What is the value of the equilibrium constant for the reaction of Na(s)with Pb2+ (aq)at 25°...

What is the value of the equilibrium constant for the reaction of Na(s)with Pb2+ (aq)at 25° C? Eº cell = 2.58 V 2Na(s) + Pb2+(aq) ↔ Pb(s) + 2Na+(aq) 3.0 x 1043 6.0 x 10200 6.0 x 10 43 1.8 x 1087

Solutions

Expert Solution

2Na(s) -----------------> 2Na^+ (aq) + 2e^- E0 = 2.71V

Pb^2+ (aq) + 2e^- ----------> Pb(s) E0 = -0.13v

--------------------------------------------------------------------------------------------

2Na(s) +Pb^2+ (aq) ---------------> Pb(s) + 2Na^+ (aq) E = 2.58v

n = 2

G⁰ = -nE0F

= -2*2.58*96500

= -497940J

G^0 = -RTlnK

-497940 = -8.314*298*2.303logK

-497940 = -5705.8483logK

logK = -497940/-5705.8483

logK = 87.27

K = 10^87.27 = 1.8*10^87 >>>>>answer

queen_honey_blossom answered 1 year ago

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