The effect of a voltmeter on the electron flow of a voltaic cell












2












$begingroup$


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?










share|improve this question







New contributor




Avalo Gu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 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
















2












$begingroup$


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?










share|improve this question







New contributor




Avalo Gu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 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














2












2








2





$begingroup$


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?










share|improve this question







New contributor




Avalo Gu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




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






share|improve this question







New contributor




Avalo Gu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question







New contributor




Avalo Gu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question






New contributor




Avalo Gu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked Apr 1 at 1:53









Avalo GuAvalo Gu

163




163




New contributor




Avalo Gu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Avalo Gu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Avalo Gu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 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




    $begingroup$
    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










2 Answers
2






active

oldest

votes


















2












$begingroup$

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.



$$V_text{measured} = iR_text{meter} - iR_text{battery} = i(R_text{meter} - R_text{battery})$$



Modern voltmeters with integrated circuits and operation amplifiers have a very large internal resistance, and thus draw a very very small current since $R_text{meter} gg R_text{battery}$ so:



$$V_text{measured} = i(R_text{meter} - R_text{battery}) approx iR_text{meter}$$



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.






share|improve this answer











$endgroup$













  • $begingroup$
    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
    yesterday










  • $begingroup$
    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
    yesterday










  • $begingroup$
    @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
    yesterday










  • $begingroup$
    @AvaloGu - Regarding your first comment. Yes a modern voltmeter has megaohms of resistance. So the electrode weight difference would be minuscule.
    $endgroup$
    – MaxW
    yesterday



















2












$begingroup$

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.






share|improve this answer











$endgroup$














    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "431"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });






    Avalo Gu is a new contributor. Be nice, and check out our Code of Conduct.










    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f111907%2fthe-effect-of-a-voltmeter-on-the-electron-flow-of-a-voltaic-cell%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    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.



    $$V_text{measured} = iR_text{meter} - iR_text{battery} = i(R_text{meter} - R_text{battery})$$



    Modern voltmeters with integrated circuits and operation amplifiers have a very large internal resistance, and thus draw a very very small current since $R_text{meter} gg R_text{battery}$ so:



    $$V_text{measured} = i(R_text{meter} - R_text{battery}) approx iR_text{meter}$$



    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.






    share|improve this answer











    $endgroup$













    • $begingroup$
      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
      yesterday










    • $begingroup$
      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
      yesterday










    • $begingroup$
      @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
      yesterday










    • $begingroup$
      @AvaloGu - Regarding your first comment. Yes a modern voltmeter has megaohms of resistance. So the electrode weight difference would be minuscule.
      $endgroup$
      – MaxW
      yesterday
















    2












    $begingroup$

    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.



    $$V_text{measured} = iR_text{meter} - iR_text{battery} = i(R_text{meter} - R_text{battery})$$



    Modern voltmeters with integrated circuits and operation amplifiers have a very large internal resistance, and thus draw a very very small current since $R_text{meter} gg R_text{battery}$ so:



    $$V_text{measured} = i(R_text{meter} - R_text{battery}) approx iR_text{meter}$$



    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.






    share|improve this answer











    $endgroup$













    • $begingroup$
      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
      yesterday










    • $begingroup$
      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
      yesterday










    • $begingroup$
      @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
      yesterday










    • $begingroup$
      @AvaloGu - Regarding your first comment. Yes a modern voltmeter has megaohms of resistance. So the electrode weight difference would be minuscule.
      $endgroup$
      – MaxW
      yesterday














    2












    2








    2





    $begingroup$

    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.



    $$V_text{measured} = iR_text{meter} - iR_text{battery} = i(R_text{meter} - R_text{battery})$$



    Modern voltmeters with integrated circuits and operation amplifiers have a very large internal resistance, and thus draw a very very small current since $R_text{meter} gg R_text{battery}$ so:



    $$V_text{measured} = i(R_text{meter} - R_text{battery}) approx iR_text{meter}$$



    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.






    share|improve this answer











    $endgroup$



    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.



    $$V_text{measured} = iR_text{meter} - iR_text{battery} = i(R_text{meter} - R_text{battery})$$



    Modern voltmeters with integrated circuits and operation amplifiers have a very large internal resistance, and thus draw a very very small current since $R_text{meter} gg R_text{battery}$ so:



    $$V_text{measured} = i(R_text{meter} - R_text{battery}) approx iR_text{meter}$$



    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.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 23 hours ago

























    answered 2 days ago









    MaxWMaxW

    15.3k22261




    15.3k22261












    • $begingroup$
      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
      yesterday










    • $begingroup$
      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
      yesterday










    • $begingroup$
      @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
      yesterday










    • $begingroup$
      @AvaloGu - Regarding your first comment. Yes a modern voltmeter has megaohms of resistance. So the electrode weight difference would be minuscule.
      $endgroup$
      – MaxW
      yesterday


















    • $begingroup$
      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
      yesterday










    • $begingroup$
      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
      yesterday










    • $begingroup$
      @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
      yesterday










    • $begingroup$
      @AvaloGu - Regarding your first comment. Yes a modern voltmeter has megaohms of resistance. So the electrode weight difference would be minuscule.
      $endgroup$
      – MaxW
      yesterday
















    $begingroup$
    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
    yesterday




    $begingroup$
    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
    yesterday












    $begingroup$
    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
    yesterday




    $begingroup$
    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
    yesterday












    $begingroup$
    @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
    yesterday




    $begingroup$
    @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
    yesterday












    $begingroup$
    @AvaloGu - Regarding your first comment. Yes a modern voltmeter has megaohms of resistance. So the electrode weight difference would be minuscule.
    $endgroup$
    – MaxW
    yesterday




    $begingroup$
    @AvaloGu - Regarding your first comment. Yes a modern voltmeter has megaohms of resistance. So the electrode weight difference would be minuscule.
    $endgroup$
    – MaxW
    yesterday











    2












    $begingroup$

    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.






    share|improve this answer











    $endgroup$


















      2












      $begingroup$

      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.






      share|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        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.






        share|improve this answer











        $endgroup$



        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.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 2 days ago

























        answered Apr 1 at 4:30









        M. FarooqM. Farooq

        1,317110




        1,317110






















            Avalo Gu is a new contributor. Be nice, and check out our Code of Conduct.










            draft saved

            draft discarded


















            Avalo Gu is a new contributor. Be nice, and check out our Code of Conduct.













            Avalo Gu is a new contributor. Be nice, and check out our Code of Conduct.












            Avalo Gu is a new contributor. Be nice, and check out our Code of Conduct.
















            Thanks for contributing an answer to Chemistry Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f111907%2fthe-effect-of-a-voltmeter-on-the-electron-flow-of-a-voltaic-cell%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            How did Captain America manage to do this?

            迪纳利

            南乌拉尔铁路局