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Which of the following reactions would have the smallest value of K at 298 K? Which of the following reactions would have the smallest value of K at 298 K? A + B → 2 C; E°cell = -0.030 V A + 2 B → C; E°cell = +0.98 V A + B → C; E°cell = +1.22 V A + B → 3 C; E°cell = +0.15 V More information is needed to determine.

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Answer:

The reaction with smallest value of K is :

A + B → 2 C; E°cell = -0.030 V

Explanation:

[tex]nFE^o_{cell}=RT\ln K[/tex]

where :

n = number of electrons transferred

F = Faraday's constant = 96500 C

[tex]E^o_{cell}[/tex] = standard electrode potential of the cell

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction = [tex]25^oC=[273+25]=298K[/tex]

[tex]K[/tex] = equilibrium constant of the reaction

As we cans see, that standard electrode potential of the cell is directly linked to the equilibrium constant of the reaction.

  • Higher [tex]E^o_{cell}[/tex] higher will be the value of K.
  • Lower [tex]E^o_{cell}[/tex] lower will be the value of K.

So, the reaction with smallest value of electrode potential will have smallest value of equilibrium constant. And that reaction is:

A + B → 2 C; [tex]E^o_{cell} =-0.030 V[/tex]

okpalawalter8

The reactions that would have the smallest value of K is

A + B → 2 C; E°cell = -0.030 V

Option A

Generally the equation for the number of electrons transferred  is mathematically given as

[tex]nFE^o_{cell}=RT\ln K[/tex]

where

T= Temperature

F=25C(298K)

R = Gas constant

R= 8.314 J/K.mol

F = Faraday's constant

F= 96500 C

We see from the equation that the E-cell is directly proportional to K(equilibrium constant of the reaction)

Hence, The reactions that would have the smallest value of K is

A + B → 2 C; E°cell = -0.030 V

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