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Question: \(\mathbf{E}^{\mathbf{0}}\)for the cell \(Zn/Z{n^{2 +}}_{(aq)}//C{u^{2 +}}_{(aq)}/Cu\)is 1.10Vat \(...

E0\mathbf{E}^{\mathbf{0}}for the cell Zn/Zn2+(aq)//Cu2+(aq)/CuZn/Z{n^{2 +}}_{(aq)}//C{u^{2 +}}_{(aq)}/Cuis 1.10Vat

25oC25^{o}C. The equilibrium constant for the reaction

Zn+Cu(aq)2+Zn + Cu_{(aq)}^{2 +}⇌Cu+Zn(aq)2+Cu + Zn_{(aq)}^{2 +}is of the order of

A

10−2810^{- 28}

B

10−3710^{- 37}

C

10+1810^{+ 18}

D

103710^{37}

Answer

103710^{37}

Explanation

Solution

Zn+Cu2+Zn + Cu^{2 +}⇌ Zn2++Cu;Eo=1.10VZn^{2 +} + Cu;E^{o} = 1.10V

E=Eo−0.05912log⁡[Zn2+][Cu2+]E = E^{o} - \frac{0.0591}{2}\log\frac{\lbrack Zn^{2 +}\rbrack}{\lbrack Cu^{2 +}\rbrack}

At equlibrium E= 0 and [Zn2+]eqm[Cu2+]eqm=K\frac{\lbrack Zn^{2 +}\rbrack eqm}{\lbrack Cu^{2 +}\rbrack eqm} = K

∴0=1.10−0.05912log⁡K\therefore 0 = 1.10 - \frac{0.0591}{2}\log K or 1.10=0.05912log⁡K1.10 = \frac{0.0591}{2}\log{}K or

K=1.94×1037K = 1.94 \times 10^{37}