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Question

The standard emf of the cell Cd(s)/CdCl2(aq)(0.1M)||AgCl(s)|Ag(s) in which the cell-reaction is Cd(s)+2AgCl(s)2Ag(s)+Cd2+(aq)+2Cl(aq)is 0.6915V at 273K and 0.6573V at 298K. Calculate enthalpy change of the reaction at 298K is?

A
176.86kJ
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B
245.6kJ
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C
+48.179kJ
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D
+223.5kJ
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Solution

The correct option is C +48.179kJ
Given, E1=0.6915V, T1=273K,
E2=0.6573V, T2=298K

Now, ΔG=nFEo cell =2×96500×0.6573
=126858.9

Δ S= nF(ET)p
Here, .ET= Temperature coefficient of e.m.f

(ET)p=E2E1T2T1=0.65730.6915298273 =0.001368

ΔS=2×96500×(0.001368)=264.024JK1mol1

As we know that,

ΔG=ΔHTΔS

126858.9=ΔH298×(264.024)

ΔH=48179.75J=48.179kJ.

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