Answer :
The question is about understanding how the cell potential of the hydrogen electrode is affected when the pH is decreased by one unit. This involves electrochemistry concepts, particularly the Nernst equation.
The Nernst equation for the hydrogen electrode can be expressed as:
[tex]E = E^0 - \frac{RT}{nF} \ln(Q)[/tex]
Where:
- [tex]E[/tex] is the cell potential.
- [tex]E^0[/tex] is the standard electrode potential. For the hydrogen electrode, [tex]E^0 = 0[/tex] V.
- [tex]R[/tex] is the universal gas constant ($8.314$ J/mol·K).
- [tex]T[/tex] is the temperature in Kelvin (at $25, ^\circ[tex]C,[/tex]T = 298$ K).
- [tex]n[/tex] is the number of moles of electrons exchanged (for H⁺/H₂, [tex]n = 2[/tex]).
- [tex]F[/tex] is Faraday's constant ($96485$ C/mol).
- [tex]Q[/tex] is the reaction quotient.
For the hydrogen electrode reaction:
[tex]2H^+(aq) + 2e^- \rightarrow H_2(g)[/tex]
The Nernst equation simplifies to:
[tex]E = - \frac{RT}{2F} \ln([H^+])[/tex]
The pH is related to the hydrogen ion concentration by [tex]pH = -\log_{10}([H^+])[/tex]. So, if the pH decreases by 1 unit, the hydrogen ion concentration [tex][H^+][/tex] increases by a factor of $10$.
Now substituting specific values at $25, ^\circ$C:
- We know [tex]\ln(10) \approx 2.303[/tex].
- Therefore, [tex]\frac{RT}{F} \ln(10) = \frac{8.314 \times 298}{2 \times 96485} \times 2.303 \approx 0.0591[/tex] V.
This means that the decrease in pH by one unit increases the cell potential by $59.1$ mV.
Thus, the correct answer is 1. increases by 59.1 mV.
