For the hydrogen electrode reaction 2H+ + 2e- ⇌ H2(g), the reduction potential becomes negative under which of the following conditions of hydrogen gas pressure (PH2) and hydrogen ion concentration [H+]? (1) PH2 = 2 atm, [H+] = 2.0 M (2) PH2 = 1 atm, [H+] = 2.0 M (3) PH2 = 1 atm, [H+] = 1.0 M (4) PH2 = 2 atm, [H+] = 1.0 M

Introduction / Context:
This electrochemistry question asks you to apply the Nernst equation to the standard hydrogen electrode reaction. The standard hydrogen electrode is defined to have a reduction potential of 0.00 V when hydrogen gas is at 1 atm and the hydrogen ion concentration is 1.0 M at 25°C. Deviations from these conditions cause the electrode potential to become positive or negative. Understanding this helps you predict the behaviour of electrodes under non standard conditions.

Given Data / Assumptions:
- Half cell reaction: 2H+ + 2e- ⇌ H2(g).
- Standard electrode potential E° is defined as 0.00 V for PH2 = 1 atm and [H+] = 1.0 M.
- Four combinations of PH2 (in atm) and [H+] (in mol L^-1) are given.
- Temperature is assumed to be 25°C (298 K) so that the usual form of the Nernst equation applies.

Concept / Approach:
The Nernst equation for this half cell at 25°C is: E = E° - (0.0591 / 2) log( PH2 / [H+]^2 ). Since E° = 0 for the standard hydrogen electrode, the expression simplifies to: E = -0.02955 log( PH2 / [H+]^2 ). The sign of E depends on the sign of the logarithm term. If the argument PH2 / [H+]^2 is greater than 1, log of that argument is positive, E is negative. If the argument is less than 1, log is negative and E is positive. If the argument equals 1, log is zero and E is zero.

Step-by-Step Solution:
Step 1: Write the simplified Nernst equation: E = -0.02955 log( PH2 / [H+]^2 ). Step 2: Evaluate the ratio PH2 / [H+]^2 for each condition. Condition 1: PH2 = 2 atm, [H+] = 2.0 M. Ratio = 2 / (2^2) = 2 / 4 = 0.5. Condition 2: PH2 = 1 atm, [H+] = 2.0 M. Ratio = 1 / (2^2) = 1 / 4 = 0.25. Condition 3: PH2 = 1 atm, [H+] = 1.0 M. Ratio = 1 / (1^2) = 1. Condition 4: PH2 = 2 atm, [H+] = 1.0 M. Ratio = 2 / (1^2) = 2. Step 3: Check whether each ratio is greater than, equal to or less than 1. Condition 1: 0.5 < 1, so log(0.5) is negative and E becomes positive. Condition 2: 0.25 < 1, so log(0.25) is negative and E is again positive. Condition 3: ratio is 1, so log(1) = 0 and E = 0 (standard conditions). Condition 4: 2 > 1, so log(2) is positive, making E negative. Step 4: Therefore, only condition 4 yields a negative electrode potential.

Verification / Alternative check:
You can also reason qualitatively. Increasing PH2 while holding [H+] constant favours the reverse reaction (formation of H+ from H2), which lowers the reduction tendency and makes the reduction potential more negative. Conversely, increasing [H+] while keeping PH2 fixed increases the reduction tendency and leads to a more positive potential. In condition 4, PH2 is higher than standard while [H+] is at standard, so you expect the electrode to be less willing to reduce H+ and thus to have a negative potential relative to the standard hydrogen electrode.

Why Other Options Are Wrong:
Condition 1: Both PH2 and [H+] are above standard, but [H+] is squared in the Nernst expression, making the denominator larger and the ratio less than 1, so E is positive, not negative.
Condition 2: Higher [H+] and standard PH2 give a ratio less than 1, again yielding a positive potential.
Condition 3: Standard conditions by definition give E = 0 V.

Common Pitfalls:
A common error is to forget the stoichiometric coefficient of H+ in the Nernst equation and fail to square [H+]. Another pitfall is to try to decide the sign of E simply by comparing PH2 and [H+] separately rather than calculating the ratio. To avoid these mistakes, always write the correct Nernst expression for the reaction and evaluate the argument of the logarithm carefully.

Final Answer:
The reduction potential of the hydrogen half cell becomes negative under condition 4 only (PH2 = 2 atm and [H+] = 1.0 M).