Question

calculate the equilibrium constant of the reaction.
Cu + 2Ag+ ---- Cu2+ +2Ag
E0cell = 0.46 v​

Answer 1

Answer:Kc = 3.92×10^15

Explanation:

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Here's ur answer:

E^0(cell) = 0.46V

n = 2

E^0(cell) = (0.059/n) log Kc

0.46 = (0.059/2) log Kc

0.46/log Kc = 0.059/2

log Kc = 2×0.46/0.059

log Kc = 0.92/0.059

log Kc = 15.57

Kc = Antilog Kc

Kc = 3.92×10^15

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Answer 2

The equilibrium constant of the reaction is $3.631\times 10^{15}$

Explanation:

$Cu+2Ag^+\rightarrow Cu^{2+}+2Ag$

$E^0=E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials.

$E^0=0.46V$

The standard emf of a cell is related to Gibbs free energy by following relation:

$\Delta G=-nFE^0$

$\Delta G$ = gibbs free energy

n= no of electrons gained or lost  = 2

F= faraday's constant  = 96500 C

$E^0$ = standard emf  = 0.46 V

$\Delta G=-2\times 96500\times (0.46)=-88780J$

The Gibbs free energy is related to equilibrium constant by following relation:

$\Delta G=-2.303RTlog K$

R = gas constant = 8.314 J/Kmol

T = temperature in kelvin =$25^0C=25+273=298K$

K = equilibrium constant

$\Delta G=-2.303RTlog K$

$-88780=-2.303\times 8.314\times 298\times logK$

$logK=15.56$

$K=3.631\times 10^{15}$

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