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:

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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

Electrochemical cells

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$The temperature coefficient of the e.m.f. of cell, (\frac{dE}{dT}) is given by :$

From equation 1 and $2, ΔS = \frac{- d (ΔG)}{dT}$
$ΔS = \frac{d (nFE)}{dT} \Rightarrow nF \frac{dE}{dT} or {(\frac{dE}{dT})}_{p} = \frac{ΔS}{nF}$

$At 300 K, ΔH for the reaction$
$Zn (s) + AgCl (s) ⟶ {ZnCl}_{2} (aq) + 2 Ag (s) is$
$- 218 kJ / mol while the e.m.f. of the cell was 1 .015 V \cdot {(\frac{dE}{dT})}_{p} of the cell is:$

From equation 2;
$\begin{array}{l} ΔH = nF {(\frac{dE}{dT})}_{p} T - {nFE}_{cell} \\ - 218 \times 1000 = 2 \times 96500 \times 300 {(\frac{dE}{dT})}_{p} - 2 x 96500 x 1.015 \\ {(\frac{dE}{dT})}_{p} = - 3 .81 \times 10^{- 4} V K^{- 1} \end{array}$

$Calculate Δ S for the given cell reaction in$
$Zn (s) + AgCl (s) ⟶ {ZnCl}_{2} (aq) + 2 Ag (s) is$
$- 218 kJ / mol while the e.m.f. of the cell was 1 .015 V \cdot {(\frac{dE}{dT})}_{p} of the cell is:$

$\begin{aligned} ΔS = nF {(\frac{dE}{dT})}_{p} = 2 \times 96500 \times - 3 .81 \times 10^{- 4} \\ = - 73 .53 J / mol - K \end{aligned}$

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Electrochemical cells

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The temperature coefficient of the e.m.f. of cell, dEdT is given by :

From equation 1 and 2,ΔS=−d(ΔG)dTΔS=d(nFE)dT⇒nFdEdT or dEdTp=ΔSnF

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:

From equation 2;ΔH=nFdEdTpT−nFEcell −218×1000=2×96500×300dEdTp-2 x 96500 x 1.015 dEdTp=−3.81×10−4 V K−1

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:

ΔS=nFdEdTp=2×96500×−3.81×10−4=−73.53J/mol−K