Question

A cell is formed by immersing two $Cd$ electrodes in $CdS{O}_4$ solutions having concentrations of $0.1M$ and $0.5M$ at ${25}^{\circ }C$.
Which of the following(s) is/are true for the given cell?

Take:
$log5\approx 0.7$

• A. $E_{cell}=0.02$
• B. $E_{cell}=0.20$
• C. Cell representation for spontaneous cell is
  $Cd\left(s\right)|C{d}^{2+}(aq,0.1M)||C{d}^{2+}(aq,0.5M)|Cd\left(s\right)$
• D. Cell representation for spontaneous cell is
  $Cd\left(s\right)|C{d}^{2+}(aq,0.5M)||C{d}^{2+}(aq,0.1M)|Cd\left(s\right)$

Answer 1

Answer: (A), (C)

Cell representation for spontaneous cell is

$Cd\left(s\right)|C{d}^{2+}(aq,0.1M)||C{d}^{2+}(aq,0.5M)|Cd\left(s\right)$
In concentration cell, for a spontaneous process, the electrode with higher concentration of electrolyte will act as cathode and lower concentration half cell will act as anode.

Thus, the cell represented as,
$Cd\left(s\right)|C{d}^{2+}(aq,0.1M)||C{d}^{2+}(aq,0.5M)|Cd\left(s\right)$

The electrode half cell reaction will be
$C{d}^{2+}(aq,0.5M)+2e^{-}\to Cd\left(s\right)$
$Cd\left(s\right)\to C{d}^{2+}(aq,0.1M)+2e^{-}$

The cell reaction will be
$C{d}^{2+}(aq,0.5M)\to C{d}^{2+}(aq,0.1M)$

By Nernst equation,

$E_{cell}=E_{cell}^0-\frac{0.059}{2}log\frac{0.1}{0.5}$

For concentration cells,
$E_{cell}^0=0$

$E_{cell}=-\frac{0.059}{2}log\frac{0.1}{0.5}$

$E_{cell}=\frac{0.059}{2}log\frac{0.5}{0.1}$

$E_{cell}=-\frac{0.059}{2}log5$

$E_{cell}=\frac{0.059}{2}\times 0.7$

$E_{cell}=0.02V$