Question

Choose the correct option(s) for the following concentration cell at $298K$:

$Zn\left(s\right)|Z{n}^{2+}(aq,C_1)||Z{n}^{2+}(aq,C_2)|Zn\left(s\right)$

$C_1\&C_2\text{are concentrations of}Z{n}^{2+}\text{ions of each half-cell.}$

• A. $E_{cell}^0=0$
• B. For a spontaneous reaction to take place, $C_1>C_2$
• C. For a spontaneous reaction to take place, $C_1<C_2$
• D. $E_{cell}^0>0$

Answer 1

Answer: (A), (C)

For a spontaneous reaction to take place, $C_1<C_2$

$Zn\left(s\right)|Z{n}^{2+}(aq,C_1)||Z{n}^{2+}(aq,C_2)|Zn\left(s\right)$

$C_1\&C_2\text{are concentrations of}Z{n}^{2+}\text{ions of each half-cell.}$
Anode:
$Zn\left(s\right)\rightleftharpoons Z{n}^{2+}\left(aq\right)+2e^{-}$

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

Net cell reaction:
$Z{n}^{2+}\left(C_2\right)\rightleftharpoons Z{n}^{2+}\left(C_1\right)$

Since in concentration cells, same metal and its solution is present.
Cathode and anode consist of same electrode.
$E_{cathode}^0=E_{anode}^0$
For all concentrationc cells
$E_{cell}^0=E_{cathode}^0-E_{anode}^0$
$E_{cell}^0=0$
Putting the value in Nernst equation at 298K :
$E_{cell}=0-\frac{0.0591}{n}\log \frac{C_1}{C_2}$

For a spontaneous reaction to take place,
$E_{cell}>0$
$-\frac{0.0591}{n}\log \frac{C_1}{C_2}>0$
For
$C_2>C_1$
the above equation is valid.
So, (b) and (d) are correct