Calculate the equilibrium constant of given reaction for Daniell cell at 298 K temperature. \(Zn(s) + Cu^{2+}(aq) ⇌ Zn^{2+}(aq) + Cu(s)\). Cell potential =1.1 volt (F=96500 coulomb)

\(1.62 × 10^{37}\)

We know,

\(E_{cell} = E^o_{cell} – \frac{RT}{nF}lnQ\)

At equilibrium,

\(Q = K\) and \(E_{cell} = 0\)

So,

\(E^o_{cell} = \frac{RT}{nF}lnK\)

\(E^o_{cell} = \frac{2.303RT}{nF}logK\)

In the given reaction, the exchange of 2 electrons takes place. Therefore, \(n=2\)

\(1.1 = \frac{2.303 × 8.314 × 298}{2 × 96500} log K \)

or, \(log K = 37.207\)

or, \(K = 1.62 × 10^{37}\)