OCR B Salters A Level Chemistry 2025 Advanced Article
Hi do those doing OCR B salters chemistry have any thoughts on the advanced article?
Reply 1
Original post by Persephonetal
Hi do those doing OCR B salters chemistry have any thoughts on the advanced article?
whats that
Reply 2
Original post by Stupidwebsiegon
whats that
It’s a prerelease in the ocr website that is included in paper 2
Reply 3
It should be found here: https://www.ocr.org.uk/Images/730677-scientific-literacy-in-chemistry-advance-notice-article-june-2025.pdf
I'm a second year undergrad chemist who did OCR physics B (and Edexcel chemistry) at A level, so I do vaguely recall how the advance notice articles work. You are encouraged by OCR to freely discuss the advance notice article with other candidates and to investigate the material individually so I see no harm in anyone sharing their thoughts here.
I'll have a look at the advance notice article in more depth when I have a bit more time on my hands and share my thoughts here. It does look like you may have more than a few equations to construct or balance and it's more than likely you'll get an E(cell) question or similar. I also wouldn't be surprised if they decided to throw a summarise the advantages of a Li-ion battery compared to a VRFB question (not too dissimilar to the hilarious last question on the 2018 paper 2 for OCR B where they asked you to summarise the evidence Napoleon was killed by his wallpaper LOL).
I'm a second year undergrad chemist who did OCR physics B (and Edexcel chemistry) at A level, so I do vaguely recall how the advance notice articles work. You are encouraged by OCR to freely discuss the advance notice article with other candidates and to investigate the material individually so I see no harm in anyone sharing their thoughts here.
I'll have a look at the advance notice article in more depth when I have a bit more time on my hands and share my thoughts here. It does look like you may have more than a few equations to construct or balance and it's more than likely you'll get an E(cell) question or similar. I also wouldn't be surprised if they decided to throw a summarise the advantages of a Li-ion battery compared to a VRFB question (not too dissimilar to the hilarious last question on the 2018 paper 2 for OCR B where they asked you to summarise the evidence Napoleon was killed by his wallpaper LOL).
Reply 4
Possible questions that spring to mind and some thoughts on how to approach them:
(a)(i) Write a pair of ionic equations for the processes used to prepare the electrolyte in the positive half cell of a VRFB. [2 marks]
(a)(ii) Construct an equation for the overall cell reaction in a VRFB. [1 mark]
e.g VO2^+ (aq) + 2H^+ (aq) + V^2+ (aq) —> VO^2+ (aq) + H2O (l) + V^3+ (aq)
(a)(iii) It is given that the standard reduction potential for the positive half cell is +1.00 V and the standard reduction potential for the negative half cell is -0.26 V. Determine the cell EMF for a VRFB and comment on the value obtained. [2 marks]
EMF = E° = (+1.00 V) - (-0.26 V) = +1.26 V
This value is positive and so the cell reaction is thermodynamically feasible under standard conditions.
(b) A student makes the following statement. “Lithium ion batteries are less suitable than VRFB for applications in cars”. Discuss this statement. [6 marks]
A bit open-ended, but if this or a similar question is asked and the general structure of a 6-marker is to be assumed, expect there to be three aspects of the question to write about, e.g summarise all the advantages and disadvantages of each cell type outlined in the sections titled “cell architecture”, “energy capacity” and “battery energy density & battery power density”.
(c) Likely some calculations involving battery energy density or battery power density. [? marks]
You could be asked to calculate an enthalpy change from this data (e.g you get the energy produced by the cell by multiplying its mass by the battery energy density and then after converting the units as required, divide by the number of moles of electrolyte blah blah), asked to do the reverse of that, or given an unfamiliar equation such as the Nernst equation (E = E° - RT/nF lnK) or lnK = nFE/RT to use in order to calculate the equilibrium constant for one cell reaction and therefore work out the actual concentrations of electrolytes used in light of the operating conditions of the cell.
(a)(i) Write a pair of ionic equations for the processes used to prepare the electrolyte in the positive half cell of a VRFB. [2 marks]
1.
V2O5 (s) + 2H^+ (aq) —> 2VO2^+ (aq) + H2O (l)
2.
VO2^+ (aq) + 2H^+ (aq) + e^- —> VO^2+ (aq) + H2O (l)
(a)(ii) Construct an equation for the overall cell reaction in a VRFB. [1 mark]
e.g VO2^+ (aq) + 2H^+ (aq) + V^2+ (aq) —> VO^2+ (aq) + H2O (l) + V^3+ (aq)
(a)(iii) It is given that the standard reduction potential for the positive half cell is +1.00 V and the standard reduction potential for the negative half cell is -0.26 V. Determine the cell EMF for a VRFB and comment on the value obtained. [2 marks]
EMF = E° = (+1.00 V) - (-0.26 V) = +1.26 V
This value is positive and so the cell reaction is thermodynamically feasible under standard conditions.
(b) A student makes the following statement. “Lithium ion batteries are less suitable than VRFB for applications in cars”. Discuss this statement. [6 marks]
A bit open-ended, but if this or a similar question is asked and the general structure of a 6-marker is to be assumed, expect there to be three aspects of the question to write about, e.g summarise all the advantages and disadvantages of each cell type outlined in the sections titled “cell architecture”, “energy capacity” and “battery energy density & battery power density”.
(c) Likely some calculations involving battery energy density or battery power density. [? marks]
You could be asked to calculate an enthalpy change from this data (e.g you get the energy produced by the cell by multiplying its mass by the battery energy density and then after converting the units as required, divide by the number of moles of electrolyte blah blah), asked to do the reverse of that, or given an unfamiliar equation such as the Nernst equation (E = E° - RT/nF lnK) or lnK = nFE/RT to use in order to calculate the equilibrium constant for one cell reaction and therefore work out the actual concentrations of electrolytes used in light of the operating conditions of the cell.
Reply 5
Original post by TypicalNerd
Possible questions that spring to mind and some thoughts on how to approach them:
(a)(i) Write a pair of ionic equations for the processes used to prepare the electrolyte in the positive half cell of a VRFB. [2 marks]
(a)(ii) Construct an equation for the overall cell reaction in a VRFB. [1 mark]
e.g VO2^+ (aq) + 2H^+ (aq) + V^2+ (aq) —> VO^2+ (aq) + H2O (l) + V^3+ (aq)
(a)(iii) It is given that the standard reduction potential for the positive half cell is +1.00 V and the standard reduction potential for the negative half cell is -0.26 V. Determine the cell EMF for a VRFB and comment on the value obtained. [2 marks]
EMF = E° = (+1.00 V) - (-0.26 V) = +1.26 V
This value is positive and so the cell reaction is thermodynamically feasible under standard conditions.
(b) A student makes the following statement. “Lithium ion batteries are less suitable than VRFB for applications in cars”. Discuss this statement. [6 marks]
A bit open-ended, but if this or a similar question is asked and the general structure of a 6-marker is to be assumed, expect there to be three aspects of the question to write about, e.g summarise all the advantages and disadvantages of each cell type outlined in the sections titled “cell architecture”, “energy capacity” and “battery energy density & battery power density”.
(c) Likely some calculations involving battery energy density or battery power density. [? marks]
You could be asked to calculate an enthalpy change from this data (e.g you get the energy produced by the cell by multiplying its mass by the battery energy density and then after converting the units as required, divide by the number of moles of electrolyte blah blah), asked to do the reverse of that, or given an unfamiliar equation such as the Nernst equation (E = E° - RT/nF lnK) or lnK = nFE/RT to use in order to calculate the equilibrium constant for one cell reaction and therefore work out the actual concentrations of electrolytes used in light of the operating conditions of the cell.
(a)(i) Write a pair of ionic equations for the processes used to prepare the electrolyte in the positive half cell of a VRFB. [2 marks]
1.
V2O5 (s) + 2H^+ (aq) —> 2VO2^+ (aq) + H2O (l)
2.
VO2^+ (aq) + 2H^+ (aq) + e^- —> VO^2+ (aq) + H2O (l)
(a)(ii) Construct an equation for the overall cell reaction in a VRFB. [1 mark]
e.g VO2^+ (aq) + 2H^+ (aq) + V^2+ (aq) —> VO^2+ (aq) + H2O (l) + V^3+ (aq)
(a)(iii) It is given that the standard reduction potential for the positive half cell is +1.00 V and the standard reduction potential for the negative half cell is -0.26 V. Determine the cell EMF for a VRFB and comment on the value obtained. [2 marks]
EMF = E° = (+1.00 V) - (-0.26 V) = +1.26 V
This value is positive and so the cell reaction is thermodynamically feasible under standard conditions.
(b) A student makes the following statement. “Lithium ion batteries are less suitable than VRFB for applications in cars”. Discuss this statement. [6 marks]
A bit open-ended, but if this or a similar question is asked and the general structure of a 6-marker is to be assumed, expect there to be three aspects of the question to write about, e.g summarise all the advantages and disadvantages of each cell type outlined in the sections titled “cell architecture”, “energy capacity” and “battery energy density & battery power density”.
(c) Likely some calculations involving battery energy density or battery power density. [? marks]
You could be asked to calculate an enthalpy change from this data (e.g you get the energy produced by the cell by multiplying its mass by the battery energy density and then after converting the units as required, divide by the number of moles of electrolyte blah blah), asked to do the reverse of that, or given an unfamiliar equation such as the Nernst equation (E = E° - RT/nF lnK) or lnK = nFE/RT to use in order to calculate the equilibrium constant for one cell reaction and therefore work out the actual concentrations of electrolytes used in light of the operating conditions of the cell.
Additionally, they could throw something on there about the colours of different vanadium ions (I think OCR A wrote a similar question on this which was on their 2019 paper 1), but I think that’s quite unlikely.
Reply 6
Original post by Persephonetal
Hi do those doing OCR B salters chemistry have any thoughts on the advanced article?
Hi, I’m sitting this exam too. I think it’s quite a nice advanced notice, especially with it more just being a comparison between two types of batteries. I was wondering if they would bring in electrode potentials, feasibility and just lots of things from DM. I’m not sure about other links as we haven’t spoken with our teachers yet, but once we do I’ll write what they thought too 🙂
Reply 7
I can’t actually remember if OCR B does anything on this, but I could definitely see them asking you to give reasons why the cell EMF in practice differs from a calculated one from standard reduction potentials.
If you get one such question, you need to consider both equilibria. Par exemple:
“In practice, a VRFB has a cell EMF of +1.50 V. A student suggests that in practice, this is the result of using a more dilute solution of V^3+ in the V^3+/V^2+ half-cell. Discuss this suggestion.”
Recalling from one of my earlier posts:
VO2^+ (aq) + 2H^+ (aq) + e^- <=> VO^2+ (aq) + H2O (l) (E° = +1.00 V)
V^3+ (aq) + e^- <=> V^2+ (aq) (E° = -0.26 V)
I would begin by stating what standard conditions are, e.g
298 K temperature, 1 mol dm^-3 concentrations of ions (and 100 kPa pressure - but there are no gases, so don’t worry about these).
Next I would comment that if the V^3+ solution is more dilute, then the V^3+/V^2+ equilibrium will have to shift left and so the electrode potential for this half cell will be more negative than the standard value of -0.26 V. Therefore the student is correct that the EMF will be more positive than expected if this is indeed the case (since EMF = E(reduction half cell) - E(oxidation half cell) and our oxidation half cell now has a more negative electrode potential).
I would then say the student has not considered other possible ways the conditions could deviate from standard (e.g using a larger concentration of VO2^+ to drive the equilibrium at the reduction half cell forward, or adjusting the temperature as appropriate - note that this isn’t an exhaustive list of the ways the EMF could be increased).
If you get one such question, you need to consider both equilibria. Par exemple:
“In practice, a VRFB has a cell EMF of +1.50 V. A student suggests that in practice, this is the result of using a more dilute solution of V^3+ in the V^3+/V^2+ half-cell. Discuss this suggestion.”
Recalling from one of my earlier posts:
VO2^+ (aq) + 2H^+ (aq) + e^- <=> VO^2+ (aq) + H2O (l) (E° = +1.00 V)
V^3+ (aq) + e^- <=> V^2+ (aq) (E° = -0.26 V)
I would begin by stating what standard conditions are, e.g
298 K temperature, 1 mol dm^-3 concentrations of ions (and 100 kPa pressure - but there are no gases, so don’t worry about these).
Next I would comment that if the V^3+ solution is more dilute, then the V^3+/V^2+ equilibrium will have to shift left and so the electrode potential for this half cell will be more negative than the standard value of -0.26 V. Therefore the student is correct that the EMF will be more positive than expected if this is indeed the case (since EMF = E(reduction half cell) - E(oxidation half cell) and our oxidation half cell now has a more negative electrode potential).
I would then say the student has not considered other possible ways the conditions could deviate from standard (e.g using a larger concentration of VO2^+ to drive the equilibrium at the reduction half cell forward, or adjusting the temperature as appropriate - note that this isn’t an exhaustive list of the ways the EMF could be increased).
(edited 1 month ago)
Reply 9
Original post by TypicalNerd
I can’t actually remember if OCR B does anything on this, but I could definitely see them asking you to give reasons why the cell EMF in practice differs from a calculated one from standard reduction potentials.
If you get one such question, you need to consider both equilibria. Par exemple:
“In practice, a VRFB has a cell EMF of +1.50 V. A student suggests that in practice, this is the result of using a more dilute solution of V^3+ in the V^3+/V^2+ half-cell. Discuss this suggestion.”
Recalling from one of my earlier posts:
VO2^+ (aq) + 2H^+ (aq) + e^- <=> VO^2+ (aq) + H2O (l) (E° = +1.00 V)
V^3+ (aq) + e^- <=> V^2+ (aq) (E° = -0.26 V)
I would begin by stating what standard conditions are, e.g
298 K temperature, 1 mol dm^-3 concentrations of ions (and 100 kPa pressure - but there are no gases, so don’t worry about these).
Next I would comment that if the V^3+ solution is more dilute, then the V^3+/V^2+ equilibrium will have to shift left and so the electrode potential for this half cell will be more negative than the standard value of -0.26 V. Therefore the student is correct that the EMF will be more positive than expected if this is indeed the case (since EMF = E(reduction half cell) - E(oxidation half cell) and our oxidation half cell now has a more negative electrode potential).
I would then say the student has not considered other possible ways the conditions could deviate from standard (e.g using a larger concentration of VO2^+ to drive the equilibrium at the reduction half cell forward, or adjusting the temperature as appropriate - note that this isn’t an exhaustive list of the ways the EMF could be increased).
If you get one such question, you need to consider both equilibria. Par exemple:
“In practice, a VRFB has a cell EMF of +1.50 V. A student suggests that in practice, this is the result of using a more dilute solution of V^3+ in the V^3+/V^2+ half-cell. Discuss this suggestion.”
Recalling from one of my earlier posts:
VO2^+ (aq) + 2H^+ (aq) + e^- <=> VO^2+ (aq) + H2O (l) (E° = +1.00 V)
V^3+ (aq) + e^- <=> V^2+ (aq) (E° = -0.26 V)
I would begin by stating what standard conditions are, e.g
298 K temperature, 1 mol dm^-3 concentrations of ions (and 100 kPa pressure - but there are no gases, so don’t worry about these).
Next I would comment that if the V^3+ solution is more dilute, then the V^3+/V^2+ equilibrium will have to shift left and so the electrode potential for this half cell will be more negative than the standard value of -0.26 V. Therefore the student is correct that the EMF will be more positive than expected if this is indeed the case (since EMF = E(reduction half cell) - E(oxidation half cell) and our oxidation half cell now has a more negative electrode potential).
I would then say the student has not considered other possible ways the conditions could deviate from standard (e.g using a larger concentration of VO2^+ to drive the equilibrium at the reduction half cell forward, or adjusting the temperature as appropriate - note that this isn’t an exhaustive list of the ways the EMF could be increased).
Reply 10
Original post by susieee768
I am confused about the direction of the half equations occurring at the anode and at the cathode during discharging and charging, does anyone have any idea what the right order us because I know that it will switch wether tis charging or discharging but I have no idea how to tell which direction each of the half equations occur and what the overall equation for those would be someone please help because I tried asking chatgpt and my teachers and I just ended up more confused
Hi I hope this helps:
Oxidation
Is
Loss of Electrons
Reduction
Is
Gain
When 2 half-cells are connected, reduction will take place in the one with the more
positive Eθ (will be in a table) so that will gain electrons, all you have to do is do a normal half equation to show this:
e.g. Ag+ + e- -> Ag
If the other half cell had a more negative Eθ value then it will be oxidised (lose electrons) so we do another half equation-
e.g. Cu -> Cu2+ + 2e-
Then if you need to combine it we times the equation Ag+ + e- -> Ag by 2 so there are 2e- so they cancel out then we have (as the overall equation):
2Ag+ + Cu -> 2Ag + Cu2+
Basically the more negative cell that is oxidised supplies the electrons to the more positively charged cell so the current moves from the more negative half cell to the more positive half cell. Ions move across the slat bridge between the two half cells.
To calculate the overall E cell value we use:
Eθ cell= Eθ reduced (more positive value) - E
θ oxidised (more negative value)
In order for the cell to work the answer should be positive (must put a + in front of your answer if you get asked to calculate it)
(Also this is just my understanding please do look into it further if you can, just in case there are any mistakes) I recommend Allery Chemistry I think this topic is in one of his DM topic questions
Reply 11
Original post by Persephonetal
Hi I hope this helps:
Oxidation
Is
Loss of Electrons
Reduction
Is
Gain
When 2 half-cells are connected, reduction will take place in the one with the more
positive Eθ (will be in a table) so that will gain electrons, all you have to do is do a normal half equation to show this:
e.g. Ag+ + e- -> Ag
If the other half cell had a more negative Eθ value then it will be oxidised (lose electrons) so we do another half equation-
e.g. Cu -> Cu2+ + 2e-
Then if you need to combine it we times the equation Ag+ + e- -> Ag by 2 so there are 2e- so they cancel out then we have (as the overall equation):
2Ag+ + Cu -> 2Ag + Cu2+
Basically the more negative cell that is oxidised supplies the electrons to the more positively charged cell so the current moves from the more negative half cell to the more positive half cell. Ions move across the slat bridge between the two half cells.
To calculate the overall E cell value we use:
Eθ cell= Eθ reduced (more positive value) - E
θ oxidised (more negative value)
In order for the cell to work the answer should be positive (must put a + in front of your answer if you get asked to calculate it)
(Also this is just my understanding please do look into it further if you can, just in case there are any mistakes) I recommend Allery Chemistry I think this topic is in one of his DM topic questions
Oxidation
Is
Loss of Electrons
Reduction
Is
Gain
When 2 half-cells are connected, reduction will take place in the one with the more
positive Eθ (will be in a table) so that will gain electrons, all you have to do is do a normal half equation to show this:
e.g. Ag+ + e- -> Ag
If the other half cell had a more negative Eθ value then it will be oxidised (lose electrons) so we do another half equation-
e.g. Cu -> Cu2+ + 2e-
Then if you need to combine it we times the equation Ag+ + e- -> Ag by 2 so there are 2e- so they cancel out then we have (as the overall equation):
2Ag+ + Cu -> 2Ag + Cu2+
Basically the more negative cell that is oxidised supplies the electrons to the more positively charged cell so the current moves from the more negative half cell to the more positive half cell. Ions move across the slat bridge between the two half cells.
To calculate the overall E cell value we use:
Eθ cell= Eθ reduced (more positive value) - E
θ oxidised (more negative value)
In order for the cell to work the answer should be positive (must put a + in front of your answer if you get asked to calculate it)
(Also this is just my understanding please do look into it further if you can, just in case there are any mistakes) I recommend Allery Chemistry I think this topic is in one of his DM topic questions
thank you so much I was just asking about the half cells specific to the actual article because form my understanding is for the VFRBS there are two half equations which are the VO]²⁺ + 2H⁺ + e⁻ ⇄ V³⁺ + H₂O at the cathode and V³⁺ + e⁻ ⇄ V²⁺ at the anode. however the direction of the equation reversed on wether the VFRB is charging or discharging and I don't know which direction it would be wether its charging or discharging for the anode and cathode
Reply 12
Original post by susieee768
thank you so much I was just asking about the half cells specific to the actual article because form my understanding is for the VFRBS there are two half equations which are the VO]²⁺ + 2H⁺ + e⁻ ⇄ V³⁺ + H₂O at the cathode and V³⁺ + e⁻ ⇄ V²⁺ at the anode. however the direction of the equation reversed on wether the VFRB is charging or discharging and I don't know which direction it would be wether its charging or discharging for the anode and cathode
When charging, the reverse takes place.
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