Question

$Zn|{Zn}^{2+}(a=0.1M)\parallel {Fe}^{2+}(a=0.01M)|Fe$. The emf of the above cell is $0.2905V$. Equilibrium constant for the cell reaction is

• A. ${10}^{0.32/0.0591}$
• B. ${10}^{0.32/0.0295}$
• C. ${10}^{0.26/0.0591}$
• D. ${10}^{0.32/0.295}$

Answer 1

The correct option is B ${10}^{0.32/0.0295}$

For cell $Zn|{Zn}^{2+}(a=0.1M)\parallel {Fe}^{2+}(a=0.01M)|Fe$

The half-cell reactions are
(i) $Zn\left(s\right)⟶{Zn}^{2+}\left(aq\right)+2e^{-}$

(ii) ${Fe}^{2+}\left(aq\right)+2e^{-}⟶Fe\left(s\right)$
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$Zn\left(s\right)+{Fe}^{2+}\left(aq\right)⟶{Zn}^{2+}\left(aq\right)+Fe\left(s\right)$
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On applying the Nernst equation

$E_{cell}=E_{cell}^o-\frac{0.0591}{n}{\log }_{10}\frac{\left[{Zn}^{2+}\right]}{\left[{Fe}^{2+}\right]}$

$0.2905=E_{cell}^o-\frac{0.0591}{n}{\log }_{10}\frac{0.1}{0.01}$

$0.2905=E_{cell}^o-0.0295\times {\log }_{10}10$

$0.2905=E_{cell}^o-0.0295\times 1$

$\therefore E_{cell}^o=0.32V$

At equilibrium $(E_{cell}=0)$

$E_{cell}=E_{cell}^o-\frac{0.0591}{n}{\log }_{10}{K}_c$

$\therefore 0=E_{cell}^o-\frac{0.0591}{n}{\log }_{10}{K}_c$

$E_{cell}^o=\frac{0.0591}{2}{\log }_{10}{K}_c$

$0.32=\frac{0.0591}{2}{\log }_{10}{K}_c$

${\log }_{10}{K}_c=\frac{0.32}{0.02955}$

$K_c={10}^{0.32/0.02955}$