelling out the F and the mol Cu which is why I got 0.0005 instead of 0.5 (1/2)

- olirushworth
**Posts:**32**Joined:**Sun Jan 12, 2014 3:07 am

For the Electrolysis Problem (10.5.1) Im unsure how Faradays =2400 C x 1F / 96500 C.

Is it always 1F or has the 1 come from somewhere?

Secondly the other sum of 0.025F x (1mol Cu/2F) x (63.5g Cu/mol Cu) = 0.72g

Im unsure how I'd get to that, is the 2F regarding the 2 electrons? I think I may have missed some sort of standard equation. Any help would be much appreciated!

Thanks

Is it always 1F or has the 1 come from somewhere?

Secondly the other sum of 0.025F x (1mol Cu/2F) x (63.5g Cu/mol Cu) = 0.72g

Im unsure how I'd get to that, is the 2F regarding the 2 electrons? I think I may have missed some sort of standard equation. Any help would be much appreciated!

Thanks

- ElleRT
**Posts:**55**Joined:**Thu Jul 09, 2015 4:50 am

" Im unsure how .

Is it always 1F or has the 1 come from somewhere? "

Hopefully from the GAMSAT Math Chapters, you are comfortable with dimensional analysis. If we are told that 1 Faraday is equal to 96500 C, and if we are given 2400 C but we want to know Faradays, then the only way to arrange the equation to get what we want is Faradays =2400 C x 1F / 96500 C because the coulombs C cancel.

"Secondly the other sum of 0.025F x (1mol Cu/2F) x (63.5g Cu/mol Cu) = 0.72g

Im unsure how I'd get to that, is the 2F regarding the 2 electrons?"

Yes, the equation Cu^2+ + 2e- ---> Cu shows that for every one mole of copper ions there are 2 moles of electrons (2 faradays) and this is where the following term comes from: 1mol Cu/2F

Recall from section 10.5 that one faraday is equal to one mole of electrons (you don't need to memorize this fact, it would be given in the problem but you need to use this fact and dimensional analysis to solve the problem).

Is it always 1F or has the 1 come from somewhere? "

Hopefully from the GAMSAT Math Chapters, you are comfortable with dimensional analysis. If we are told that 1 Faraday is equal to 96500 C, and if we are given 2400 C but we want to know Faradays, then the only way to arrange the equation to get what we want is Faradays =2400 C x 1F / 96500 C because the coulombs C cancel.

"Secondly the other sum of 0.025F x (1mol Cu/2F) x (63.5g Cu/mol Cu) = 0.72g

Im unsure how I'd get to that, is the 2F regarding the 2 electrons?"

Yes, the equation Cu^2+ + 2e- ---> Cu shows that for every one mole of copper ions there are 2 moles of electrons (2 faradays) and this is where the following term comes from: 1mol Cu/2F

Recall from section 10.5 that one faraday is equal to one mole of electrons (you don't need to memorize this fact, it would be given in the problem but you need to use this fact and dimensional analysis to solve the problem).

- goldstanda3269
**Posts:**1673**Joined:**Wed Aug 25, 2010 10:59 pm

Ah okay I understand the first bit but I was just wondering why do we

do 0.025 x (1/2) x 63.5?

Sorry I just cant seem to get my head around it.

Thanks

do 0.025 x (1/2) x 63.5?

Sorry I just cant seem to get my head around it.

Thanks

- ElleRT
**Posts:**55**Joined:**Thu Jul 09, 2015 4:50 am

The key to dimensional analysis is assessing the units you have, and the units you need to end up with. That helps you know what goes in the denominator or numerator, when you multiply or divide.

The previous step concluded that the number of faradays is 0.025F. But the question is asking about grams of copper so we need to find a relationship between the two, and then using dimensional analysis to cancel units so that faradays are converted to grams. The first thing to come to mind is the conversion of moles of copper to faradays based on the equation we mentioned before, and shown in 10.5.1: 1mol Cu/2F.

Now that means that faradays will cancel and we'll end up with moles of copper. To convert moles to grams requires the atomic (or molecular if it is a molecule) weight: 63.5g Cu/mol Cu.

Now it is written so that units cancel and grams become the result.

0.025F x (1mol Cu/2F) x (63.5g Cu/mol Cu) = 0.72g

The previous step concluded that the number of faradays is 0.025F. But the question is asking about grams of copper so we need to find a relationship between the two, and then using dimensional analysis to cancel units so that faradays are converted to grams. The first thing to come to mind is the conversion of moles of copper to faradays based on the equation we mentioned before, and shown in 10.5.1: 1mol Cu/2F.

Now that means that faradays will cancel and we'll end up with moles of copper. To convert moles to grams requires the atomic (or molecular if it is a molecule) weight: 63.5g Cu/mol Cu.

Now it is written so that units cancel and grams become the result.

0.025F x (1mol Cu/2F) x (63.5g Cu/mol Cu) = 0.72g

- goldstanda3269
**Posts:**1673**Joined:**Wed Aug 25, 2010 10:59 pm

"E(cell) = E(red)+E(ox) so What is reduction and What is Oxidation?"

LEO is a GERC mnemonic (PHY 10.2 and videos: Electrolysis and Electrochemical Cell):

This reaction is the Gain Electron Reduction at Cathode which is spontaneous as written because E is positive, so we don't need to switch the reaction around.

Cl2(g) + 2e- ↔ 2Cl-

+ 1.36 V

This reaction has a negative Eo so does not occur as written, so we must switch it:

Na+ + e- ↔ Na(s)

-2.71 V

The switch is a Loss Electron Oxidation is Anode:

Na(s) ↔ Na+ + e-

+2.71 V

E(cell) = E(red)+E(ox)

E(cell) = 1.36 + 2.71 = 4.07 which is spontaneous due to positive Eo value (i.e. this is a galvanic/voltaic cell AKA a battery which spontaneously makes electricity from chemical energy, a form of electrochemical cell).

LEO is a GERC mnemonic (PHY 10.2 and videos: Electrolysis and Electrochemical Cell):

This reaction is the Gain Electron Reduction at Cathode which is spontaneous as written because E is positive, so we don't need to switch the reaction around.

Cl2(g) + 2e- ↔ 2Cl-

+ 1.36 V

This reaction has a negative Eo so does not occur as written, so we must switch it:

Na+ + e- ↔ Na(s)

-2.71 V

The switch is a Loss Electron Oxidation is Anode:

Na(s) ↔ Na+ + e-

+2.71 V

E(cell) = E(red)+E(ox)

E(cell) = 1.36 + 2.71 = 4.07 which is spontaneous due to positive Eo value (i.e. this is a galvanic/voltaic cell AKA a battery which spontaneously makes electricity from chemical energy, a form of electrochemical cell).

- goldstanda3269
**Posts:**1673**Joined:**Wed Aug 25, 2010 10:59 pm

A. No, that is not correct. Electronegatively refers to the pull on an electron in a molecular environment. Ions are not necessarily molecules.

B. Yes, the tendency will be provided (or you need to calculate it) by suggesting the sign of the Eo value. This is explained in detail in CHM 10.1. The more positive the Eo value, the more likely the reaction will occur spontaneously as written. There are many GS chapter review questions on this issue (Gen Chem Chapter 10), ACER's red booklet has 2 units on this issue, and several GS exams review this issue.

B. Yes, the tendency will be provided (or you need to calculate it) by suggesting the sign of the Eo value. This is explained in detail in CHM 10.1. The more positive the Eo value, the more likely the reaction will occur spontaneously as written. There are many GS chapter review questions on this issue (Gen Chem Chapter 10), ACER's red booklet has 2 units on this issue, and several GS exams review this issue.

- goldstanda3269
**Posts:**1673**Joined:**Wed Aug 25, 2010 10:59 pm

Yes, that is correct.

CHM 10.1, 10.2, 10.4 and Electro. videos.

More than what is required for the GAMSAT: http://chemwiki.ucdavis.edu/Analytical_ ... lytic_cell

CHM 10.1, 10.2, 10.4 and Electro. videos.

More than what is required for the GAMSAT: http://chemwiki.ucdavis.edu/Analytical_ ... lytic_cell

- goldstanda3269
**Posts:**1673**Joined:**Wed Aug 25, 2010 10:59 pm

"1. We generally call 'cell' as battery which generate electric power."

- That is not accurate.

- A battery is a galvanic cell. The opposite (i.e. NOT a battery) is an electrolytic cell.

"But the term electrolytic 'cell' seems not have any battery role, it just waste electric power. It would be appropriate being called electrolytic coating..Is there any reason that being called electrolytic 'cell'??"

- Coating and cell are not the same.

- Please see the video on electrolysis so you can see a cell or look at Chap 10.

"2. Is there any condition for rode? Any metal is right?"

- Please rephrase the question.

"3. The solution should be 'ionic bond compound' right?

If organic compound like sucrose, it will not happen. Right?"

- Correct. Electrochemistry requires electrolytes (ionic compounds which can conduct electricity). Sucrose is not an electrolyte.

- CHM 5.3.2, CHM 10.4

- That is not accurate.

- A battery is a galvanic cell. The opposite (i.e. NOT a battery) is an electrolytic cell.

"But the term electrolytic 'cell' seems not have any battery role, it just waste electric power. It would be appropriate being called electrolytic coating..Is there any reason that being called electrolytic 'cell'??"

- Coating and cell are not the same.

- Please see the video on electrolysis so you can see a cell or look at Chap 10.

"2. Is there any condition for rode? Any metal is right?"

- Please rephrase the question.

"3. The solution should be 'ionic bond compound' right?

If organic compound like sucrose, it will not happen. Right?"

- Correct. Electrochemistry requires electrolytes (ionic compounds which can conduct electricity). Sucrose is not an electrolyte.

- CHM 5.3.2, CHM 10.4

- goldstanda3269
**Posts:**1673**Joined:**Wed Aug 25, 2010 10:59 pm

Just to be clear, your question has absolutely nothing to do with the GAMSAT.

Nonetheless, the answer is 'no' because the metal must be chosen so that it does not interact with the liquid in which it is immersed, examples: platinum, stainless steel, titanium, graphite (a form of carbon), etc.

Nonetheless, the answer is 'no' because the metal must be chosen so that it does not interact with the liquid in which it is immersed, examples: platinum, stainless steel, titanium, graphite (a form of carbon), etc.

- goldstanda3269
**Posts:**1673**Joined:**Wed Aug 25, 2010 10:59 pm

I have tried several browsers, but they all failed to open the picture outlining the basic elements of the ETC.

Could you fix it?

Could you fix it?

- li.zhang96890
**Posts:**36**Joined:**Tue Aug 18, 2015 9:19 pm

We will have that image updated within 24 h.

- goldstanda3269
**Posts:**1673**Joined:**Wed Aug 25, 2010 10:59 pm

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