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For the reaction
2AgCl (s) + H2 (g) (1 atm) → 2 Ag (s) + 2H+ (0.1 M) + 2Cl- (0.1 M),∆Go = - 43600 J at 25o C
Calculate the e.m.f. of the cell.[log 10–n = – n]
Given:2AgCl (s) + H2 (g) (1 atm) → 2 Ag (s) + 2H+ + 2Cl-∆Go = - 43600 JT = 25o C = 298 KWe know,∆Go = -nFEcell°n = 2∴ Ecell° = ∆Go nF = -(-43600)2 x 96500 = 0.226 VEcell = Ecello -0.059n log [H+]2[Cl-]2[Ag][AgCl]2[H2]= 0.226 -0.0592 log H+2Cl-2= 0.226 -0.0592 log (0.1)2(0.1)2= 0.226 -0.0592 log (0.1 x 0.1 x 0.1 x 0.1)= 0.226 -0.0592 log 10-4= 0.226 - 0.0592 (-4) = 0.226 + 0.118Ecell = 0.344 V
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