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What is the relationship between Gibbs free energy of the cell reaction in a galvanic cell and the emf of the cell ? When will the maximum work be obtained froma galvanic cell ? |
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Answer» Solution :If concentration of all reacting species is unity, then `E_("cell")=E_("Cell")^(@)` and `DeltaG^(@)=-nFE_("cell")^(@)` where, `DeltaG_(r)G^(@)` is STAN dard GIBBS energy of the reaction `E_("cell")^(@)=` emf of the cell `nF=`charge PASSED If we want to obtain m aximum WORK from a galvanic cell then charge has to be passed reversible. The reversibly work done by a galvanic cell is equal to decrease in its Gibbs energy `Delta_(r)G=-nFE_("cell")` As `E_("cell")`is an intensive parameter but `Delta_(f)=G` is an EXTENSIVE thermodynamic property and thevalue depends on n. For reaction, `Zn(s)+Cu^(2+)(aq)rarrZn^(2+)(aq)+Cu(s)` in a galvanic cell. `Delta_(r)G=-2FE_("cell")` [Here,n=2] |
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