Calculate the pH for the hydrogen half-cell of the following voltaic cell if the observed cell voltage is 1.00 V.
Mn|Mn2+(1.00 M)||H+(pH =?), H2(1.00 atm)|Pt
To calculate the pH for the hydrogen half-cell in the given voltaic cell, you need to use the Nernst equation. The Nernst equation relates the pH of a solution to the standard cell potential (E°) and the observed cell potential (E) of the cell.
The Nernst equation for the hydrogen half-cell is:
E = E° - (0.0592/n) * log [H+]
Where:
E = observed cell potential
E° = standard cell potential (for the complete reaction)
n = number of electrons transferred in the half-reaction
[H+] = concentration of H+ ions (in mol/L)
In this case, the observed cell potential (E) is given as 1.00 V. The standard cell potential (E°) can be determined from a reference table or calculated using the standard electrode potentials. For the hydrogen half-cell, E° is 0 V.
The hydrogen half-cell reaction is:
2H+(aq) + 2e- -> H2(g)
From the balanced reaction, we see that two electrons (n=2) are transferred.
Now, let's plug in the values into the Nernst equation:
1.00 V = 0 V - (0.0592/2) * log [H+]
We can rearrange the equation:
0.0592 * log [H+] = -1.00 V
Now, solve for the concentration of H+ by taking the antilog of both sides:
[H+] = 10^(-1.00/0.0592)
[H+] = 0.072 M
To calculate the pH, we need to take the negative logarithm of the H+ concentration:
pH = -log [H+]
pH = -log (0.072)
pH ≈ 1.14
Therefore, the pH for the hydrogen half-cell in the given voltaic cell is approximately 1.14.