3. Emf of cell given Ag(s), AgCl(s)...
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3. Emf of cell given Ag(s), AgCl(s)||KCl(aq)|Hg, Cl_(s)/Hg(s) is 0.05 V at 300 K and temperature coefficient of the cell is 3.34 x 10-4 VK-1 Calculate the change in enthalpy of the cell. (1) 965 (2) 9650 (4) 96.5 (3) 96500 A F = 0.799 V

JEE/Engineering Exams
Chemistry
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( 2 A g rightarrow 2 A g^{+}+2 e^{-} quad:(text { anode }) ) ( H g_{2}^{2+}+2 e^{-} rightarrow 2 H g quad ) : ( quad(text { cathode }) ) ( therefore quad n=2 ) [ begin{array}{l}Delta H=-n F E_{text {coll }}+n F Tleft(frac{partial E_{text {cell }}}{partial T}right)_{p} =2 times 96500left(300 times 3.34 times 10^{-4}-0.05right) =9650 mathrm{J} mathrm{mol}^{-1}end{array} ]
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