The equilibrium constant for the reaction given below at 298 K is : ${\rm{Zn}}\left( s \right) + {\rm{F}}{{\rm{e}}^{2 + }}\left( {aq} \right) \to {\rm{Z}}{{\rm{n}}^{2 + }}\left( {aq} \right) + {\rm{Fe}}\left( s \right)$ ; $E_{cell}^ \circ $ = 2905 V at 298 K

NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

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The equilibrium constant for the reaction given below at 298 K is :

\mathrm{Zn}\left( s \right) + \mathrm{F}{\mathrm{e}^{2 + }}\left( {aq} \right) \to \mathrm{Z}{\mathrm{n}^{2 + }}\left( {aq} \right) + \mathrm{Fe}\left( s \right)

;

E_{cell}^ \circ

= 2905 V at 298 K

View answer and explanation

Correct option: D

{10^{0.32/0.295}}

Option D is correct because it uses the relationship between the standard cell potential and the equilibrium constant:

E_{cell}^\circ = \frac{{0.0591}}{n}logK

at 298 K.

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