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I'm trying to figure out how batteries and voltage work in actuality.

Statements I assume to be true.

  1. At the negative pole of the battery a chemical reaction occurs releasing electrons that repel each other creating a electric force. Because of this Force electrons move into a conducting circuit. At the positive pole a chemical reaction occur which picks up the electrons.

  2. By convention the negative pole is said to have low potential and the positive pole to have high potential. The potential difference is called voltage with the negative side or ground said to be at 0 potential as a reference.

  3. The electrons at the negative pole create an E-field and when the negative pole is connected to a conductor the electrons enter the conductor and are distributed equally, if the conductor is connected to the positive pole(through a load) the electrons instead are consumed by that positive pole chemical reaction which in turn free up electrons at the minuspole so that the chemical reaction at the minus pole can continue releasing electrons.

  4. As the chemical agents are used up the rate of chemical reaction goes down and the battery gets weaker and then dies.

Are these statements accurate?

QUESTION 1. In Fig 2 in the image below, how does connecting batteries in series produce a higher voltage, meaning a higher potential difference? The connecting of the batteries does not produce more electrons at the minus pole connected to the conducting wire. Meaning point E electrons doesnt move to point C.

Is it related to magnetic fields? That when eĺectrons from D move to E those electrons in motion creates a magnetic field affecting the entire E field thus increaseing the potential difference?

QUESTION 2. In Fig 3 why doesn't electrons still go from one negative pole on one battery to the positive pole on the other?

Edit: I mean the electrons should still get consumed by the chemical reaction there continuing the process?

enter image description here

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    \$\begingroup\$ Does this answer your question? Can't for the life of me figure out voltage of batteries in series \$\endgroup\$
    – JRE
    Dec 29, 2020 at 19:40
  • \$\begingroup\$ For 2: notice that the electrons will flow from E to D and then cancel out. And then each battery only has half the p.d. that it wants, so it creates more \$\endgroup\$
    – user253751
    Dec 29, 2020 at 19:40
  • \$\begingroup\$ For 3: notice that the electrons will flow from C to F and then cancel out. And then each battery only has half the p.d. that it wants, so it creates more. But this time, the charge builds up on the inside. \$\endgroup\$
    – user253751
    Dec 29, 2020 at 19:41
  • \$\begingroup\$ electrons cannot be consumed by chemical reactions \$\endgroup\$
    – jsotola
    Dec 30, 2020 at 3:56

2 Answers 2

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how does connecting batteries in series produce a higher voltage, meaning a higher potential difference? The connecting of the batteries does not produce more electrons at the minus pole connected to the conducting wire. Meaning point E electrons doesnt move to point C.

They do, though. So long as the circuit is completed (by the "R" element), the electrons at E will flow into the left cell and be transported by the chemical processes insdide it to point C.

Okay, not literally the same electrons. But these electrons will flow into the left cell, combine with positive ions at the cathode, enabling the chemical reaction in the cell to proceed and produce more electrons at its anode.

In Fig 3 why doesn't electrons still go from one negative pole on one battery to the positive pole on the other?

Because it takes a voltage gradient of 1000's of volts per centimeter to get electrons to fly through the air, and here you have only 3 V difference. If you moved the two ends of the batteries within about 10 um of each other you might produce a spark allowing the electrons to flow from one to the other.

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  • \$\begingroup\$ In 3. I mean why doesn't current continue to flow through the wire at a voltage of 1.5V ? \$\endgroup\$
    – user272397
    Dec 29, 2020 at 20:00
  • \$\begingroup\$ "They do, though. So long as the circuit is completed (by the "R" element), the electrons at E will flow into the left cell and be transported by the chemical processes insdide it to point C." If that's true wouldn't the battery go on forever? If when electrons reach the positive pole they just return by an ion to the negatvie pole and again go through the wire what is being consumed in the process, depleting the battery? \$\endgroup\$
    – user272397
    Dec 29, 2020 at 20:14
  • \$\begingroup\$ @user272397 The chemicals involved in the reaction are depleted. \$\endgroup\$
    – Hearth
    Dec 29, 2020 at 20:36
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I might have figured it out. I studied galvanic cells a bit. So there is a chemical agent at the anode(negative) releasing electrons, producing positively charged ions, the reverse happens at the cathode(positive) reducing a chemical agent producing a neutral atom from the ion. In the battery there are also other ions both positive and negative that simply move from one end to the other of the battery to maintain the potential difference but doesn't release or pick up electrons, they simply maintain the potential "pressure" on the agents that can actually release electrons.

If I'm right I guess it turned out to be a chemistry question.

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