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For an electrochemical cell Sn s Sn 2 + aq ,   1 M Pb 2 + aq ,   1 M Pb s the ratio Sn 2 + Pb 2 + when this cell attains equilibrium is _______ (Given: E Sn 2 + Sn 0 = - 0 .14   V , E Pb 2 + Pb 0 = - 0 .13   V ,    2 .303 RT F = 0 .06 )

Correct answer

0

Step-by-step 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 = 0.01 V At equilibrium : E cell  = 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

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