Determine at 298 for cell : Pt|Q,QH2,H+||1MKCl|Hg2Cl2|Hg(I)|Pt the pH when Ecell=0 given E∘RP(RHS)=0.28,E∘RP(LHS)=0.699(write the value to the nearest integer)
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Solution
E0cell=E0R−E0L=0.28−0.694=−0.419 Ecll=E0cell−0.0592nlogQ At equilibrium, Ecell=0 and Q=K 0=−0.419−0.05922logK logK=−14.155 K=6.99×10−15 The net cell reaction is Hg2+2+QH2→2Hg+Q+2H+ K=[H+]2=6.99×10−15 [H+]=8.36×10−8 pH=−log[H+]=−log8.36×10−8=7.1