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Set up Nernst equation for the standard dry cell. Using this eqation show that the voltage of a dry cell has to decrease with use. |
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Answer» Solution :The voltage of an electrochemical cell is related to the concentration of the solutions involved in the cell. A change in concentration of cell solutions results in a change in voltage that can be quantitatively described by the Nernst equation. The voltage of an electrochemical cell involves an oxidation reaction and a reduction reaction. In the case of a copper-zinc cell the two half reactions are WRITTEN as REDUCTIONS , and their STANDARD cell potentials are : `Cu^(2+)(aq) + 2e to Cu(s)` reduction =`+0.34V` `Zn^(2+)(aq) + 2e to Zn`(s) reduction =-0.76 V Copper will be reduction species in the copper-zinc cell due it's larger standard reduction potential. The voltage then for a Zn, `Zn^(2+)"||"Cu^(2+) + Zn^(2+)(aq)` Equation 1 : ex. `Cu^(2+)(aq) + 2e to Cu(s) E^(@)` reduction =+0.34 V `Zn(s) + Zn^(2+)(aq) to2e^(-)E^(@)` oxidation =-0.76 V `E_("cell") =E_("reduction")-E_("oxidation")` `+1.10 V =+0.34 V-(-0.76 V)` The voltage fo any electrochemical cell is a function of the molar concentrations of the compounds involved in the cell, as described by the Nernst equation : `aA(s) + bB^(+) (aq) to cC(s)+ DD^(+)(aq)` Equation 2:`E_("cell")=E^(@)-(0.06//n)log{D^(+]d)//[B^(+)]^(b)}` Wquation 3: Using equation 1 as the balanced chemical equation for the `Zn, Zn^(2+)||Cu^(2+) , Cu` cell. Equation 3 becomes : `E_("cell") =E^(@)-(*0.03)log{Zn^(2+]//[Cu^(2+)]` As the `Cu^(2+)` concentration decreases the coltage fo the cell changes . The `Cu^(2+)` concentration can change due to dilution, or due to the formation of a complex ion, `Cu(NH_(3))_(4)^(2+)` , upon addition of `NH_(3)` |
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