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

Step-by-step explanation:

It is known that relation between
`E_(cell)`,
`E^(o)_(cell)`, and pH is as follows.

`E_(cell) = E^(o)_(cell) - ((0.0591)/(n)) * 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) - ((0.0591)/(n)) * 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.