The reduction potential of hydrogen half-cell will be negative if
(a) p(H2) = 1 atm and [H+] = 2.0 M

Q: Why is the standard reduction potential ($E^\circ$) for hydrogen exactly zero?

By IUPAC convention, the Standard Hydrogen Electrode (SHE) is chosen as the universal reference electrode. Its potential is arbitrarily assigned a value of exactly $0 \text{ V}$ at all temperatures so that the potentials of all other electrodes can be measured relative to it.

Q: Why did we use $\frac{1}{2}H_2$ instead of $H_2$ in the reaction?

We used $H^+ + e^- \rightarrow \frac{1}{2}H_2$ to make the electron transfer $n=1$. However, if you use $2H^+ + 2e^- \rightarrow H_2$, where $n=2$, the Nernst equation becomes $E = -\frac{0.0591}{2} \log \frac{p_{H_2}}{[H^+]^2}$. Solving for $E [H^+]^2$, which yields the exact same final result!

Q: What happens if the log argument is exactly 1?

If the reaction quotient $Q$ is exactly 1, then $\log(1) = 0$. This makes the entire term zero, and the electrode potential $E$ will be equal to its standard potential $E^\circ$, which is $0 \text{ V}$ for hydrogen.

Q: Does the pressure unit have to be in atm?

Yes, in the standard Nernst equation, the activities of gases are expressed in terms of their partial pressures in atm (or bar) relative to the standard pressure of $1 \text{ atm}$ (or $1 \text{ bar}$).

Q: Why does increasing the $H^+$ concentration make the potential more positive?

According to Le Chatelier's principle, increasing the concentration of the reactant ($H^+$) drives the reduction reaction forward. This increases the electrode's tendency to gain electrons, thereby making its reduction potential more positive.

Question

The reduction potential of hydrogen half-cell will be negative if

Options:

(A) $p(H_2) = 1 	ext{ atm}$ and $[H^+] = 2.0 	ext{ M}$

(B) $p(H_2) = 1 	ext{ atm}$ and $[H^+] = 1.0 	ext{ M}$

(C) $p(H_2) = 2 	ext{ atm}$ and $[H^+] = 1.0 	ext{ M}$

(D) $p(H_2) = 2 	ext{ atm}$ and $[H^+] = 2.0 	ext{ M}$

Sigma Percentile · Accepted Answer

### The Standard Hydrogen Electrode

Imagine you are looking at a Standard Hydrogen Electrode (SHE). It is the universal reference point in electrochemistry. The setup is elegant: a platinum foil is submerged in an acidic solution containing $H^+$ ions, and hydrogen gas ($H_2$) is continuously bubbled over it.

Because it is the reference, the standard reduction potential of the hydrogen electrode, $E^\circ_{red}$, is defined to be exactly **$0 	ext{ V}$** at all temperatures.

### The Master Equation

To find the electrode potential under non-standard conditions (when concentrations aren't $1 	ext{ M}$ or pressures aren't $1 	ext{ atm}$), we rely on the Nernst equation. First, let's write the reduction half-reaction for the hydrogen electrode:

$$H^+(aq) + e^- ightarrow \frac{1}{2}H_2(g)$$

Notice that exactly one electron is transferred, so $n = 1$. The Nernst equation for this half-cell is:

$$E_{red} = E^\circ_{red} - \frac{0.0591}{n} \log_{10} Q$$

Here, the reaction quotient $Q$ is the ratio of the product to the reactant. Since the product is a gas, we use its partial pressure, and for the reactant, we use its molar concentration:

$$Q = \frac{(p_{H_2})^{1/2}}{[H^+]}$$

### Finding the Golden Condition

Let's substitute our known values ($E^\circ_{red} = 0$ and $n = 1$) into the Nernst equation:

$$E_{red} = -0.0591 \log_{10} \left( \frac{(p_{H_2})^{1/2}}{[H^+]} ight)$$

The question asks us to find the condition where the reduction potential $E_{red}$ is **negative**. Look closely at the equation above. There is a negative sign right in front of the logarithm.

For the entire expression to be negative, the logarithmic term itself must be positive. Mathematically, $\log_{10}(x) > 0$ only when $x > 1$. Therefore, our reaction quotient must be strictly greater than one:

$$\frac{(p_{H_2})^{1/2}}{[H^+]} > 1$$

Rearranging this gives us our golden condition:

$$(p_{H_2})^{1/2} > [H^+]$$

### Evaluating the Options

Now, it's just a matter of testing the given options against our golden condition:

**Option (a):** $p(H_2) = 1 	ext{ atm}$, $[H^+] = 2.0 	ext{ M}$
$1^{1/2} = 1$. Is $1 > 2.0$? No. The potential here would be positive.

**Option (b):** $p(H_2) = 1 	ext{ atm}$, $[H^+] = 1.0 	ext{ M}$
$1^{1/2} = 1$. Is $1 > 1.0$? No, it is equal. The potential here would be exactly zero.

**Option (c):** $p(H_2) = 2 	ext{ atm}$, $[H^+] = 1.0 	ext{ M}$
$2^{1/2} \approx 1.414$. Is $1.414 > 1.0$? **Yes!** This satisfies our condition perfectly.

**Option (d):** $p(H_2) = 2 	ext{ atm}$, $[H^+] = 2.0 	ext{ M}$
$2^{1/2} \approx 1.414$. Is $1.414 > 2.0$? No.

Thus, the reduction potential of the hydrogen half-cell will be negative when the pressure of hydrogen gas is $2 	ext{ atm}$ and the concentration of $H^+$ ions is $1.0 	ext{ M}$.

Animated Solution for Chemistry - Electrochemistry: The reduction potential of hydrogen half-cell will be negative if

Select Answer:

Visualized Solution

\text{Standard Hydrogen Electrode}

\text{Reduction Half-Reaction}

\text{Nernst Equation}

\text{Substituting Values}

\text{Condition for } E_{red} < 0

\text{Solving the Inequality}

\text{Evaluating Options}

\text{Conclusion}

The Sigma Insight: Electrochemical Cells

The Standard Hydrogen Electrode

The Master Equation

Finding the Golden Condition

Evaluating the Options

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