Question

A voltaic cell is constructed in which the cathode is a standard hydrogen electrode and the anode is a hydrogen electrode ()= 1atm) immersed in a solution of unknown [H+]. If the cell potential is 0.238 V, what is the pH of the unknown solution at 298 K?

Answer 1

Explanation:

It is known that relation between $E_{cell}$, $E^{o}_{cell}$, and pH is as follows.

          $E_{cell} = E^{o}_{cell} - (\frac{0.0591}{n}) \times log[H^{+}]$

Also, it is known that $E^{o}_{cell}$ for hydrogen is equal to zero.

Hence, substituting the given values into the above equation as follows.

     $E_{cell} = E^{o}_{cell} - (\frac{0.0591}{n}) \times log[H^{+}]$

         0.238 V = 0 - (\frac{0.0591}{1}) \times log[H^{+}] [/tex]

                  $-log[H^{+}]$ = 4.03

                         $[H^{+}]$ = antilog 4.03

                                           = 3.5

As, pH = $-log[H^{+}]$.

Thus, we can conclude that pH of the given unknown solution at 298 K is 3.5.