The cell potential (Ecell) of a reaction is related as Δ$$G=$$−nFEcell , where ΔG represents max. useful electrical work n $$=$$ no. of moles of electrons exchanged during the reactionfor reversible cell reaction $$d($$Δ$$G)=$$ΔrVdp−ΔrS⋅dTat constant pressure $$d($$Δ$$G)=$$−ΔrS⋅dT ∵ At constant pressure Δ$$G=$$ΔH−T⋅ΔS . . . . (1)∴ Δ$$G=$$Δ$$H+Td($$Δ$$G)dTp$$ . . . . .(2)dEcell$$ dTp is known as temperature coefficient of the e.m.f of the cell The temperature coefficient of the e.m.f. of cell, dEdT is given by : At $$300K$$,ΔH for the reaction $$Zn(s)+AgCl(s)⟶$$ZnCl2(aq)+2Ag(s)$$ is −$$218kJ/mol$$ while the e.m.f. of the cell was $$1.015V$$⋅dEdTp of the cell is: Calculate ΔS for the given cell reaction in $$Zn(s)+AgCl(s)$$⟶$$ZnCl2(aq)+2Ag(s)$$ is −$$218kJ/mol$$ while the e.m.f. of the cell was $$1.015V$$⋅dEdTp of the cell is:

Question

The cell potential (Ecell) of a reaction is related as Δ$$G=$$−nFEcell , where ΔG represents max. useful electrical work n $$=$$ no. of moles of electrons exchanged during the reactionfor reversible cell reaction $$d($$Δ$$G)=$$ΔrVdp−ΔrS⋅dTat constant pressure $$d($$Δ$$G)=$$−ΔrS⋅dT ∵ At constant pressure  Δ$$G=$$ΔH−T⋅ΔS            . . . . (1)∴ Δ$$G=$$Δ$$H+Td($$Δ$$G)dTp$$                                      . . . . .(2)dEcell$$ dTp is known as temperature coefficient of the e.m.f of the cell  The temperature coefficient of the e.m.f. of cell, dEdT is given by :  At $$300K$$,ΔH for the reaction $$Zn(s)+AgCl(s)⟶$$ZnCl2(aq)+2Ag(s)$$ is −$$218kJ/mol$$ while the e.m.f. of the cell was $$1.015V$$⋅dEdTp of the cell is:  Calculate ΔS for the given cell reaction in $$Zn(s)+AgCl(s)$$⟶$$ZnCl2(aq)+2Ag(s)$$ is −$$218kJ/mol$$ while the e.m.f. of the cell was $$1.015V$$⋅dEdTp of the cell is:

Infinity Learn · Accepted Answer

From equation $$1$$ and $$2$$,Δ$$S=$$−$$d($$Δ$$G)dT$$Δ$$S=d(nFE)dT$$⇒nFdEdT or $$dEdTp=$$ΔSnFFrom equation $$2$$;Δ$$H=nFdEdTpT$$−nFEcell −$$218$$×$$1000=2$$×$$96500$$×$$300dEdTp-2$$ x $$96500$$ x $$1.015$$ $$dEdTp=$$−$$3.81$$×$$10$$−$$4$$ V K−$$1$$Δ$$S=nFdEdTp=2$$×$$96500$$×−$$3.81$$×$$10$$−$$4=$$−$$73.53J/mol$$−K

Q.

Unlock the Full Solution and Master the Concept