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The effect of a voltmeter on the electron flow of a voltaic cell
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Electrochemical CellElectrode potentials at interfaces?Cathode + Anode + Rechargeable batteryHalf cell method of voltage calculation in an electrochemical cellIn a galvanic cell where the two electrodes are in the same electrolyte solution, why do reduction and oxidation occur separately?Iron/Iron(III) Concentration CellSomething I don't understand about batteriesAre my intuitions about how batteries work right?Why does increasing the concentration of oxidizing agent decrease the E° in a galvanic cell?
$begingroup$
Due to a voltmeter’s strong resistance nature, how could electrons flow from anode to cathod? It is more reasonable to assume that a voltmeter would stop the redox reaction because the electrons could never reach the positive electrode. When I measure the potential energy between two electrodes with a voltmeter, is the reaction happening? If the reaction is not taking place, how come the potential difference exists?
redox
asked Apr 1 at 1:53
Avalo GuAvalo Gu
164
$endgroup$
add a comment |
$begingroup$
Due to a voltmeter’s strong resistance nature, how could electrons flow from anode to cathod? It is more reasonable to assume that a voltmeter would stop the redox reaction because the electrons could never reach the positive electrode. When I measure the potential energy between two electrodes with a voltmeter, is the reaction happening? If the reaction is not taking place, how come the potential difference exists?
redox
asked Apr 1 at 1:53
Avalo GuAvalo Gu
164
$endgroup$
1
$begingroup$
Hi and welcome to Chemistry SE! If the circuit is complete without the voltmeter, the reaction is happening. Even with the voltmeter (if I assume it to be a non ideal one), the reaction is still happening, (provided the circuit is complete). You just register a very small current, which does not necessarily imply a very small potential difference as there is also a high resistance.
$endgroup$
– Akari
Apr 1 at 2:21
add a comment |
$begingroup$
Due to a voltmeter’s strong resistance nature, how could electrons flow from anode to cathod? It is more reasonable to assume that a voltmeter would stop the redox reaction because the electrons could never reach the positive electrode. When I measure the potential energy between two electrodes with a voltmeter, is the reaction happening? If the reaction is not taking place, how come the potential difference exists?
redox
asked Apr 1 at 1:53
Avalo GuAvalo Gu
164
$endgroup$
Due to a voltmeter’s strong resistance nature, how could electrons flow from anode to cathod? It is more reasonable to assume that a voltmeter would stop the redox reaction because the electrons could never reach the positive electrode. When I measure the potential energy between two electrodes with a voltmeter, is the reaction happening? If the reaction is not taking place, how come the potential difference exists?
redox
redox
asked Apr 1 at 1:53
Avalo GuAvalo Gu
164
asked Apr 1 at 1:53
Avalo GuAvalo Gu
164
asked Apr 1 at 1:53
Avalo GuAvalo Gu
164
asked Apr 1 at 1:53
Avalo GuAvalo Gu
164
asked Apr 1 at 1:53
Avalo GuAvalo Gu
164
164
1
$begingroup$
Hi and welcome to Chemistry SE! If the circuit is complete without the voltmeter, the reaction is happening. Even with the voltmeter (if I assume it to be a non ideal one), the reaction is still happening, (provided the circuit is complete). You just register a very small current, which does not necessarily imply a very small potential difference as there is also a high resistance.
$endgroup$
– Akari
Apr 1 at 2:21
add a comment |
1
$begingroup$
Hi and welcome to Chemistry SE! If the circuit is complete without the voltmeter, the reaction is happening. Even with the voltmeter (if I assume it to be a non ideal one), the reaction is still happening, (provided the circuit is complete). You just register a very small current, which does not necessarily imply a very small potential difference as there is also a high resistance.
$endgroup$
– Akari
Apr 1 at 2:21
1
1
$begingroup$
Hi and welcome to Chemistry SE! If the circuit is complete without the voltmeter, the reaction is happening. Even with the voltmeter (if I assume it to be a non ideal one), the reaction is still happening, (provided the circuit is complete). You just register a very small current, which does not necessarily imply a very small potential difference as there is also a high resistance.
$endgroup$
– Akari
Apr 1 at 2:21
$begingroup$
Hi and welcome to Chemistry SE! If the circuit is complete without the voltmeter, the reaction is happening. Even with the voltmeter (if I assume it to be a non ideal one), the reaction is still happening, (provided the circuit is complete). You just register a very small current, which does not necessarily imply a very small potential difference as there is also a high resistance.
$endgroup$
– Akari
Apr 1 at 2:21
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, the reaction must be taking place for the voltmeter to measure the difference in potential. Remember
$$V = iR$$
thus a voltmeter works by knowing its internal resistance, $R$, and measuring $i$. Also notice that if $i=0$ then $V = 0$, so a voltmeter must have at least some tiny current to work.
The problem with a voltmeter is that you can imagine that a battery has an internal resistance. So the more current flowing through the battery the less voltage will be measured by the voltmeter. Thus the cell voltage measured by a voltmeter would be low, at least theoretically.
$$V_textmeasured = iR_textmeter - iR_textbattery = i(R_textmeter - R_textbattery)$$
Modern voltmeters with integrated circuits and operation amplifiers have a very large internal resistance, and thus draw a very very small current since $R_textmeter gg R_textbattery$ so:
$$V_textmeasured = i(R_textmeter - R_textbattery) approx iR_textmeter$$
In the not to distance past voltmeters drew too much current to give an accurate reading for a cell, so a Wheatstone bridge was used. But even though the galvanometer of the Wheatstone bridge is used to detect "no" current, it still has some finite sensitivity, so the ideal of absolutely no current flow is certainly not reached. The other weakness of a Wheatstone bridge is the need for high precision resisters. So to measure the galvanic cell potential a Wheatstone bridge no longer has any practical usefulness.
answered Apr 1 at 6:56
MaxWMaxW
15.8k22261
$endgroup$
$begingroup$
I ran my voltaic cell with a voltemeter attached for more than one hour, the weight of electrode did not change before and after the reaction. Is such result also due to the negligible amount of current passing through?
$endgroup$
– Avalo Gu
Apr 3 at 0:32
$begingroup$
Also, what do you think about M.Farooq’s answer? His answer makes a lot of sense as well because electrode potential does exist between the metal strip and its relevent sulfate solution.
$endgroup$
– Avalo Gu
Apr 3 at 1:26
$begingroup$
@AvaloGu - His first sentence (An ideal voltmeter has an infinite resistance, so does not let any current flow in the system.) is just wrong. Current must flow to measure voltage.
$endgroup$
– MaxW
Apr 3 at 5:00
$begingroup$
@AvaloGu - Regarding your first comment. Yes a modern voltmeter has megaohms of resistance. So the electrode weight difference would be minuscule.
$endgroup$
– MaxW
Apr 3 at 5:03
add a comment |
$begingroup$
An ideal voltmeter has an infinite resistance, so does not let any current flow in the system. However nothing is ideal in real life, it consumes a very small amount of current to move the magnet and needle arrangement or electronics. An ideal ammeter has zero resistance. Assume that you have a water tank on your ceiling which supplies water to your kitchen tap. If I add a pressure gauge before your closed kitchen tap, it will read a positive pressure, much greater than the atmospheric pressure. Note that there is no water flow. Forget about the textbook picture of measuring galvanic cell voltages using a voltmeter (as shown in chemistry textbooks), as nobody every measures cell potential this way. A reliable approach to measure cell potential is called null-point detection (you can search more about it). As the name indicates, it draws zero current at the point of balance condition. It is one of the most precise and accurate techniques in classical potentiometry.
Just think a little deeper, the two electrodes in a galvanic cell do not "know" each other and they are oblivious to each other's presence. So current flow is not a requirement to measure potential difference. Each single electrode is in contact with its ion in the solution. If you have a zinc and copper galvanic cell, each has its own electrode potential e.g. a Zn rod dips in Zn2+ soln and Cu rod dips in Cu2+ solution. Electrode potential is the thermodynamic tendency of ions (in solution) to reduce themselves without any current flow, in this case Zn2+ --> Zn and Cu2+ --> Cu. The voltmeter simply reads the difference of potential between zinc and copper, however it does not know the individual electrode potential of zinc and copper.
answered Apr 1 at 4:30
M. FarooqM. Farooq
1,755111
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
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$begingroup$
Yes, the reaction must be taking place for the voltmeter to measure the difference in potential. Remember
$$V = iR$$
thus a voltmeter works by knowing its internal resistance, $R$, and measuring $i$. Also notice that if $i=0$ then $V = 0$, so a voltmeter must have at least some tiny current to work.
The problem with a voltmeter is that you can imagine that a battery has an internal resistance. So the more current flowing through the battery the less voltage will be measured by the voltmeter. Thus the cell voltage measured by a voltmeter would be low, at least theoretically.
$$V_textmeasured = iR_textmeter - iR_textbattery = i(R_textmeter - R_textbattery)$$
Modern voltmeters with integrated circuits and operation amplifiers have a very large internal resistance, and thus draw a very very small current since $R_textmeter gg R_textbattery$ so:
$$V_textmeasured = i(R_textmeter - R_textbattery) approx iR_textmeter$$
In the not to distance past voltmeters drew too much current to give an accurate reading for a cell, so a Wheatstone bridge was used. But even though the galvanometer of the Wheatstone bridge is used to detect "no" current, it still has some finite sensitivity, so the ideal of absolutely no current flow is certainly not reached. The other weakness of a Wheatstone bridge is the need for high precision resisters. So to measure the galvanic cell potential a Wheatstone bridge no longer has any practical usefulness.
answered Apr 1 at 6:56
MaxWMaxW
15.8k22261
$endgroup$
$begingroup$
I ran my voltaic cell with a voltemeter attached for more than one hour, the weight of electrode did not change before and after the reaction. Is such result also due to the negligible amount of current passing through?
$endgroup$
– Avalo Gu
Apr 3 at 0:32
$begingroup$
Also, what do you think about M.Farooq’s answer? His answer makes a lot of sense as well because electrode potential does exist between the metal strip and its relevent sulfate solution.
$endgroup$
– Avalo Gu
Apr 3 at 1:26
$begingroup$
@AvaloGu - His first sentence (An ideal voltmeter has an infinite resistance, so does not let any current flow in the system.) is just wrong. Current must flow to measure voltage.
$endgroup$
– MaxW
Apr 3 at 5:00
$begingroup$
@AvaloGu - Regarding your first comment. Yes a modern voltmeter has megaohms of resistance. So the electrode weight difference would be minuscule.
$endgroup$
– MaxW
Apr 3 at 5:03
add a comment |
$begingroup$
Yes, the reaction must be taking place for the voltmeter to measure the difference in potential. Remember
$$V = iR$$
thus a voltmeter works by knowing its internal resistance, $R$, and measuring $i$. Also notice that if $i=0$ then $V = 0$, so a voltmeter must have at least some tiny current to work.
The problem with a voltmeter is that you can imagine that a battery has an internal resistance. So the more current flowing through the battery the less voltage will be measured by the voltmeter. Thus the cell voltage measured by a voltmeter would be low, at least theoretically.
$$V_textmeasured = iR_textmeter - iR_textbattery = i(R_textmeter - R_textbattery)$$
Modern voltmeters with integrated circuits and operation amplifiers have a very large internal resistance, and thus draw a very very small current since $R_textmeter gg R_textbattery$ so:
$$V_textmeasured = i(R_textmeter - R_textbattery) approx iR_textmeter$$
In the not to distance past voltmeters drew too much current to give an accurate reading for a cell, so a Wheatstone bridge was used. But even though the galvanometer of the Wheatstone bridge is used to detect "no" current, it still has some finite sensitivity, so the ideal of absolutely no current flow is certainly not reached. The other weakness of a Wheatstone bridge is the need for high precision resisters. So to measure the galvanic cell potential a Wheatstone bridge no longer has any practical usefulness.
answered Apr 1 at 6:56
MaxWMaxW
15.8k22261
$endgroup$
$begingroup$
I ran my voltaic cell with a voltemeter attached for more than one hour, the weight of electrode did not change before and after the reaction. Is such result also due to the negligible amount of current passing through?
$endgroup$
– Avalo Gu
Apr 3 at 0:32
$begingroup$
Also, what do you think about M.Farooq’s answer? His answer makes a lot of sense as well because electrode potential does exist between the metal strip and its relevent sulfate solution.
$endgroup$
– Avalo Gu
Apr 3 at 1:26
$begingroup$
@AvaloGu - His first sentence (An ideal voltmeter has an infinite resistance, so does not let any current flow in the system.) is just wrong. Current must flow to measure voltage.
$endgroup$
– MaxW
Apr 3 at 5:00
$begingroup$
@AvaloGu - Regarding your first comment. Yes a modern voltmeter has megaohms of resistance. So the electrode weight difference would be minuscule.
$endgroup$
– MaxW
Apr 3 at 5:03
add a comment |
$begingroup$
Yes, the reaction must be taking place for the voltmeter to measure the difference in potential. Remember
$$V = iR$$
thus a voltmeter works by knowing its internal resistance, $R$, and measuring $i$. Also notice that if $i=0$ then $V = 0$, so a voltmeter must have at least some tiny current to work.
The problem with a voltmeter is that you can imagine that a battery has an internal resistance. So the more current flowing through the battery the less voltage will be measured by the voltmeter. Thus the cell voltage measured by a voltmeter would be low, at least theoretically.
$$V_textmeasured = iR_textmeter - iR_textbattery = i(R_textmeter - R_textbattery)$$
Modern voltmeters with integrated circuits and operation amplifiers have a very large internal resistance, and thus draw a very very small current since $R_textmeter gg R_textbattery$ so:
$$V_textmeasured = i(R_textmeter - R_textbattery) approx iR_textmeter$$
In the not to distance past voltmeters drew too much current to give an accurate reading for a cell, so a Wheatstone bridge was used. But even though the galvanometer of the Wheatstone bridge is used to detect "no" current, it still has some finite sensitivity, so the ideal of absolutely no current flow is certainly not reached. The other weakness of a Wheatstone bridge is the need for high precision resisters. So to measure the galvanic cell potential a Wheatstone bridge no longer has any practical usefulness.
answered Apr 1 at 6:56
MaxWMaxW
15.8k22261
$endgroup$
Yes, the reaction must be taking place for the voltmeter to measure the difference in potential. Remember
$$V = iR$$
thus a voltmeter works by knowing its internal resistance, $R$, and measuring $i$. Also notice that if $i=0$ then $V = 0$, so a voltmeter must have at least some tiny current to work.
The problem with a voltmeter is that you can imagine that a battery has an internal resistance. So the more current flowing through the battery the less voltage will be measured by the voltmeter. Thus the cell voltage measured by a voltmeter would be low, at least theoretically.
$$V_textmeasured = iR_textmeter - iR_textbattery = i(R_textmeter - R_textbattery)$$
Modern voltmeters with integrated circuits and operation amplifiers have a very large internal resistance, and thus draw a very very small current since $R_textmeter gg R_textbattery$ so:
$$V_textmeasured = i(R_textmeter - R_textbattery) approx iR_textmeter$$
In the not to distance past voltmeters drew too much current to give an accurate reading for a cell, so a Wheatstone bridge was used. But even though the galvanometer of the Wheatstone bridge is used to detect "no" current, it still has some finite sensitivity, so the ideal of absolutely no current flow is certainly not reached. The other weakness of a Wheatstone bridge is the need for high precision resisters. So to measure the galvanic cell potential a Wheatstone bridge no longer has any practical usefulness.
answered Apr 1 at 6:56
MaxWMaxW
15.8k22261
edited Apr 3 at 6:48
answered Apr 1 at 6:56
MaxWMaxW
15.8k22261
answered Apr 1 at 6:56
MaxWMaxW
15.8k22261
answered Apr 1 at 6:56
MaxWMaxW
15.8k22261
15.8k22261
$begingroup$
I ran my voltaic cell with a voltemeter attached for more than one hour, the weight of electrode did not change before and after the reaction. Is such result also due to the negligible amount of current passing through?
$endgroup$
– Avalo Gu
Apr 3 at 0:32
$begingroup$
Also, what do you think about M.Farooq’s answer? His answer makes a lot of sense as well because electrode potential does exist between the metal strip and its relevent sulfate solution.
$endgroup$
– Avalo Gu
Apr 3 at 1:26
$begingroup$
@AvaloGu - His first sentence (An ideal voltmeter has an infinite resistance, so does not let any current flow in the system.) is just wrong. Current must flow to measure voltage.
$endgroup$
– MaxW
Apr 3 at 5:00
$begingroup$
@AvaloGu - Regarding your first comment. Yes a modern voltmeter has megaohms of resistance. So the electrode weight difference would be minuscule.
$endgroup$
– MaxW
Apr 3 at 5:03
add a comment |
$begingroup$
I ran my voltaic cell with a voltemeter attached for more than one hour, the weight of electrode did not change before and after the reaction. Is such result also due to the negligible amount of current passing through?
$endgroup$
– Avalo Gu
Apr 3 at 0:32
$begingroup$
Also, what do you think about M.Farooq’s answer? His answer makes a lot of sense as well because electrode potential does exist between the metal strip and its relevent sulfate solution.
$endgroup$
– Avalo Gu
Apr 3 at 1:26
$begingroup$
@AvaloGu - His first sentence (An ideal voltmeter has an infinite resistance, so does not let any current flow in the system.) is just wrong. Current must flow to measure voltage.
$endgroup$
– MaxW
Apr 3 at 5:00
$begingroup$
@AvaloGu - Regarding your first comment. Yes a modern voltmeter has megaohms of resistance. So the electrode weight difference would be minuscule.
$endgroup$
– MaxW
Apr 3 at 5:03
$begingroup$
I ran my voltaic cell with a voltemeter attached for more than one hour, the weight of electrode did not change before and after the reaction. Is such result also due to the negligible amount of current passing through?
$endgroup$
– Avalo Gu
Apr 3 at 0:32
$begingroup$
I ran my voltaic cell with a voltemeter attached for more than one hour, the weight of electrode did not change before and after the reaction. Is such result also due to the negligible amount of current passing through?
$endgroup$
– Avalo Gu
Apr 3 at 0:32
$begingroup$
Also, what do you think about M.Farooq’s answer? His answer makes a lot of sense as well because electrode potential does exist between the metal strip and its relevent sulfate solution.
$endgroup$
– Avalo Gu
Apr 3 at 1:26
$begingroup$
Also, what do you think about M.Farooq’s answer? His answer makes a lot of sense as well because electrode potential does exist between the metal strip and its relevent sulfate solution.
$endgroup$
– Avalo Gu
Apr 3 at 1:26
$begingroup$
@AvaloGu - His first sentence (An ideal voltmeter has an infinite resistance, so does not let any current flow in the system.) is just wrong. Current must flow to measure voltage.
$endgroup$
– MaxW
Apr 3 at 5:00
$begingroup$
@AvaloGu - His first sentence (An ideal voltmeter has an infinite resistance, so does not let any current flow in the system.) is just wrong. Current must flow to measure voltage.
$endgroup$
– MaxW
Apr 3 at 5:00
$begingroup$
@AvaloGu - Regarding your first comment. Yes a modern voltmeter has megaohms of resistance. So the electrode weight difference would be minuscule.
$endgroup$
– MaxW
Apr 3 at 5:03
$begingroup$
@AvaloGu - Regarding your first comment. Yes a modern voltmeter has megaohms of resistance. So the electrode weight difference would be minuscule.
$endgroup$
– MaxW
Apr 3 at 5:03
add a comment |
$begingroup$
An ideal voltmeter has an infinite resistance, so does not let any current flow in the system. However nothing is ideal in real life, it consumes a very small amount of current to move the magnet and needle arrangement or electronics. An ideal ammeter has zero resistance. Assume that you have a water tank on your ceiling which supplies water to your kitchen tap. If I add a pressure gauge before your closed kitchen tap, it will read a positive pressure, much greater than the atmospheric pressure. Note that there is no water flow. Forget about the textbook picture of measuring galvanic cell voltages using a voltmeter (as shown in chemistry textbooks), as nobody every measures cell potential this way. A reliable approach to measure cell potential is called null-point detection (you can search more about it). As the name indicates, it draws zero current at the point of balance condition. It is one of the most precise and accurate techniques in classical potentiometry.
Just think a little deeper, the two electrodes in a galvanic cell do not "know" each other and they are oblivious to each other's presence. So current flow is not a requirement to measure potential difference. Each single electrode is in contact with its ion in the solution. If you have a zinc and copper galvanic cell, each has its own electrode potential e.g. a Zn rod dips in Zn2+ soln and Cu rod dips in Cu2+ solution. Electrode potential is the thermodynamic tendency of ions (in solution) to reduce themselves without any current flow, in this case Zn2+ --> Zn and Cu2+ --> Cu. The voltmeter simply reads the difference of potential between zinc and copper, however it does not know the individual electrode potential of zinc and copper.
answered Apr 1 at 4:30
M. FarooqM. Farooq
1,755111
$endgroup$
add a comment |
$begingroup$
An ideal voltmeter has an infinite resistance, so does not let any current flow in the system. However nothing is ideal in real life, it consumes a very small amount of current to move the magnet and needle arrangement or electronics. An ideal ammeter has zero resistance. Assume that you have a water tank on your ceiling which supplies water to your kitchen tap. If I add a pressure gauge before your closed kitchen tap, it will read a positive pressure, much greater than the atmospheric pressure. Note that there is no water flow. Forget about the textbook picture of measuring galvanic cell voltages using a voltmeter (as shown in chemistry textbooks), as nobody every measures cell potential this way. A reliable approach to measure cell potential is called null-point detection (you can search more about it). As the name indicates, it draws zero current at the point of balance condition. It is one of the most precise and accurate techniques in classical potentiometry.
Just think a little deeper, the two electrodes in a galvanic cell do not "know" each other and they are oblivious to each other's presence. So current flow is not a requirement to measure potential difference. Each single electrode is in contact with its ion in the solution. If you have a zinc and copper galvanic cell, each has its own electrode potential e.g. a Zn rod dips in Zn2+ soln and Cu rod dips in Cu2+ solution. Electrode potential is the thermodynamic tendency of ions (in solution) to reduce themselves without any current flow, in this case Zn2+ --> Zn and Cu2+ --> Cu. The voltmeter simply reads the difference of potential between zinc and copper, however it does not know the individual electrode potential of zinc and copper.
answered Apr 1 at 4:30
M. FarooqM. Farooq
1,755111
$endgroup$
add a comment |
$begingroup$
An ideal voltmeter has an infinite resistance, so does not let any current flow in the system. However nothing is ideal in real life, it consumes a very small amount of current to move the magnet and needle arrangement or electronics. An ideal ammeter has zero resistance. Assume that you have a water tank on your ceiling which supplies water to your kitchen tap. If I add a pressure gauge before your closed kitchen tap, it will read a positive pressure, much greater than the atmospheric pressure. Note that there is no water flow. Forget about the textbook picture of measuring galvanic cell voltages using a voltmeter (as shown in chemistry textbooks), as nobody every measures cell potential this way. A reliable approach to measure cell potential is called null-point detection (you can search more about it). As the name indicates, it draws zero current at the point of balance condition. It is one of the most precise and accurate techniques in classical potentiometry.
Just think a little deeper, the two electrodes in a galvanic cell do not "know" each other and they are oblivious to each other's presence. So current flow is not a requirement to measure potential difference. Each single electrode is in contact with its ion in the solution. If you have a zinc and copper galvanic cell, each has its own electrode potential e.g. a Zn rod dips in Zn2+ soln and Cu rod dips in Cu2+ solution. Electrode potential is the thermodynamic tendency of ions (in solution) to reduce themselves without any current flow, in this case Zn2+ --> Zn and Cu2+ --> Cu. The voltmeter simply reads the difference of potential between zinc and copper, however it does not know the individual electrode potential of zinc and copper.
answered Apr 1 at 4:30
M. FarooqM. Farooq
1,755111
$endgroup$
An ideal voltmeter has an infinite resistance, so does not let any current flow in the system. However nothing is ideal in real life, it consumes a very small amount of current to move the magnet and needle arrangement or electronics. An ideal ammeter has zero resistance. Assume that you have a water tank on your ceiling which supplies water to your kitchen tap. If I add a pressure gauge before your closed kitchen tap, it will read a positive pressure, much greater than the atmospheric pressure. Note that there is no water flow. Forget about the textbook picture of measuring galvanic cell voltages using a voltmeter (as shown in chemistry textbooks), as nobody every measures cell potential this way. A reliable approach to measure cell potential is called null-point detection (you can search more about it). As the name indicates, it draws zero current at the point of balance condition. It is one of the most precise and accurate techniques in classical potentiometry.
Just think a little deeper, the two electrodes in a galvanic cell do not "know" each other and they are oblivious to each other's presence. So current flow is not a requirement to measure potential difference. Each single electrode is in contact with its ion in the solution. If you have a zinc and copper galvanic cell, each has its own electrode potential e.g. a Zn rod dips in Zn2+ soln and Cu rod dips in Cu2+ solution. Electrode potential is the thermodynamic tendency of ions (in solution) to reduce themselves without any current flow, in this case Zn2+ --> Zn and Cu2+ --> Cu. The voltmeter simply reads the difference of potential between zinc and copper, however it does not know the individual electrode potential of zinc and copper.
answered Apr 1 at 4:30
M. FarooqM. Farooq
1,755111
edited Apr 1 at 13:52
answered Apr 1 at 4:30
M. FarooqM. Farooq
1,755111
answered Apr 1 at 4:30
M. FarooqM. Farooq
1,755111
answered Apr 1 at 4:30
M. FarooqM. Farooq
1,755111
1,755111
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Hi and welcome to Chemistry SE! If the circuit is complete without the voltmeter, the reaction is happening. Even with the voltmeter (if I assume it to be a non ideal one), the reaction is still happening, (provided the circuit is complete). You just register a very small current, which does not necessarily imply a very small potential difference as there is also a high resistance.
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– Akari
Apr 1 at 2:21