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Electrode potentials and electrochemical cells

Electrode potentials and electrochemical cells. An electrochemical cell. Torch bulb. Acidified potassium dichromate solution. A carbon electrode. Magnesium pencil sharpener.

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Electrode potentials and electrochemical cells

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  1. Electrode potentials and electrochemical cells

  2. An electrochemical cell Torch bulb Acidified potassium dichromate solution. A carbon electrode Magnesium pencil sharpener

  3. When the magnesium and carbon electrodes are dipped into the acidified dichromate solution, the bulb immediately glows.

  4. After the cell has been run for a time, the dichromate solution has darkened: Cr2O72– has been reduced to Cr3+. After Before

  5. Cr2O72- + 14H+ + 6e–→ 2Cr3+ + 7H2O The dichromate has gained electrons. Where did they come from? The magnesium: Mg → Mg2+ + 2e– We provided a wire for the electrons to move from the magnesium to the carbon electrode where reduction took place.

  6. In the previous system all the chemicals were in the one beaker, but it is possible to have the oxidation and reduction reactions in separate containers, as long as we provide: • a wire for the electrons to travel from one half-cell to the other • a salt bridge to allow ions to move from one half-cell to the other (otherwise the transfer of electrons would result in the beakers becoming charged).

  7. The voltage is zero because the circuit is not yet complete. Zn(s) Cu(s) 1 mol L-1 Cu2+(aq) 1 mol L-1 Zn2+(aq) The Zn2+/Zn, Cu2+/Cu system

  8. As soon as the salt bridge is added the voltmeter shows that the circuit is now complete. The cell voltage is 1.124 V The salt bridge is a U-tube filled with agar jelly saturated with potassium nitrate. Ions from the salt bridge enter the beakers to balance the charge caused by the transfer of electrons.

  9. The Cu2+/Cu cell is replaced with a smaller beaker and smaller electrode. The cell voltage remains at 1.124 V.

  10. Although the amount of copper in contact with the solution has been halved, and the volume of solution was halved, the voltage produced was unchanged. Cell potentials do not change with the amount of reagent used.

  11. The Cu2+ solution is replaced with a much more dilute solution. The cell voltage has changed. It is now 1.047 V The voltage produced depends on the concentration of the solution.

  12. A little of the original Cu2+ solution is heated. (We already know that the voltage is not affected by the amount of solution used or the size of the electrode). The voltage has changed. It is now 1.182 V. The voltage produced depends on the temperature of the solution.

  13. The Zn2+/Zn half-cell is combined with a Mg2+/Mg half-cell. The new voltage is 0.853 V.

  14. Finally, the Mg2+/Mg cell is combined with the original Cu2+/Cu cell. This voltage is 1.977 V.

  15. 0.853 V 1.124 V Mg2+/Mg Zn2+/Zn Cu2+/Cu 1.977 V Cell voltages (cell potentials) can be added up.

  16. To compare the oxidising and reducing power of the various half-cells, we measure the voltage of each one with respect to the H+/H2 half-cell. As demonstrated earlier, the cell voltage changes with temperature and concentration. In cells involving gases (such as the H+/H2 system) the gaseous equivalent of concentration — pressure — is also important. So the cell voltages are measured under standard temperature, pressure and concentration of reagents in each standard half-cell. Standard concentration = 1.0 mol L–1. Standard temperature = 25 °C (not 20 °C, which is the common room temperature in NZ, but 25 °C which is standard room temperature in the US). Standard pressure = 101.5 kPa (1 atmosphere)

  17. When we calculate theoretical cell voltages using E° values from data books, we assume these standard conditions. The voltages measured in real life are likely to be slightly different from these theoretical figures, because we seldom have solutions of exactly 1.0 mol L–1, our room temperature is seldom 25 °C, and we probably aren’t using platinum electrodes.

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