I know this stack exchange may not be the most appropriate to ask this question and it is better to ask it at chemistry stack exchange. However, I got no satisfying answers there so I figured I'll just ask it here

If we look at system at constant temperature and volume which is galvanic cell (battery), first law of thermodynamics states: $$ dU = dQ + dW'$$

Where W' is electrical work exchanged between galvanic cell and surroundings and Q is heat exchanged with surroundings.

I am using chemistry sign convention for work ,which can be seen in the way first law is written, that is work is positive if surroundings does it on system. In our example, electrical work is going to be negative because we have galvanic cell (battery) which by definition does electrical work on surroundings.

We know that adding heat to the system increases internal energy of the system because it increases mostly average kinetic energy of the molecules. It can also affect average potential energy of interaction (intermolecular and chemical bonds).

What about electrical work in context of galvanic cells? Electrical work is work done by electric field when charge moves certain potential difference. What does that have to do with internal energy changes in context of galvanic cells? With heat exchanged, I do understand how it affects internal energy, but with electrical work I am less sure.

For sake of simplicity we look at galvanic cell working reversibly, so there aren't any losses due to irreversibility. In what way does galvanic cell give work to the surroundings if there is no some electric motor which converts electrical energy to mechanical since electrical work does exist even if there is no any motor which uses electrical energy or current to produce some other form of energy? In case of electric motor it is clear how electrical work is done on surroundings, but without it I don't understand how is it given to surroundings?


2 Answers 2


You are asking two different questions: 1. How does the internal energy in a galvanic cell change? and 2. How does a galvanic cell do work on its surroundings?

Consider the reaction $$Zn+Cu^{2+}\rightarrow Cu+Zn^{2+}$$in a Daniell cell. Do you think that the internal energy change for 1 mole of pure zinc (s) plus a 1 molal aqueous solution of CuSO4 going to 1 mole of pure copper (s) plus a 1 molal aqueous solution ZnSO4 is zero?

In the case of electrical work, consider a resistor that current flows through. If 1 mole of electrons flow through the resistor, EI work is done, after which the resistor cools to room temperature. So the change in internal energy of the resistor is zero, and the work done to force the electrons through the resistor is equal to the heat given off to the room.

  • $\begingroup$ 1) I am not sure will internal energy change because of electron transfer between electrodes and species participating in reaction (I know it should due to first law, but I don't understand it) I know that internal energy is sum of microscopic kinetic and potential energy and during electrochemical reaction, there is transfer of electrons between species through external circuit where electrons flow due to potential difference (redox reaction). I am not sure how does electron transfer influence microscopic kinetic and potential energy of the system? $\endgroup$ Commented Jun 3, 2021 at 11:35
  • $\begingroup$ 2) Yes, this is a good explanation for electrical work generally. However, in my question I asked how is electrical work given to surroundings in context of galvanic cells working REVERSIBLY. Explanation you gave is good, but it isn't applicable to my example because galvanic cell works reversibly, so there is no joule heating and other irreversible losses. $\endgroup$ Commented Jun 3, 2021 at 11:41
  • $\begingroup$ 2) If the cell is what you define as your system, then it can definitely experience a reversible change while the resistor, as part of the surroundings, experiences an irreversible change. Have you never heard of "internally reversible processes?" $\endgroup$ Commented Jun 3, 2021 at 11:50
  • $\begingroup$ 1) It is much more than electron transfer. The zinc goes from a solid to ions in an aqueous solution (including non-ideal mixing with the water molecules) while the opposite happens to the copper. $\endgroup$ Commented Jun 3, 2021 at 11:53
  • $\begingroup$ 2) I didn't, but I did check it out. Yes, if we define system as galvanic cell, it can change reversibly between two states if there are no irreversible processes happening inside a cell. I am not sure what irreversible processes can happen inside a cell? If we balance cell EMF by outer voltage and change outer voltage infinitesimally between two states so that current tends to zero which is reversible operation of a cell than there is no joule heating of a resistor and we also have external reversibility if I did understand this correctly. Also, heat transfer can cause external irrever. $\endgroup$ Commented Jun 5, 2021 at 16:26

What about electrical work in context of galvanic cells? Electrical work is work done by electric field when charge moves certain potential difference. What does that have to do with internal energy changes in context of galvanic cells?

It might be helpful to consider the change in internal energy as a consequence of the combination of the following:

  1. The galvanic cell first converts chemical potential energy to electrical potential energy within the cell. For the Daniell cell the electrical potential energy is, I understand, 1.108 Joules per coulomb of charge (1.108 volts between the terminals with no load). At this stage there is no change in the internal energy of the cell, just a conversion of chemical potential energy to electrical potential energy.

  2. When the cell is connected to a circuit, the cell produces an electric field in the circuit exerting a force on the mobile electrons in the circuit doing work moving the charge through the circuit. The electrical potential energy of the cell is transferred to the circuit where it is either dissipated as resistance heating or stored in the electric and magnetic fields of capacitance and inductance, where it can be converted to other forms of work.

Taken together, the overall effect is when the cell does work moving charge through a circuit, there is a decrease in the chemical potential energy of the cell, and thus a decrease in its internal energy.

With regard to galvanic cells working reversibly I don't see how that is relevant with respect to its change in internal energy. Only conservation of energy is relevant. If it works reversibly (no internal losses) more energy is available to do work than if it doesn't work reversibly. The change in internal energy only depends on the work done on the surroundings.

Hope this helps;.

  • $\begingroup$ Yes, thank you for a good answer. If we have only resistor in a circuit than when charges move certain potential difference between electrodes, their electric potential energy is converted to kinetic energy which is than lost into heat due to collisions with atoms in a metal lattice. $\endgroup$ Commented Jun 21, 2021 at 15:10
  • $\begingroup$ It seems that all available potential energy is turned into heat since joule heat is equal U*I which is basically electric power, so charges don't gain kinetic energy when moving through circuit. So, cell losses energy to surroundings because joule heat generated is eventually given to surroundings which causes decrease in internal energy and this is one example how electrical work causes decrease in internal energy in galvanic cells. Did I explain/understand this correctly? $\endgroup$ Commented Jun 21, 2021 at 15:12
  • $\begingroup$ @DarioMirić Yes you did. For a purely resistive circuit, the internal energy of the cell is lost at a rate equal to the power dissipated in the resistance (Joule heating). $\endgroup$
    – Bob D
    Commented Jun 21, 2021 at 15:25
  • $\begingroup$ @DarioMirić I would add to my answer that there would also be some loss of internal energy of the battery due to heat transfer directly from the battery case to its surroundings. That's because all real batteries have some internal resistance in which power is dissipated. That raises the temperature of the battery above the surroundings and heat transfer to the surroundings. These heat "losses" of internal energy should be much less than that due to work the battery does on the charge. $\endgroup$
    – Bob D
    Commented Jun 21, 2021 at 16:32
  • $\begingroup$ Yes, if internal resistance is small compared to outer resistance which should be in well designed battery or cell. $\endgroup$ Commented Jun 21, 2021 at 20:00

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