Question

the equlibrium constant of the given cell reaction at 298K will be
Mg(s) + Cu^2+(aq) ----------> Mg2+(aq) +cu(s)
Given: E^0 cell= 2.7 V)

a) 10^72.1
b) 10^82.5
c) 10^55.6
d) 10^91.5
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Answer 1

Answer:

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Answer 2

Answer:

The equilibrium constant  at 298K of the given cell reaction at 298K will be equal to $10^{91.3}$.

Therefore, the option (d) is correct.

Explanation:

Given, the cell reaction at 298K is :

Mg(s)   +   Cu²⁺(aq.) →  Mg²⁺(aq.)   +   Cu(s)

We know that the relation between electrode potential and equilibrium constant of the cell:-

$nFE^{o}_{cell}=RTlnK_{eq}$                                                      ............(1)

where F is faraday constant, F = 96500Cmol⁻¹

and, R is the gas constant has value, R= 8.314JK⁻¹mol¹

The number of electrons, n = 2

The value of standard electrode potential of the cell, $E^{o}_{cell}=2.7V$

Put all the values R, T , n, F and E⁰ in equation (1):

$(8.314)(298)lnK_{eq}=(2)(96500)(2.7)$

$2.303*logK_{eq}=210.32$

$logK_{eq}=91.3$

$K_{eq}=10^{91.3}$

Therefore, the equilibrium constant is equal to $10^{91.3}$.