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Derive an expression for Nernst equation. |
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Answer» Solution :NERNST equation : Nernst equation is the one which relates the cell potential and the concentration of the SPECIES involved in an electrochemical reaction. Let us consider an electrochemical cell for which the overall redox reaction is, `""xA+xB iff1C+mD` The reaction quotient Q for the above reaction is given below `Q=([C]^(l)[D]^(m))/([A]^(x)[B]^(y)) "...(1)"` We have already learnt that, `DeltaG=DeltaG+RTInQ"...(2)"` The Gibbs free energy can be related to the cell emf as follows [`therefore` equation (1) and (2)] `DeltaG=-nFE_("cell"), DeltaG^(@)=-nFE_("cell")^(@)` SUBSITUTE these values and Q from (1) in the equation (2) `(2) rArr -nFE_("cell")=-nFE_("cell")^(@)+RTln""([C]^(l)[D]^(m))/([A]^(x)[B]^(y)]"..(3)"` Divide the whole equation (3) by (-nF) `(4) rArr E_("cell")=E_("cell")-(RT)/(nF)ln""([C]^(l)[D]^(m))/([A]^(x)[B]^(y))` (or) `E_("cell")=E_("cell")-(2.303RT)/(nF)log""([C]^(l)[D]^(m))/([A]^(x)[B]^(y))"...(4)"` The above equation (4) is called the Nernst equation At `25^(@)C` (298K), the above equation (4) becomes, `E_(cell)=E_(cell)^(@)-(2.303 times 8.314 times 298)/(n(96500))log""([C]^(l)[D]^(m))/([A]^(x)[B]^(y))` `E_(cell)=E_(cell)^(@)-(0.0591)/(n)log""([C]^(l)[D]^(m))/([A]^(x)[B]^(y))"...(5)"` |
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