barry68v10
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From VT engineering dept. "The electrical force produced by an electrochemical cell is measured by the cell voltage. Cell voltage depends on the redox reactions occuring in the cell and the concentration of the reactants, but not on the number of electrons passing through the cell: it is an intensive property. The standard voltage E0 of a cell is measured when all reactants are at 25oC and at 1M concentration or 1 atm pressure.
For a copper-zinc cell like
Zn|Zn+2||Cu+2|Cu the standard cell voltage E0 = 1.101 V. Since we can split a redox reaction into two parts, we can also define standard voltages for both the oxidation and reduction parts of the reaction, Eox0 and Ered0. We arbitrarily pick the hydrogen reduction half reaction
2H+(aq) + 2e- -> H2(g) to have Ered0 = 0, and measure all other half reaction voltages in relationship to it. The oxidation half reactions are just the reaction run in reverse, and the half cell oxidation voltage is just the negative of the reduction voltage. For example:
Reduction: Zn+2(aq) + 2e- -> Zn(s) Ered0 = -0.762 V Oxidation: Zn(s) -> Zn+2(aq) + 2e- Eox0 = 0.762 V The standard voltage of a cell, E0, is just the sum of the standard voltages of the oxidation and reduction half reactions. This is independent of the number of electrons in the half reaction: if you need to multiply a half reaction by 3 to balance the redox equation, you do not multiply the cell voltage by 3. For example, the zinc-copper cell above is made up of the reactions
Oxidation: Zn(s) -> Zn+2(aq) + 2e- Eox0 = 0.762 V Reduction: Cu+2(aq) + 2e- -> Cu(s) Ered0 = 0.339 V E0 for the cell is Ered0 + Eox0 = 0.339 + 0.762 = 1.101 V Example: What is the standard voltage for the cell
Mn|Mn+2||Fe+3|Fe+2 Solution: First, seperate the cell into two half reactions
Oxidation: Mn(s) -> Mn+2(aq) + 2e- Reduction: Fe+3(aq) + e- -> Fe+2(aq) The iron reduction has a voltage of 0.769 V. The oxidation of manganese is not listed, but the reduction (reverse reaction) is listed with a reduction voltage of -1.182 V. The oxidation thus has a voltage of +1.182 V To compute the cell voltage, simply add the two voltages. It doesn't matter that the manganese reaction has two electrons and the iron only one
E0 = Ered0 + Eox0 = 0.769 + 1.182 = 1.951 V"
So, my 0.9-1.2 volt number was the oxidation portion of the reaction, didn't account for the reduction portion...
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