Question

In the cell $Pt\left(s\right),H_2\left(g\right)|1barHCl\left(aq\right)|Ag\left(s\right)|Pt\left(s\right)$ the cell potential is $0.92$ when ${10}^{-6}$ molal $HCl$ solution is used. The standard electrode potential of $(AgCl/Ag,C{l}^{-})$ electrode is:

[Given: $\frac{2.303RT}{F}=0.06V$ at $298K$]

• A. 0.20V
• B. 0.076V
• C. 0.040V
• D. 0.94V

Answer 1

The correct option is A 0.20V

The half-cell reactions are,

At Anode:$\frac{1}{2}{H}_2\left(g\right)\to H^{+}\left(aq\right)+e^{-}$

At Cathode:$AgCl\left(s\right)+e^{-}\to Ag\left(s\right)+C{l}^{-}\left(aq\right)$

Complete reaction:$AgCl\left(s\right)+\frac{1}{2}{H}_2\left(g\right)\to Ag\left(s\right)+C{l}^{-}\left(aq\right)+H^{+}\left(aq\right)$

We know,

$E_{cell}^0=E_{cathode}^0-E_{anode}^0=(SRP{)}_{cathode}-(SRP{)}_{anode}$

We know standard hydrogen potential is assumed to be zero.

So,$(SRP{)}_{anode}=0$

Let, $(SRP{)}_{cathode}=x$

So,

$E_{cell}^0=x$

Now we use the Nernst equation,

$E_{cell}=E_{cell}^0-\frac{2.303RT}{nF}log\left(Q\right)$

$⟹E_{cell}=E_{cell}^0-0.06\times log\left(\right[C{l}^{-}\left]\right[H^{+}\left]\right)$

$n=1$;

$0.92=x-\frac{0.06}{1}log({10}^{-6}\times {10}^{-6})$

$⟹x=0.20V$