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Question

For an electrochemical cell, Sn(s)/Sn2+(aq),1M)//Pb2+(aq,1M)/Pb(s). The ratio [Sn2+]/[Pb2+] when this cell attains equilibrium is ______.

Given:E°Sn(s)/Sn2+=0.14V;E°Pb2+/Pb)/Sn2+=0.13V,2.303RT/F=0.06V


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Solution

For an electrochemical cell, Sn(s)/Sn2+(aq,1M)//Pb2+(aq,1M)/Pb(s). The ratio [Sn2+]/[Pb2+] when this cell attains equilibrium is2.15

Step 1: Given:E°Sn(s)/Sn2+=0.14V;E°Pb2+/Pb)/Sn2+=0.13V,2.303RT/F=0.06V

Step 2: Half anodic reaction involved : SnSn2++2e-

Half Cathodic reaction involved : Pb2++2e-Pb

Net reaction: Sn+Pb2++2e-Pb+Sn2+

E°cell=E°cathodeE°anode

Step 3: According to the Nernst equation,E°cell=E°cathode(0.06/2)log[Sn2+]/[Pb2+]

Ecell=E°cell(0.06/2)log[Sn2+]/[Pb2+]

E°cell=0.01V

At equilibrium state Ecell=0

0=0.01(0.06/2)log[Sn2+]/[Pb2+]0.01=(0.06/2)log[Sn2+]/[Pb2+]log[Sn2+]/[Pb2+]=1/3[Sn2+]/[Pb2+]=2.15

Hence, The ratio[Sn2+]/[Pb2+] when this cell attains equilibrium is 2.15


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