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$E^{\Theta }_{Cell} = 1.1V$ for Daniel cell. Which of the following expressions are correct description of state of equilibrium in this cell?

$(i) \; 1.1 = K_{c}$

$(ii) \;\frac{2.303RT}{2 F} logK_{c} = 1.1$

$(iii)\; log K_{c} = 2.2/0.059$

$(iv) log K_{c} = 1.1$

Answers (1)

The answer is the option (ii,iii)

$E_{cell}^{0}=\frac{2.303\; RT}{nF} logK_{c}=\frac{0.059}{2}logK_{c}$

$1.1=\frac{0.059}{2}logK_{c}$

$logK_{c}=\frac{2.2}{0.059}$

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