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Give Nernst equation for following reaction M_((aq))^(n+)+n e^(-)to M_((S)) |
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Answer» Solution :* It is assumed that the concentration of all the species involved in the electrode reaction is unity. This need not be ALWAYS true. * Nernst showed that for the electrode reaction: `M_((aq))^(n+)+n E^(-) to M_((S))` * The electrode potential at any concentration MEASURED with respect to STANDARD hydrogen electrode can be represented by: `E_(("M^(n+)|M))=E_((M^(n+)|M))^(Theta)-(RT)/(nF)LN([M])/([M^(n+)])` * But concentration of solid M is taken as unity and we have `E_((M^(n+)|M))=E_((M^(n+)|M))^(Theta)-(RT)/(nF)ln([1])/([M^(n+)])` Where, n=number of electrons taking part in the reaction . `E_((M^(n+)|M))`=potential of electrode at non standard concentration. `E_((M^(n+)|M))^(Theta)`=standard reduction potential of electrode. R=Gas constant=`8.314KJ^(-1)" "mol^(-1)` F=Feraday constant=96487 and T=TemperatureinKelvin `[M^(n+)]`=Concentration of species `M^(n+)` in molarity. |
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