I made a battery like this.


By piling up 6 battery of this(actually I made this using 10 yen and 1 yen coins:10yen coin is Cu and 1 yen coin is Al.) I could turn on an LED light. The plausible reactions taking place at the cathode and the anode are as follows.

Anode: $\ce{Al ->Al^3+ + 3e^-}$
Cathode:$\ce{Cu^2+ +2e^- -> Cu}$ and $\ce{2H^+ + 2e^- -> H2}$ enter image description here

I have an question here. Why doesn't all the electrons released from Al reduce $\ce{Cu^2+}$ or $\ce{H^+}$ immediately at the anode site?

This url shows the battery on which mine based. But I used paper wetted with HCl which contains CuSO4 , not NaCl saturated water like shown in the URL and picture above.

  • 1
    $\begingroup$ Be aware that attempts of theoretical evaluation of voltage and potentials of galvanic cells involving Al anodes is opening the Pandora's box . $\endgroup$
    – Poutnik
    Commented May 13, 2023 at 13:54
  • $\begingroup$ Could you draw a diagram of your cell and battery (collection of cells)? Also "10 yes and 1 yen coins" is not clear. What do you mean by this? $\endgroup$
    – ananta
    Commented May 13, 2023 at 13:54
  • $\begingroup$ And aluminum metal reacts with both hydrochloric acid and copper sulfate solution. Your cell notation does not show any salt bridge, so I am guessing you used the different coins and suitably wetted paper or cardboard dividers. $\endgroup$
    – Ed V
    Commented May 13, 2023 at 13:56
  • 1
    $\begingroup$ I assume 10 yen coins are from copper and 1 yen coins from aluminium. // Some electrons released by anodic Al reduce H+ to H2 even directly at the anode. $\endgroup$
    – Poutnik
    Commented May 13, 2023 at 13:56
  • 1
    $\begingroup$ I originally missed that there is a single electrolyte and cells have therefore wrong design, allowing Al to react with H+ and Cu^2+ directly, regardless being part of the galvanic cell settings. $\endgroup$
    – Poutnik
    Commented May 14, 2023 at 4:11


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