For an electrochemical cell Sn s Sn 2 + aq ,   1 M Pb 2 + aq ,   1 M Pb s the ratio Sn 2 + Pb 2…

For an electrochemical cell SnsSn2+aq, 1MPb2+aq, 1MPbs the ratio Sn2+Pb2+ when this cell attains equilibrium is _______
(Given: ESn2+Sn0=-0.14 V,EPb2+Pb0=-0.13 V,  2.303RTF=0.06)

Solution

Cell reaction is:

Sn(s)+ Pb +2 (aq)→ Sn +2 (aq)+Pb(s)

Apply Nernst equation:

E cell  = E cell  0 − 0.06 2 log Sn +2 Pb +2 .....  1

E cell  0 =−0.13+0.14=0.01V

At equilibrium : Ecell =0 Substituting in (1)

0=0.01− 0.06 2 log Sn +2 Pb +2

⇒  1 3 =log Sn +2 Pb +2

⇒  Sn +2 Pb +2 =2.15

Asked in: JEE Main 2020 (08 Jan Shift 2)

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