Understanding Reduction Potential of Hydrogen Half-Cell

The reduction potential for a half-cell reaction is given by the Nernst equation. For the hydrogen half-cell, the reduction reaction is:

$2H^{+}\left(\mathrm{aq}\right)+2e^{-}\to H_2\left(g\right)$

The standard reduction potential, E°, for this reaction is defined as 0 V under standard conditions: p(H₂) = 1 atm and [H⁺] = 1.0 M.

The Nernst equation for the hydrogen electrode is:

$E=E^{°}-\frac{0.059}{2}\log \frac{p\left(H_2\right)}{[H^{+}{]}^2}$ at 298 K

Since E° = 0, this simplifies to:

$E=-\frac{0.059}{2}\log \frac{p\left(H_2\right)}{[H^{+}{]}^2}$

The reduction potential E will be negative when the log term is positive. This happens when the argument of the log is greater than 1:

$\frac{p\left(H_2\right)}{[H^{+}{]}^2}>1$

Let's evaluate each option to see which one satisfies this condition.

Step-by-Step Evaluation

Step 1: Analyze Option 1

p(H₂) = 2 atm, [H⁺] = 1.0 M

Calculate the ratio: $\frac{p\left(H_2\right)}{[H^{+}{]}^2}=\frac{2}{(1{)}^2}=\frac{2}{1}=2$

Since 2 > 1, the log is positive, and E will be negative.

Step 2: Analyze Option 2

p(H₂) = 1 atm, [H⁺] = 2.0 M

Calculate the ratio: $\frac{p\left(H_2\right)}{[H^{+}{]}^2}=\frac{1}{(2{)}^2}=\frac{1}{4}=0.25$

Since 0.25 < 1, the log is negative, and E will be positive.

Step 3: Analyze Option 3

p(H₂) = 2 atm, [H⁺] = 2.0 M

Calculate the ratio: $\frac{p\left(H_2\right)}{[H^{+}{]}^2}=\frac{2}{(2{)}^2}=\frac{2}{4}=0.5$

Since 0.5 < 1, the log is negative, and E will be positive.

Step 4: Analyze Option 4

p(H₂) = 1 atm, [H⁺] = 1.0 M

Calculate the ratio: $\frac{p\left(H_2\right)}{[H^{+}{]}^2}=\frac{1}{(1{)}^2}=\frac{1}{1}=1$

Since log(1) = 0, E = 0 V (the standard state).

Final Answer: The reduction potential will be negative for the first option: p(H₂) = 2 atm and [H⁺] = 1.0 M.

Related Topics & Formulae

Key Concept: Nernst Equation

The Nernst equation is used to calculate the electrode potential of a half-cell under non-standard conditions. The general form is:

$E=E^{°}-\frac{RT}{nF}\ln Q$

Where:
E = Cell potential under non-standard conditions
E° = Standard cell potential
R = Universal gas constant (8.314 J/mol·K)
T = Temperature in Kelvin
n = Number of moles of electrons transferred in the reaction
F = Faraday's constant (96485 C/mol)
Q = Reaction quotient

At 298 K (25°C), this simplifies to:

$E=E^{°}-\frac{0.059}{n}\log Q$

Standard Hydrogen Electrode (SHE)

The Standard Hydrogen Electrode is the primary reference electrode with a defined potential of 0 V. It consists of a platinum electrode immersed in a 1 M H⁺ solution, with hydrogen gas bubbled at 1 atm pressure.

Reaction Quotient (Q)

For a general reduction reaction: $aA+bB+n{e}^{-}\to cC+dD$

The reaction quotient Q is given by: $Q=\frac{\left[C{]}^c\right[D{]}^d}{\left[A{]}^a\right[B{]}^b}$

For gases, concentrations are replaced by partial pressures.