HSC Chemistry Question Thread

[Mathew587]

Hi can someone help me with this question.
Write overall Redox equations for the following reaction in an acidic environment and identify the oxidising and reducing agent in each case
A) MNO4^-1 (aq) + H2S (aq) -> Mn^2+ (aq) + S(s)

B) ClO^-1 (aq) + SO2 (g) -> Cl^-1(aq) + SO4^2- (aq)

Grt help btw :/

[jakesilove]

Hey guys i'm having trouble in determining whether these equations are redox reactions. How can i tell?

A) Mg + H₂SO⁴ --> MgSO₄ + H₂ (for thisone i said yes b/c Mg is oxidising ang H₂ is reducing???)
B) 2MnO₄^- + 16H⁺ + 10I^- --> 2Mn²⁺+ 5I₂ + 8H₂O

Hey! For the first one you're absolutely correct! Great job

Let's take a closer look at the second one. Hydrogen looks like it stays exactly the same, as does Oxygen, so we can attack the equation by looking at the Magnesium and the Iodine.

What is the initial charge of Magnesium? Well,




So, the initial charge of magnesium must have been 7+. Clearly, the final charge on Magnesium is now 2+, therefore it has REDUCED (gained electrons, or decreased in oxidation number). If one species reduces, another MUST oxidise. In this case, Iodine oxidises (in this case, goes from -1 to 0).

Does that make sense? It's definitely worth being able to write out the half equations for these reactions, leaving out spectator ions.

[jakesilove]

Grt help btw :/

Sorry Mathew, must have missed that question! Will get on it right now

[jakesilove]

Hi can someone help me with this question.
Write overall Redox equations for the following reaction in an acidic environment and identify the oxidising and reducing agent in each case
A) MNO4^-1 (aq) + H2S (aq) -> Mn^2+ (aq) + S(s)

B) ClO^-1 (aq) + SO2 (g) -> Cl^-1(aq) + SO4^2- (aq)

Now, I'm not entirely sure what the 'overall redox equation... in an acidic environment' means. Perhaps it means that there is an excess of Hydrogen ions? Regardless, the equation is hopelessly wrong, so let's address that first.

Let's keep the reactants the same, and see what products really are



Okay, I'm almost certain that the 'acidic environment' part of the question means we should ADD Hydrogen ions to the system



That's looking a lot nicer! It's balanced, everything is happy, as expected.

Now, let's look at what is oxidising, and what is reducing. The Hydrogen ions are staying the same, as is the Oxygen atoms. Generally, it's going to be the metals/solids that oxidise/reduce anyway, so it's a fair guess that we should look to the Manganese and the Sulfer.

What is the initial charge on the Manganese? Well,




Great! So, the initial charge was +7, and it goes to +2. The oxidation number has decreased, therefore the substance has reduced! Accordingly, Sulfur has oxidised (gone from -2 to 0).

Working on the second question now

Okay, we have



Whilst I suppose we could balance the equation like this, I'm more comfortable making everything as stable as possible, and introducing an 'acidic environment'. As such, the equation will likely look more like this



Beautiful, balanced, awesome. Again, Hydrogen and Oxygen are unlikely to have oxidised or reduced. So, let's look at the Chlorine and the Sulfur.
Sulfur clearly starts as +4, and becomes +6. Therefore, it oxidises (increases in Oxidation number). Chlorine starts as +1, and ends as -1. Therefore, it reduces!

Great! Difficult question, and I'm sure the chemical formulas could have been completed in a number of ways, but hey this worked. Did all that make sense? Sorry for the delayed response!

[Bubbly_bluey]

Hello again!
Two quick questions: in the chemistry data sheet, there is a list of standard potentials, some with negative and positive voltages.
1) what does the sign mean (like chemically what is happening to determine the sign of voltage)?
2) when you work out the redox reaction (like choosing two half equations but flipping one of them) we can calculate the voltage when you add them. So with a voltaic cell, assuming we have a perfect experiment, is the reading on the volt meter suppose to match the value we worked out from the equations? Hope this makes sense.
Thank you 

[jakesilove]

Hello again!
Two quick questions: in the chemistry data sheet, there is a list of standard potentials, some with negative and positive voltages.
1) what does the sign mean (like chemically what is happening to determine the sign of voltage)?
2) when you work out the redox reaction (like choosing two half equations but flipping one of them) we can calculate the voltage when you add them. So with a voltaic cell, assuming we have a perfect experiment, is the reading on the volt meter suppose to match the value we worked out from the equations? Hope this makes sense.
Thank you 

Hey! Essentially, if the sign is positive, the species will reduce very readily. If the sign is negative, the species will be be difficult to reduce (and therefore oxidise very readily). If the overall reaction has a positive sign, the cell will move forward (ie. react, causing a current to flow). If the overall reaction as a negative sign, the cell will not move forward (ie. no reaction, no current).

Yep, you're absolutely right about the second bit! Assuming one mol/L concentration of solution, perfect transmission of current, no energy loss etc. that it was you would expect a voltmeter to read

[Mathew587]

Now, I'm not entirely sure what the 'overall redox equation... in an acidic environment' means. Perhaps it means that there is an excess of Hydrogen ions? Regardless, the equation is hopelessly wrong, so let's address that first.

Let's keep the reactants the same, and see what products really are



Okay, I'm almost certain that the 'acidic environment' part of the question means we should ADD Hydrogen ions to the system



That's looking a lot nicer! It's balanced, everything is happy, as expected.

Now, let's look at what is oxidising, and what is reducing. The Hydrogen ions are staying the same, as is the Oxygen atoms. Generally, it's going to be the metals/solids that oxidise/reduce anyway, so it's a fair guess that we should look to the Manganese and the Sulfer.

What is the initial charge on the Manganese? Well,




Great! So, the initial charge was +7, and it goes to +2. The oxidation number has decreased, therefore the substance has reduced! Accordingly, Sulfur has oxidised (gone from -2 to 0).

Working on the second question now

Okay, we have



Whilst I suppose we could balance the equation like this, I'm more comfortable making everything as stable as possible, and introducing an 'acidic environment'. As such, the equation will likely look more like this



Beautiful, balanced, awesome. Again, Hydrogen and Oxygen are unlikely to have oxidised or reduced. So, let's look at the Chlorine and the Sulfur.
Sulfur clearly starts as +4, and becomes +6. Therefore, it oxidises (increases in Oxidation number). Chlorine starts as +1, and ends as -1. Therefore, it reduces!

Great! Difficult question, and I'm sure the chemical formulas could have been completed in a number of ways, but hey this worked. Did all that make sense? Sorry for the delayed response!

yup that helped. ty

[Bubbly_bluey]

Hi! i need a little help with this question.
Citiric acid, the predominant acid in lemon juice, is a triprotic acid. A student titrated 25mL samples of lemon juice with 0.55mol/L NaOH.The mean titration volum was 29.50mL. The molar mass of citric acid is 192.12g/mol. What was the concentration of the citric acid in the  lemon?

I went to calulate the moles of NaOH but i dont know how to use the fact that citric acid is a triprotic acid. The answer says i have to divide by 3 but why do you have to do that?
Thanks

[kiwiberry]

Hi! i need a little help with this question.
Citiric acid, the predominant acid in lemon juice, is a triprotic acid. A student titrated 25mL samples of lemon juice with 0.55mol/L NaOH.The mean titration volum was 29.50mL. The molar mass of citric acid is 192.12g/mol. What was the concentration of the citric acid in the  lemon?

I went to calulate the moles of NaOH but i dont know how to use the fact that citric acid is a triprotic acid. The answer says i have to divide by 3 but why do you have to do that?
Thanks

Because citric acid is triprotic, it is able to donate 3 H⁺ in solution. So for every mole of citric acid, 3 moles of OH^- are needed to neutralise the H⁺. Therefore 3 moles of NaOH are needed to completely neutralise 1 mole of citric acid, meaning that you have to divide n(NaOH) by 3 to get n(citric acid)

[J.B]

I was wondering if someone would be able to explain this question (12) to me?
Thanks

 

[1tankengine]

I know it sounds stupid but I can't get this question right-

A student was investigating the heat of combustion of ethanol. She used an ethanol
burner that had an initial mass of 68.0 g. She then lit the burner and placed it under a
beaker containing 500 g of water. After a few minutes, she noticed that the water
temperature had risen from 24°C to 38°C and the burner now weighed 66.5 g.
She made the assumption that only the water was heated.
What would be the student’s value for the heat of combustion for ethanol?

It is from a past half yearly paper at my school and it says the answer is 859 kJ mol–1 but I'm getting a different answer.

[J.B]

I know it sounds stupid but I can't get this question right-

A student was investigating the heat of combustion of ethanol. She used an ethanol
burner that had an initial mass of 68.0 g. She then lit the burner and placed it under a
beaker containing 500 g of water. After a few minutes, she noticed that the water
temperature had risen from 24°C to 38°C and the burner now weighed 66.5 g.
She made the assumption that only the water was heated.
What would be the student’s value for the heat of combustion for ethanol?

It is from a past half yearly paper at my school and it says the answer is 859 kJ mol–1 but I'm getting a different answer.

I received 897KJ mol-1
But I could be wrong.

[kiwiberry]

I was wondering if someone would be able to explain this question (12) to me?
Thanks

 (Image removed from quote.)

I think it's A! All Bronsted-Lowry acids must contain a H⁺ to donate. When this proton is donated, the acid's conjugate base is formed, which will have an overall negative charge since a positive charge (H⁺) was removed from it. eg. HCl/Cl^-, HNO₃/NO₃^- etc.

[1tankengine]

I received 897KJ mol-1
But I could be wrong.

That's closer than I got.. Would you be able to tell me how you worked it out?

[Ellie__]

Hey! Check out this formula sheet that I made a while ago. In my opinion, it has everything you need to memorise! Let me know if you have any questions

THANKYOUUU!!