emf of cell is given ag(s),agcl(s)||Kcl(aq)Hg2cl2||Hg(s) is 0.05V at 300k and temperature coefficient of the cell is 3.34×10^-4 calculate the change in enthalpy of cell
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Explanation:
free energy equation
dg=vdp-sdt
at constant pressure
dg/dt = -s
nf de/dt = dels
nfTcell = dels
delh=delg+tdels
delh = -nef+tdels
delh = nf(tTcell - Ecell )
delh = 2*96500*(0.1012 - 0.05 )
delh = 2*96500*0.0502
delh = 9.6886 *10^3 vk^-1
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