Still taught to reverse oxidation half cells in electrochemistry?Oxidation of metals/halogens by oxygen gas in acidic aqueous solutionDeriving a reduction potential from two other reduction potentialsIf given a half reaction, how do you determine if an element can exist in a acidic conditions?Electrolysis of dilute and concentrated sodium chloride and the Nernst equationRedox - concentration cellHow to calculate the electrochemical potential of a cell using known half-cell potentials?What is the purpose of the electrolyte in the half-cell where oxidation is taking place?Calculating the standard reduction potential for the oxidation of waterNo sign flipping while figuring out the emf of voltaic cell?What are the products in the reaction of sulphuric acid and Group 2 metals?Oxidation of metals/halogens by oxygen gas in acidic aqueous solution
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Still taught to reverse oxidation half cells in electrochemistry?
Oxidation of metals/halogens by oxygen gas in acidic aqueous solutionDeriving a reduction potential from two other reduction potentialsIf given a half reaction, how do you determine if an element can exist in a acidic conditions?Electrolysis of dilute and concentrated sodium chloride and the Nernst equationRedox - concentration cellHow to calculate the electrochemical potential of a cell using known half-cell potentials?What is the purpose of the electrolyte in the half-cell where oxidation is taking place?Calculating the standard reduction potential for the oxidation of waterNo sign flipping while figuring out the emf of voltaic cell?What are the products in the reaction of sulphuric acid and Group 2 metals?Oxidation of metals/halogens by oxygen gas in acidic aqueous solution
$begingroup$
In a question, Oxidation of metals/halogens by oxygen gas in acidic aqueous solution, there was a point made that reduction half-cells should not be reversed.
I was taught ($approx 1970$) in the following manner. Given the reduction half cells:
$$
beginalign
ceO2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l) &quad E^circ &= pu+1.23 V \
ceAg+ + e- &→ Ag(s) &quad E^circ &= pu+0.799 V \
ceBr2(l) + 2 e- &→ 2 Br-(aq) &quad E^circ &= pu+1.065 V
endalign
$$
To get the the standard oxidation potentials you reverse the products and reactants and you must also flip the sign of the reaction. So:
$$
beginalign
ceAg(s) &→ Ag+ + e- &quad E^circ &= pu-0.799 V \
ce2 Br-(aq) &→ Br2(l) + 2 e- &quad E^circ &= pu-1.065 V
endalign
$$
Now you can write a balanced chemical reaction as such:
$$
beginalign
ceO2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l) &quad E^circ &= pu+1.23 V \
ce4Ag(s) &→ 4Ag+ + 4e- &quad E^circ &= pu-0.799 V \
ceO2(g) + 4 H+(aq) + 4 Ag(s)&→ 4Ag+ + 2 H2O(l) &quad E_textTotal^circ &= pu+1.23 - 0.799 = +0.43 V
endalign
$$
and to get the cell potential the two half cells are added, one being positive and the other negative. If the EMF is positive the forward reaction is spontaneous, if negative the reverse reaction is spontaneous.
Is flipping the reduction reaction not the standard method taught these days to balance redox reactions?
electrochemistry
asked Apr 13 at 16:04
MaxWMaxW
16.3k22363
$endgroup$
add a comment |
$begingroup$
In a question, Oxidation of metals/halogens by oxygen gas in acidic aqueous solution, there was a point made that reduction half-cells should not be reversed.
I was taught ($approx 1970$) in the following manner. Given the reduction half cells:
$$
beginalign
ceO2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l) &quad E^circ &= pu+1.23 V \
ceAg+ + e- &→ Ag(s) &quad E^circ &= pu+0.799 V \
ceBr2(l) + 2 e- &→ 2 Br-(aq) &quad E^circ &= pu+1.065 V
endalign
$$
To get the the standard oxidation potentials you reverse the products and reactants and you must also flip the sign of the reaction. So:
$$
beginalign
ceAg(s) &→ Ag+ + e- &quad E^circ &= pu-0.799 V \
ce2 Br-(aq) &→ Br2(l) + 2 e- &quad E^circ &= pu-1.065 V
endalign
$$
Now you can write a balanced chemical reaction as such:
$$
beginalign
ceO2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l) &quad E^circ &= pu+1.23 V \
ce4Ag(s) &→ 4Ag+ + 4e- &quad E^circ &= pu-0.799 V \
ceO2(g) + 4 H+(aq) + 4 Ag(s)&→ 4Ag+ + 2 H2O(l) &quad E_textTotal^circ &= pu+1.23 - 0.799 = +0.43 V
endalign
$$
and to get the cell potential the two half cells are added, one being positive and the other negative. If the EMF is positive the forward reaction is spontaneous, if negative the reverse reaction is spontaneous.
Is flipping the reduction reaction not the standard method taught these days to balance redox reactions?
electrochemistry
asked Apr 13 at 16:04
MaxWMaxW
16.3k22363
$endgroup$
$begingroup$
It looks like silver-silver cell would be even better than silver zinc cell, with the added value of perpetuum mobile. :-)
$endgroup$
– Poutnik
Apr 13 at 17:05
$begingroup$
For the standard silver cell vs a standard silver cell the EMF would be 0 and thus no current flow. Because of the Nernst equation the potential of a non-standard silver cell would vary if the $ceA_Ag+ ne 1.000 $.
$endgroup$
– MaxW
Apr 13 at 17:09
$begingroup$
I know it well. But it does not make sense to write the silver electrode potential like if silver had been better reduction agent than zinc.
$endgroup$
– Poutnik
Apr 13 at 17:12
$begingroup$
@Poutnik - Huh? I wrote the Ag reaction with an EMF as -0.799V as an oxidation not a reduction so that I could balance the equation.
$endgroup$
– MaxW
Apr 13 at 17:17
$begingroup$
The point is, electrode potential is for 2-way redox reaction equilibrium, holding this potential with respect to the standard hydrogen electrode. For $ceAg/Ag+$, it is positive, not negative. With the open voltage, the oxidation and reduction is ongoing with the same rate.
$endgroup$
– Poutnik
Apr 13 at 17:20
add a comment |
$begingroup$
In a question, Oxidation of metals/halogens by oxygen gas in acidic aqueous solution, there was a point made that reduction half-cells should not be reversed.
I was taught ($approx 1970$) in the following manner. Given the reduction half cells:
$$
beginalign
ceO2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l) &quad E^circ &= pu+1.23 V \
ceAg+ + e- &→ Ag(s) &quad E^circ &= pu+0.799 V \
ceBr2(l) + 2 e- &→ 2 Br-(aq) &quad E^circ &= pu+1.065 V
endalign
$$
To get the the standard oxidation potentials you reverse the products and reactants and you must also flip the sign of the reaction. So:
$$
beginalign
ceAg(s) &→ Ag+ + e- &quad E^circ &= pu-0.799 V \
ce2 Br-(aq) &→ Br2(l) + 2 e- &quad E^circ &= pu-1.065 V
endalign
$$
Now you can write a balanced chemical reaction as such:
$$
beginalign
ceO2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l) &quad E^circ &= pu+1.23 V \
ce4Ag(s) &→ 4Ag+ + 4e- &quad E^circ &= pu-0.799 V \
ceO2(g) + 4 H+(aq) + 4 Ag(s)&→ 4Ag+ + 2 H2O(l) &quad E_textTotal^circ &= pu+1.23 - 0.799 = +0.43 V
endalign
$$
and to get the cell potential the two half cells are added, one being positive and the other negative. If the EMF is positive the forward reaction is spontaneous, if negative the reverse reaction is spontaneous.
Is flipping the reduction reaction not the standard method taught these days to balance redox reactions?
electrochemistry
asked Apr 13 at 16:04
MaxWMaxW
16.3k22363
$endgroup$
In a question, Oxidation of metals/halogens by oxygen gas in acidic aqueous solution, there was a point made that reduction half-cells should not be reversed.
I was taught ($approx 1970$) in the following manner. Given the reduction half cells:
$$
beginalign
ceO2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l) &quad E^circ &= pu+1.23 V \
ceAg+ + e- &→ Ag(s) &quad E^circ &= pu+0.799 V \
ceBr2(l) + 2 e- &→ 2 Br-(aq) &quad E^circ &= pu+1.065 V
endalign
$$
To get the the standard oxidation potentials you reverse the products and reactants and you must also flip the sign of the reaction. So:
$$
beginalign
ceAg(s) &→ Ag+ + e- &quad E^circ &= pu-0.799 V \
ce2 Br-(aq) &→ Br2(l) + 2 e- &quad E^circ &= pu-1.065 V
endalign
$$
Now you can write a balanced chemical reaction as such:
$$
beginalign
ceO2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l) &quad E^circ &= pu+1.23 V \
ce4Ag(s) &→ 4Ag+ + 4e- &quad E^circ &= pu-0.799 V \
ceO2(g) + 4 H+(aq) + 4 Ag(s)&→ 4Ag+ + 2 H2O(l) &quad E_textTotal^circ &= pu+1.23 - 0.799 = +0.43 V
endalign
$$
and to get the cell potential the two half cells are added, one being positive and the other negative. If the EMF is positive the forward reaction is spontaneous, if negative the reverse reaction is spontaneous.
Is flipping the reduction reaction not the standard method taught these days to balance redox reactions?
electrochemistry
electrochemistry
asked Apr 13 at 16:04
MaxWMaxW
16.3k22363
asked Apr 13 at 16:04
MaxWMaxW
16.3k22363
edited Apr 13 at 16:40
MaxW
asked Apr 13 at 16:04
MaxWMaxW
16.3k22363
asked Apr 13 at 16:04
MaxWMaxW
16.3k22363
asked Apr 13 at 16:04
MaxWMaxW
16.3k22363
16.3k22363
$begingroup$
It looks like silver-silver cell would be even better than silver zinc cell, with the added value of perpetuum mobile. :-)
$endgroup$
– Poutnik
Apr 13 at 17:05
$begingroup$
For the standard silver cell vs a standard silver cell the EMF would be 0 and thus no current flow. Because of the Nernst equation the potential of a non-standard silver cell would vary if the $ceA_Ag+ ne 1.000 $.
$endgroup$
– MaxW
Apr 13 at 17:09
$begingroup$
I know it well. But it does not make sense to write the silver electrode potential like if silver had been better reduction agent than zinc.
$endgroup$
– Poutnik
Apr 13 at 17:12
$begingroup$
@Poutnik - Huh? I wrote the Ag reaction with an EMF as -0.799V as an oxidation not a reduction so that I could balance the equation.
$endgroup$
– MaxW
Apr 13 at 17:17
$begingroup$
The point is, electrode potential is for 2-way redox reaction equilibrium, holding this potential with respect to the standard hydrogen electrode. For $ceAg/Ag+$, it is positive, not negative. With the open voltage, the oxidation and reduction is ongoing with the same rate.
$endgroup$
– Poutnik
Apr 13 at 17:20
add a comment |
$begingroup$
It looks like silver-silver cell would be even better than silver zinc cell, with the added value of perpetuum mobile. :-)
$endgroup$
– Poutnik
Apr 13 at 17:05
$begingroup$
For the standard silver cell vs a standard silver cell the EMF would be 0 and thus no current flow. Because of the Nernst equation the potential of a non-standard silver cell would vary if the $ceA_Ag+ ne 1.000 $.
$endgroup$
– MaxW
Apr 13 at 17:09
$begingroup$
I know it well. But it does not make sense to write the silver electrode potential like if silver had been better reduction agent than zinc.
$endgroup$
– Poutnik
Apr 13 at 17:12
$begingroup$
@Poutnik - Huh? I wrote the Ag reaction with an EMF as -0.799V as an oxidation not a reduction so that I could balance the equation.
$endgroup$
– MaxW
Apr 13 at 17:17
$begingroup$
The point is, electrode potential is for 2-way redox reaction equilibrium, holding this potential with respect to the standard hydrogen electrode. For $ceAg/Ag+$, it is positive, not negative. With the open voltage, the oxidation and reduction is ongoing with the same rate.
$endgroup$
– Poutnik
Apr 13 at 17:20
$begingroup$
It looks like silver-silver cell would be even better than silver zinc cell, with the added value of perpetuum mobile. :-)
$endgroup$
– Poutnik
Apr 13 at 17:05
$begingroup$
It looks like silver-silver cell would be even better than silver zinc cell, with the added value of perpetuum mobile. :-)
$endgroup$
– Poutnik
Apr 13 at 17:05
$begingroup$
For the standard silver cell vs a standard silver cell the EMF would be 0 and thus no current flow. Because of the Nernst equation the potential of a non-standard silver cell would vary if the $ceA_Ag+ ne 1.000 $.
$endgroup$
– MaxW
Apr 13 at 17:09
$begingroup$
For the standard silver cell vs a standard silver cell the EMF would be 0 and thus no current flow. Because of the Nernst equation the potential of a non-standard silver cell would vary if the $ceA_Ag+ ne 1.000 $.
$endgroup$
– MaxW
Apr 13 at 17:09
$begingroup$
I know it well. But it does not make sense to write the silver electrode potential like if silver had been better reduction agent than zinc.
$endgroup$
– Poutnik
Apr 13 at 17:12
$begingroup$
I know it well. But it does not make sense to write the silver electrode potential like if silver had been better reduction agent than zinc.
$endgroup$
– Poutnik
Apr 13 at 17:12
$begingroup$
@Poutnik - Huh? I wrote the Ag reaction with an EMF as -0.799V as an oxidation not a reduction so that I could balance the equation.
$endgroup$
– MaxW
Apr 13 at 17:17
$begingroup$
@Poutnik - Huh? I wrote the Ag reaction with an EMF as -0.799V as an oxidation not a reduction so that I could balance the equation.
$endgroup$
– MaxW
Apr 13 at 17:17
$begingroup$
The point is, electrode potential is for 2-way redox reaction equilibrium, holding this potential with respect to the standard hydrogen electrode. For $ceAg/Ag+$, it is positive, not negative. With the open voltage, the oxidation and reduction is ongoing with the same rate.
$endgroup$
– Poutnik
Apr 13 at 17:20
$begingroup$
The point is, electrode potential is for 2-way redox reaction equilibrium, holding this potential with respect to the standard hydrogen electrode. For $ceAg/Ag+$, it is positive, not negative. With the open voltage, the oxidation and reduction is ongoing with the same rate.
$endgroup$
– Poutnik
Apr 13 at 17:20
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I have done a significant amount of research over the past ten years to trace to origins of these electrochemical conventions and luckily got a chance to discuss these with some top electrochemists. I have been planning to write an article on this issue since it is a perpetual confusion. Basically, the origin of these "signs" issues originated in Germany and USA. The Ostwald school of thought, wrote the electrode potentials in accordance with the electrostatic sign of the electrode with reference to the standard hydrogen electrode. This was the so-called European convention. Keep in mind that electrostatic signs of the electrodes are invariant. It does not matter how you write them. A silver electrode in Ag+/Ag half-cell is always positively charged with respect to the hydrogen electrode under standard conditions.
The American school of thought, with its own influence such as Gibbs, Lewis and Randall, Ligane, Latimer, had a thermodynamic view of the signs relating electrode potentials to the Gibbs free energy. Latimer wrote a very famous book "Oxidation Potentials" which is available from Internet archives. Yes, under those conditions we can flip the signs back and forth while keeping Gibbs free energy in mind. The tug of war between European convention vs. American convention went on until the 50s-70s. Textbooks are usually 20 years behind current research...these issues lingered on.
Finally, electrochemists decided, including Allan J. Bard (from US who wrote a very influential electrochemistry book taught all over the world), that let us go back to the original notion- keep the electrostatic signs associated with the electrode potentials which are invariant with respect to the way we write them. This has "almost" terminated the sign convention issue of US vs. European signs of electrode potentials. The plus and minuses took almost a century to resolve. It is amazing to see the insight of those early scientists such as Ostwald, that world finally accepted their views.
answered Apr 13 at 16:41
M. FarooqM. Farooq
3,000316
$endgroup$
$begingroup$
So how do you balance the chemical equation if you don't reverse the silver half cell? The whole point in flipping the silver reaction was to get electrons on both sides of the chemical reaction so that the electrons are the same on both sides.
$endgroup$
– MaxW
Apr 13 at 17:00
$begingroup$
Certainly we need to write the silver half equation as Ag -> Ag+ e, but I would not write electrode potentials next to them and flip signs. This confuses students because in Ecell= Ecathode-Eanode, they may wonder whether they should change Eanode sign or not. The negative sign already accounts for the oxidation.
$endgroup$
– M. Farooq
Apr 13 at 17:29
add a comment |
$begingroup$
I still teach switching the sign. I find it easier to remember adding up reduction potential and oxidation potential. The half reactions are written as reduction in a table of reduction potentials, so it makes sense that you have to treat the oxidation half reaction differently.
If the cell potential is calculated from reduction potential of the cathode half reaction minus that of the anode half reaction, there are more points of possible errors, including trouble with arithmetic.
You might argue that students learn more and gain deeper insight when given more opportunity to make mistakes, so maybe my approach is inferior. For example, students can use the approach I am using without knowing which electrode is the anode and which is the cathode.
answered Apr 13 at 17:00
Karsten TheisKarsten Theis
6,3961048
$endgroup$
$begingroup$
Just curious - Does the text book you use also employ that technique for balancing redox reactions?
$endgroup$
– MaxW
Apr 13 at 17:29
2
$begingroup$
I use multiple textbooks, and my students might use other sources entirely. Of the textbooks I use, D. M. Hanson's Foundations of Chemistry uses $E_textred + E_textox$ while OpenStax Chemistry uses $E_textcathode - E_textanode$.
$endgroup$
– Karsten Theis
Apr 13 at 17:39
add a comment |
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2 Answers
2
active
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votes
2 Answers
2
active
oldest
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active
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$begingroup$
I have done a significant amount of research over the past ten years to trace to origins of these electrochemical conventions and luckily got a chance to discuss these with some top electrochemists. I have been planning to write an article on this issue since it is a perpetual confusion. Basically, the origin of these "signs" issues originated in Germany and USA. The Ostwald school of thought, wrote the electrode potentials in accordance with the electrostatic sign of the electrode with reference to the standard hydrogen electrode. This was the so-called European convention. Keep in mind that electrostatic signs of the electrodes are invariant. It does not matter how you write them. A silver electrode in Ag+/Ag half-cell is always positively charged with respect to the hydrogen electrode under standard conditions.
The American school of thought, with its own influence such as Gibbs, Lewis and Randall, Ligane, Latimer, had a thermodynamic view of the signs relating electrode potentials to the Gibbs free energy. Latimer wrote a very famous book "Oxidation Potentials" which is available from Internet archives. Yes, under those conditions we can flip the signs back and forth while keeping Gibbs free energy in mind. The tug of war between European convention vs. American convention went on until the 50s-70s. Textbooks are usually 20 years behind current research...these issues lingered on.
Finally, electrochemists decided, including Allan J. Bard (from US who wrote a very influential electrochemistry book taught all over the world), that let us go back to the original notion- keep the electrostatic signs associated with the electrode potentials which are invariant with respect to the way we write them. This has "almost" terminated the sign convention issue of US vs. European signs of electrode potentials. The plus and minuses took almost a century to resolve. It is amazing to see the insight of those early scientists such as Ostwald, that world finally accepted their views.
answered Apr 13 at 16:41
M. FarooqM. Farooq
3,000316
$endgroup$
$begingroup$
So how do you balance the chemical equation if you don't reverse the silver half cell? The whole point in flipping the silver reaction was to get electrons on both sides of the chemical reaction so that the electrons are the same on both sides.
$endgroup$
– MaxW
Apr 13 at 17:00
$begingroup$
Certainly we need to write the silver half equation as Ag -> Ag+ e, but I would not write electrode potentials next to them and flip signs. This confuses students because in Ecell= Ecathode-Eanode, they may wonder whether they should change Eanode sign or not. The negative sign already accounts for the oxidation.
$endgroup$
– M. Farooq
Apr 13 at 17:29
add a comment |
$begingroup$
I have done a significant amount of research over the past ten years to trace to origins of these electrochemical conventions and luckily got a chance to discuss these with some top electrochemists. I have been planning to write an article on this issue since it is a perpetual confusion. Basically, the origin of these "signs" issues originated in Germany and USA. The Ostwald school of thought, wrote the electrode potentials in accordance with the electrostatic sign of the electrode with reference to the standard hydrogen electrode. This was the so-called European convention. Keep in mind that electrostatic signs of the electrodes are invariant. It does not matter how you write them. A silver electrode in Ag+/Ag half-cell is always positively charged with respect to the hydrogen electrode under standard conditions.
The American school of thought, with its own influence such as Gibbs, Lewis and Randall, Ligane, Latimer, had a thermodynamic view of the signs relating electrode potentials to the Gibbs free energy. Latimer wrote a very famous book "Oxidation Potentials" which is available from Internet archives. Yes, under those conditions we can flip the signs back and forth while keeping Gibbs free energy in mind. The tug of war between European convention vs. American convention went on until the 50s-70s. Textbooks are usually 20 years behind current research...these issues lingered on.
Finally, electrochemists decided, including Allan J. Bard (from US who wrote a very influential electrochemistry book taught all over the world), that let us go back to the original notion- keep the electrostatic signs associated with the electrode potentials which are invariant with respect to the way we write them. This has "almost" terminated the sign convention issue of US vs. European signs of electrode potentials. The plus and minuses took almost a century to resolve. It is amazing to see the insight of those early scientists such as Ostwald, that world finally accepted their views.
answered Apr 13 at 16:41
M. FarooqM. Farooq
3,000316
$endgroup$
$begingroup$
So how do you balance the chemical equation if you don't reverse the silver half cell? The whole point in flipping the silver reaction was to get electrons on both sides of the chemical reaction so that the electrons are the same on both sides.
$endgroup$
– MaxW
Apr 13 at 17:00
$begingroup$
Certainly we need to write the silver half equation as Ag -> Ag+ e, but I would not write electrode potentials next to them and flip signs. This confuses students because in Ecell= Ecathode-Eanode, they may wonder whether they should change Eanode sign or not. The negative sign already accounts for the oxidation.
$endgroup$
– M. Farooq
Apr 13 at 17:29
add a comment |
$begingroup$
I have done a significant amount of research over the past ten years to trace to origins of these electrochemical conventions and luckily got a chance to discuss these with some top electrochemists. I have been planning to write an article on this issue since it is a perpetual confusion. Basically, the origin of these "signs" issues originated in Germany and USA. The Ostwald school of thought, wrote the electrode potentials in accordance with the electrostatic sign of the electrode with reference to the standard hydrogen electrode. This was the so-called European convention. Keep in mind that electrostatic signs of the electrodes are invariant. It does not matter how you write them. A silver electrode in Ag+/Ag half-cell is always positively charged with respect to the hydrogen electrode under standard conditions.
The American school of thought, with its own influence such as Gibbs, Lewis and Randall, Ligane, Latimer, had a thermodynamic view of the signs relating electrode potentials to the Gibbs free energy. Latimer wrote a very famous book "Oxidation Potentials" which is available from Internet archives. Yes, under those conditions we can flip the signs back and forth while keeping Gibbs free energy in mind. The tug of war between European convention vs. American convention went on until the 50s-70s. Textbooks are usually 20 years behind current research...these issues lingered on.
Finally, electrochemists decided, including Allan J. Bard (from US who wrote a very influential electrochemistry book taught all over the world), that let us go back to the original notion- keep the electrostatic signs associated with the electrode potentials which are invariant with respect to the way we write them. This has "almost" terminated the sign convention issue of US vs. European signs of electrode potentials. The plus and minuses took almost a century to resolve. It is amazing to see the insight of those early scientists such as Ostwald, that world finally accepted their views.
answered Apr 13 at 16:41
M. FarooqM. Farooq
3,000316
$endgroup$
I have done a significant amount of research over the past ten years to trace to origins of these electrochemical conventions and luckily got a chance to discuss these with some top electrochemists. I have been planning to write an article on this issue since it is a perpetual confusion. Basically, the origin of these "signs" issues originated in Germany and USA. The Ostwald school of thought, wrote the electrode potentials in accordance with the electrostatic sign of the electrode with reference to the standard hydrogen electrode. This was the so-called European convention. Keep in mind that electrostatic signs of the electrodes are invariant. It does not matter how you write them. A silver electrode in Ag+/Ag half-cell is always positively charged with respect to the hydrogen electrode under standard conditions.
The American school of thought, with its own influence such as Gibbs, Lewis and Randall, Ligane, Latimer, had a thermodynamic view of the signs relating electrode potentials to the Gibbs free energy. Latimer wrote a very famous book "Oxidation Potentials" which is available from Internet archives. Yes, under those conditions we can flip the signs back and forth while keeping Gibbs free energy in mind. The tug of war between European convention vs. American convention went on until the 50s-70s. Textbooks are usually 20 years behind current research...these issues lingered on.
Finally, electrochemists decided, including Allan J. Bard (from US who wrote a very influential electrochemistry book taught all over the world), that let us go back to the original notion- keep the electrostatic signs associated with the electrode potentials which are invariant with respect to the way we write them. This has "almost" terminated the sign convention issue of US vs. European signs of electrode potentials. The plus and minuses took almost a century to resolve. It is amazing to see the insight of those early scientists such as Ostwald, that world finally accepted their views.
answered Apr 13 at 16:41
M. FarooqM. Farooq
3,000316
edited Apr 13 at 21:29
answered Apr 13 at 16:41
M. FarooqM. Farooq
3,000316
answered Apr 13 at 16:41
M. FarooqM. Farooq
3,000316
answered Apr 13 at 16:41
M. FarooqM. Farooq
3,000316
3,000316
$begingroup$
So how do you balance the chemical equation if you don't reverse the silver half cell? The whole point in flipping the silver reaction was to get electrons on both sides of the chemical reaction so that the electrons are the same on both sides.
$endgroup$
– MaxW
Apr 13 at 17:00
$begingroup$
Certainly we need to write the silver half equation as Ag -> Ag+ e, but I would not write electrode potentials next to them and flip signs. This confuses students because in Ecell= Ecathode-Eanode, they may wonder whether they should change Eanode sign or not. The negative sign already accounts for the oxidation.
$endgroup$
– M. Farooq
Apr 13 at 17:29
add a comment |
$begingroup$
So how do you balance the chemical equation if you don't reverse the silver half cell? The whole point in flipping the silver reaction was to get electrons on both sides of the chemical reaction so that the electrons are the same on both sides.
$endgroup$
– MaxW
Apr 13 at 17:00
$begingroup$
Certainly we need to write the silver half equation as Ag -> Ag+ e, but I would not write electrode potentials next to them and flip signs. This confuses students because in Ecell= Ecathode-Eanode, they may wonder whether they should change Eanode sign or not. The negative sign already accounts for the oxidation.
$endgroup$
– M. Farooq
Apr 13 at 17:29
$begingroup$
So how do you balance the chemical equation if you don't reverse the silver half cell? The whole point in flipping the silver reaction was to get electrons on both sides of the chemical reaction so that the electrons are the same on both sides.
$endgroup$
– MaxW
Apr 13 at 17:00
$begingroup$
So how do you balance the chemical equation if you don't reverse the silver half cell? The whole point in flipping the silver reaction was to get electrons on both sides of the chemical reaction so that the electrons are the same on both sides.
$endgroup$
– MaxW
Apr 13 at 17:00
$begingroup$
Certainly we need to write the silver half equation as Ag -> Ag+ e, but I would not write electrode potentials next to them and flip signs. This confuses students because in Ecell= Ecathode-Eanode, they may wonder whether they should change Eanode sign or not. The negative sign already accounts for the oxidation.
$endgroup$
– M. Farooq
Apr 13 at 17:29
$begingroup$
Certainly we need to write the silver half equation as Ag -> Ag+ e, but I would not write electrode potentials next to them and flip signs. This confuses students because in Ecell= Ecathode-Eanode, they may wonder whether they should change Eanode sign or not. The negative sign already accounts for the oxidation.
$endgroup$
– M. Farooq
Apr 13 at 17:29
add a comment |
$begingroup$
I still teach switching the sign. I find it easier to remember adding up reduction potential and oxidation potential. The half reactions are written as reduction in a table of reduction potentials, so it makes sense that you have to treat the oxidation half reaction differently.
If the cell potential is calculated from reduction potential of the cathode half reaction minus that of the anode half reaction, there are more points of possible errors, including trouble with arithmetic.
You might argue that students learn more and gain deeper insight when given more opportunity to make mistakes, so maybe my approach is inferior. For example, students can use the approach I am using without knowing which electrode is the anode and which is the cathode.
answered Apr 13 at 17:00
Karsten TheisKarsten Theis
6,3961048
$endgroup$
$begingroup$
Just curious - Does the text book you use also employ that technique for balancing redox reactions?
$endgroup$
– MaxW
Apr 13 at 17:29
2
$begingroup$
I use multiple textbooks, and my students might use other sources entirely. Of the textbooks I use, D. M. Hanson's Foundations of Chemistry uses $E_textred + E_textox$ while OpenStax Chemistry uses $E_textcathode - E_textanode$.
$endgroup$
– Karsten Theis
Apr 13 at 17:39
add a comment |
$begingroup$
I still teach switching the sign. I find it easier to remember adding up reduction potential and oxidation potential. The half reactions are written as reduction in a table of reduction potentials, so it makes sense that you have to treat the oxidation half reaction differently.
If the cell potential is calculated from reduction potential of the cathode half reaction minus that of the anode half reaction, there are more points of possible errors, including trouble with arithmetic.
You might argue that students learn more and gain deeper insight when given more opportunity to make mistakes, so maybe my approach is inferior. For example, students can use the approach I am using without knowing which electrode is the anode and which is the cathode.
answered Apr 13 at 17:00
Karsten TheisKarsten Theis
6,3961048
$endgroup$
$begingroup$
Just curious - Does the text book you use also employ that technique for balancing redox reactions?
$endgroup$
– MaxW
Apr 13 at 17:29
2
$begingroup$
I use multiple textbooks, and my students might use other sources entirely. Of the textbooks I use, D. M. Hanson's Foundations of Chemistry uses $E_textred + E_textox$ while OpenStax Chemistry uses $E_textcathode - E_textanode$.
$endgroup$
– Karsten Theis
Apr 13 at 17:39
add a comment |
$begingroup$
I still teach switching the sign. I find it easier to remember adding up reduction potential and oxidation potential. The half reactions are written as reduction in a table of reduction potentials, so it makes sense that you have to treat the oxidation half reaction differently.
If the cell potential is calculated from reduction potential of the cathode half reaction minus that of the anode half reaction, there are more points of possible errors, including trouble with arithmetic.
You might argue that students learn more and gain deeper insight when given more opportunity to make mistakes, so maybe my approach is inferior. For example, students can use the approach I am using without knowing which electrode is the anode and which is the cathode.
answered Apr 13 at 17:00
Karsten TheisKarsten Theis
6,3961048
$endgroup$
I still teach switching the sign. I find it easier to remember adding up reduction potential and oxidation potential. The half reactions are written as reduction in a table of reduction potentials, so it makes sense that you have to treat the oxidation half reaction differently.
If the cell potential is calculated from reduction potential of the cathode half reaction minus that of the anode half reaction, there are more points of possible errors, including trouble with arithmetic.
You might argue that students learn more and gain deeper insight when given more opportunity to make mistakes, so maybe my approach is inferior. For example, students can use the approach I am using without knowing which electrode is the anode and which is the cathode.
answered Apr 13 at 17:00
Karsten TheisKarsten Theis
6,3961048
answered Apr 13 at 17:00
Karsten TheisKarsten Theis
6,3961048
answered Apr 13 at 17:00
Karsten TheisKarsten Theis
6,3961048
answered Apr 13 at 17:00
Karsten TheisKarsten Theis
6,3961048
6,3961048
$begingroup$
Just curious - Does the text book you use also employ that technique for balancing redox reactions?
$endgroup$
– MaxW
Apr 13 at 17:29
2
$begingroup$
I use multiple textbooks, and my students might use other sources entirely. Of the textbooks I use, D. M. Hanson's Foundations of Chemistry uses $E_textred + E_textox$ while OpenStax Chemistry uses $E_textcathode - E_textanode$.
$endgroup$
– Karsten Theis
Apr 13 at 17:39
add a comment |
$begingroup$
Just curious - Does the text book you use also employ that technique for balancing redox reactions?
$endgroup$
– MaxW
Apr 13 at 17:29
2
$begingroup$
I use multiple textbooks, and my students might use other sources entirely. Of the textbooks I use, D. M. Hanson's Foundations of Chemistry uses $E_textred + E_textox$ while OpenStax Chemistry uses $E_textcathode - E_textanode$.
$endgroup$
– Karsten Theis
Apr 13 at 17:39
$begingroup$
Just curious - Does the text book you use also employ that technique for balancing redox reactions?
$endgroup$
– MaxW
Apr 13 at 17:29
$begingroup$
Just curious - Does the text book you use also employ that technique for balancing redox reactions?
$endgroup$
– MaxW
Apr 13 at 17:29
2
2
$begingroup$
I use multiple textbooks, and my students might use other sources entirely. Of the textbooks I use, D. M. Hanson's Foundations of Chemistry uses $E_textred + E_textox$ while OpenStax Chemistry uses $E_textcathode - E_textanode$.
$endgroup$
– Karsten Theis
Apr 13 at 17:39
$begingroup$
I use multiple textbooks, and my students might use other sources entirely. Of the textbooks I use, D. M. Hanson's Foundations of Chemistry uses $E_textred + E_textox$ while OpenStax Chemistry uses $E_textcathode - E_textanode$.
$endgroup$
– Karsten Theis
Apr 13 at 17:39
add a comment |
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$begingroup$
It looks like silver-silver cell would be even better than silver zinc cell, with the added value of perpetuum mobile. :-)
$endgroup$
– Poutnik
Apr 13 at 17:05
$begingroup$
For the standard silver cell vs a standard silver cell the EMF would be 0 and thus no current flow. Because of the Nernst equation the potential of a non-standard silver cell would vary if the $ceA_Ag+ ne 1.000 $.
$endgroup$
– MaxW
Apr 13 at 17:09
$begingroup$
I know it well. But it does not make sense to write the silver electrode potential like if silver had been better reduction agent than zinc.
$endgroup$
– Poutnik
Apr 13 at 17:12
$begingroup$
@Poutnik - Huh? I wrote the Ag reaction with an EMF as -0.799V as an oxidation not a reduction so that I could balance the equation.
$endgroup$
– MaxW
Apr 13 at 17:17
$begingroup$
The point is, electrode potential is for 2-way redox reaction equilibrium, holding this potential with respect to the standard hydrogen electrode. For $ceAg/Ag+$, it is positive, not negative. With the open voltage, the oxidation and reduction is ongoing with the same rate.
$endgroup$
– Poutnik
Apr 13 at 17:20