Question

$Pt|H_2(g,P_1)|HCl\left(0.1M\right)|H_2(g,P_2)|Pt$

A concentration cell is formed by two hydrogen electrodes having hydrogen gas at different pressures $P_1\&P_2$, dipped in the same electrolytic solution $\left(0.1MHCl\right)$ at $298K$.
Choose the correct option :

• A. $E_{cell}^0<0$
• B. Higher the concentration of electrolyte, more is the EMF generated.
• C. For spontaneous reaction to take place,
  $P_1>P_2$
• D. $E_{cell}^0>0$

Answer 1

The correct option is C For spontaneous reaction to take place,

$P_1>P_2$
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
$E_{cell}^0=E_{cathode}^0-E_{anode}^0$
$E_{cell}^0=0$

$Pt|H_2(g,P_1)|HCl\left(0.1M\right)|H_2(g,P_2)|Pt$
Anode:
$H_2(g,P_1)\rightleftharpoons 2H^{+}(aq,xM)+2e^{-}$
Cathode:
$2H^{+}(aq,xM)+2e^{-}\rightleftharpoons H_2(g,P_1)$
Net cell reaction is :
$H_2\left(P_1\right)\rightleftharpoons H_2\left(P_2\right)$
Putting the value in Nernst equation at 298K :
$E_{cell}=0-\frac{0.0591}{2}\log \frac{P_2}{P_1}$
EMF generated is independent of the concentration of the electrolyte $\left(HCl\right)$.

For a spontaneous reaction to take place,

$E_{cell}>0$

$-\frac{0.0591}{2}\log \frac{P_2}{P_1}>0$
For ,
$P_1>P_2$
the above equation is valid.