Question

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

• A. $\log K_c=\frac{2.2}{0.059}$
• B. $\frac{2.303RT}{2F}\log K_c=1.1$
• C. $1.1=K_c$
• D. $\log K_c=1.1$

Answer 1

Answer: (A), (B)

$\frac{2.303RT}{2F}\log K_c=1.1$

Nernst equation for Daniel cell

For Daniel cell, $E_{cell}^{\Theta }=1.1V$

Reactions taking place in Daniel cell:

At anode:

$Zn\left(s\right)\to Z{n}^{2+}\left(aq\right)+2e^{-}$

At cathode:

$C{u}^{2+}\left(aq\right)+2e^{-}\to Cu\left(s\right)$

Overall cell reaction:

$C{u}^{2+}\left(aq\right)+Zn\left(s\right)\to Cu\left(s\right)+Z{n}^{2+}\left(aq\right)$

Thus, Nernst equation for the cell will be:

$E_{cell}=E_{cell}^{\Theta }-\frac{2.303RT}{2F}\log \frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$

Calculation of $\log K_c$ at equilibrium

At equilibrium, $E_{cell}=0$ and $K_c=\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$

$\therefore 0=E_{cell}^{\Theta }-\frac{2.303RT}{2F}\log K_c$
Given, $E_{cell}^{\Theta }=1.1$
$\therefore \frac{2.303RT}{2F}\log K_c=E_{cell}^{\Theta }=1.1$

Considering;
$R=8.314Jmo{l}^{-1}{K}^{-1}andT=298K$

Putting the values in the Nernst equation:
$\frac{2.303RT}{2F}\log K_c=E_{cell}^{\Theta }$
$\Rightarrow 2.303\times 8.314\times \frac{298}{2F}\log K_c=1.1$
$\Rightarrow \frac{2.303\times 8.314\times 298}{2\times 96500}\log K_c=1.1$
$\therefore \frac{0.059}{2}\log K_c=1.1$

$\therefore \log K_c=\frac{2.2}{0.059}$

Thus,
$\frac{2.303RT}{2F}\log K_c=1.1$ and
$\log K_c=\frac{2.2}{0.059}$

Hence, the correct options are (B) and (C).