Question

The equilibrium constant for the reaction is ______ × 10²⁶.

\[\ce{Fe + CuSO4 <=> FeSO4 + Cu}\] at 25°C.

Given `"E"_("Fe"//"Fe"^(2+))^0` = 0.44 V

`"E"_("Cu"//"Cu"^(2+))^0` = - 0.337 V

विकल्प

• 5.29

• 6.03

• 1.71

• 0.56

Answer 1

The equilibrium constant for the reaction is 1.71 × 10²⁶.

Explanation:

\[\ce{Fe -> Fe^{2+}  + 2e-}\]

\[\ce{Cu^{+2} + 2e- -> Cu}\]

`"E"_"cell"^0` = 0.44 + 0.337 = 0.777

`"E"_"cell"^0 = 0.0591/n` log K

`(0.777 xx 2)/0.0591` = log K

log K = 26.06

K = 1.71 × 10²⁶