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Consider the cell described below at 285 K: Sn | Sn2+ (1.01 M) || Pb2+ (2.07 M) Pb Given EºPb2+ + 2e-Pb = -0.131 V and Eºsn2+

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When cell coll dead Ecell 0 From nemsteauatioh Eett 2.303 RT hF tett where Potential of cell standanad poterntalef cell Ecellohen cell dead Eol= un eactish aulibrum constaut h= 2 দ RP Redcug petantal Oxdes pateataa4 Oxidatcon half ceacteon Sn s Sn2te2.3D3 RT nF 2.303 RT HF ৭ Snlsl +Pha4) +Phls 2Sn) (1.01+2M Initha Concontyatcon ofF. Pb42] CSnt21 L.D M 1.D1 M sn42] CPb121 .9.658 1.01-+24 2.04-2 5.5-9.65g ze = 1.01+ 24 4.493= 3.658 22 22= 1.228 M fnal lontentrateon of [Pht1 2.Dt 24 Q.D7-T228 LPh2

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