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If I connect a zinc half cell (Left hand side) and a copper half cell (right hand side), I get an E cell value of 1.1V taken from the voltmeter. If I replace the voltmeter with a light bulb, I will get light and heat energy of the bulb.

My question: Will the voltage across the cell remain the same? I mean, the potential difference arises due to the difference in the amount of charge build up on the electrodes... so an electron flows from the zinc electrode to the copper electrode, will the voltage across the bulb decrease?

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The electrons are generated by a redox reaction. The zinc metal is oxidised to form a solution of Zn$^{2+}$ ions, and Cu$^{2+}$ ions are reduced to form metallic copper.

$$ Zn \rightarrow Zn^{2+} + 2e $$

$$ Cu^{2+} + 2e \rightarrow Cu $$

So as long as some metallic zinc remains at the anode the voltage will stay the same. Once all the metallic zinc has been dissolved your cell will stop producing a potential difference.

However the voltage will fall when you connect a load, like a light bulb, because the cell has a non-zero internal resistance. If the internal resistance is $R_{int}$ then when you draw a current $I$ the measured voltage will fall by $IR_{int}$.

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  • $\begingroup$ So, if I connect a light bulb, there will be a constant voltage across it but the value will be (V-IRint)? $\endgroup$
    – Eliza
    Commented Jan 2, 2014 at 13:11
  • $\begingroup$ Yes. The voltage will remain constant until all the zinc has dissolved. $\endgroup$
    – John Rennie
    Commented Jan 2, 2014 at 13:48
  • $\begingroup$ Doesn't the potential difference arise due to the difference in the amount of charge build up on the electrodes? Then as the electrons flow from the zinc electrode and zinc is used up, won't the charge on the zinc electrode decrease... and the potential difference should change... I am still confused how the constant voltage arises $\endgroup$
    – Eliza
    Commented Jan 2, 2014 at 14:00
  • $\begingroup$ @Eliza: there is no charge buildup. At the cathode metallic Zn is oxidised to Zn$^{2+}$ ions and releases 2 electrons per zinc atom. These electrons flow out of the battery through your bulb and to the anode. There they react with a Cu$^{2+}$ ion and reduce it to form metallic copper. So electrons are produced at the zinc cathode and consumed at the copper anode. The net effect is to dissolve the zinc cathode and precipitate copper at the anode. There is no buildup of electrons. $\endgroup$
    – John Rennie
    Commented Jan 2, 2014 at 16:45
  • $\begingroup$ then what causes a potential difference between the electrodes? (i.e. when I connect a voltmeter, there is a reading) $\endgroup$
    – Eliza
    Commented Jan 2, 2014 at 16:48
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The theoretical voltage is given by the Nernst Equation $$\Delta G=-nFE=-RT\ln K_\mathrm{eq}+RT\ln Q$$ E is the cell potential Keq is the equilibrium constant and Q is the reaction quotient(the ratio of the concentrations of the products raised to their stoichiometric coefficients divided by the concentration of the reactants raised to their stoichiometric coefficients) So yes the theoretical cell potential is more when Q is small(more reactant than product) and when Keq is large(equilibrium falls further in products). The concentrations of Cu2+ and Zn2+ thus play a part in the cell potential (we want higher reactant concentrations for higher potential). We note the cell potential does deviate from the theoretical cell potential and this phenomenon is known as overpotential.

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