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The cell potential (E("cell")) of a reac...

The cell potential `(E_("cell"))` of a reaction is related as `DeltaG = -nFE_("cell")` , where `DeltaG` represents max. useful electrical work .
n=no. of moles of electrons exchanged during the reactino of reversible cell reaction `d(DeltaG) = (DeltaV)dP-(DeltaS).dT`
at constant pressure ` d(DeltaG) = - (DeltaS).dT`
`:'` At constant pressure `DeltaG = DeltaG = DeltaH-TDeltaS`.......(1)
`:. DeltaG = DeltaH + T((d(DeltaG))/(dT))_P` ......... (B)
` ((dE_("cell"))/(dT))_(P)` is known as temperature coefficient of the e.m.f of the cell
When `DeltaS` increases, temperature coefficient of the emf of cell

A

`-4.2 xx 10^(-4) VK^(-)`

B

`-3.81 xx 10^(-4) VK^(-1)`

C

`0.11 VK^(-) `

D

`7.62 xx 10^(-4)VK^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B

`DeltaG = DeltaH - DeltaS.T , DeltaS = ((DeltaH -DeltaG)/(T)) = nF (dE_("cell"))/(dt))_(P) implies ((dE_("cell"))/(dt))_(P) = ((DeltaA - DeltaG)/(nFT))`
`= (DeltaA - (-nFE_("cell")))/(nFT) = ((DeltaH)/(nFT) + (nFE_("cell"))/(nFT)) = ((DeltaH)/(nFT) + (E_("cell"))/(T)) = ((-218 xx 1000)/( 2 xx 96000 xx 300) + (1.015)/(300))`
`=[-0.003765 + 0.003383] = -0.00038168 = -3.8167 xx 10^(-4) V//K`
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