Consider the following cell reaction: Mn(s) + 2 H+(? M) -> Mn2+(1.00 M) + H2(g)(1.00 atm)If the cell potential at 298 K is 0.881 volts, what is the pH of the hydrogen electrode? 

Expert Answers
gsenviro eNotes educator| Certified Educator

We can use the Nernst equation to determine the pH of the anode (which is the hydrogen electrode). The Nernst equation can be written as:

`E = E^0 - (0.0592/n) log_10 ([H^+]^2/p_H_2)`

We also know that hydrogen anode is the reference electrode and hence, 

`E - E^0 = 0.881 V`

Also, n = 2 and `p_H_2` = 1 atm.

Thus, substituting the values in the equation, we get:

`0.881 = -(0.0592/2) log_10([H^+]^2/1)`

solving this equation, we get the concentration of protons as 1.313 x 10^-15 M.

This can be converted to pH by using the following relation:

`pH = -log_10[H^+]`

That is, pH = -log(1.313 x 10^-15) = 14.88

A pH of 14.88 is not possible, as the range of pH is between 0 and 14.

Kindly check the value of cell potential and try it using the example solved here. Suppose the cell potential is 0.511 V. Then, using the example case solved here, we can determine the anode pH to be 8.63.

Hope this helps. 

Further Reading: