Q.Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

The key idea is the Nernst equation applied to the hydrogen electrode, where the potential depends on the concentration (activity) of $H^+$ ions, which is directly related to pH.

For the hydrogen electrode reaction: $2H^+ + 2e^- \rightarrow H_2(g)$, the Nernst equation at 298 K is:

Given $E^\circ = 0 \, \text{V}$ (standard hydrogen electrode) and $P_{H_2} = 1 \, \text{atm}$ (standard condition), this simplifies to:

Substituting pH = 10:

The hydrogen electrode is the reference point for all electrochemistry — its potential under standard conditions (1 M H⁺, 1 atm H₂, 298 K) is defined as exactly 0 V. But when the solution isn't acidic, the H⁺ concentration changes, and so does the electrode's potential. The Nernst equation tells us exactly how.

For the half‑cell reaction

the Nernst equation at 298 K is:

Here $E^\circ = 0$ V, $n = 2$, and we take $P_{\text{H}_2} = 1$ atm (standard pressure). That simplifies things beautifully.

1. Substitute the known values

2. Simplify the log term

So

3. Connect to pH By definition, $\text{pH} = -\log [\text{H}^+]$, so $\log [\text{H}^+] = -\text{pH}$. Therefore

This is a clean, linear relationship: every increase of 1 pH unit makes the hydrogen electrode potential more negative by 59.1 mV.

4. Plug in pH = 10

Method: Nernst Equation for a Hydrogen Electrode

This is a direct application of the Nernst equation to a single electrode (half-cell), not a full cell.

Step 1: Write the half-cell reaction

For a hydrogen electrode:

Step 2: Write the Nernst equation for this half-cell

The general Nernst equation at 298 K (25°C) is:

Where:

• $E^\circ$ = standard electrode potential = 0 V (by definition for SHE)
• $n$ = number of electrons transferred = 2
• $Q$ = reaction quotient = $\frac{P_{H_2}}{[H^+]^2}$

Step 3: Substitute conditions

Given pH = 10, so:

For a standard hydrogen electrode, $P_{H_2} = 1 \text{ atm}$.

Therefore:

Step 4: Calculate the potential

Final Answer

The potential of the hydrogen electrode at pH 10 is $\mathbf{-0.591\ V}$.

Key Concept Check

• The negative potential makes sense: at pH 10 (basic), $H^+$ concentration is very low, so the reduction of $H^+$ to $H_2$ is less favourable than under standard conditions — hence the potential is lower (more negative).
• If you ever forget the formula, remember: $E = -0.0591 \times \text{pH}$ for a hydrogen electrode at 25°C (since $n=1$ per $H^+$ in the simplified form). Here: $E = -0.0591 \times 10 = -0.591\ V$ ✓

Common Mistakes: Potential of Hydrogen Electrode at pH 10

The Correct Approach First

For a hydrogen electrode, the half-cell reaction is:

The Nernst equation for this electrode (at 298 K) is:

Since $E^\circ = 0 \, \text{V}$ and $P_{H_2} = 1 \, \text{atm}$ (standard conditions), this simplifies to:

At pH = 10: $E = -0.591 \, \text{V}$

✗ Mistake 1: Forgetting the Negative Sign

What students do: They calculate $E = +0.0591 \times \text{pH}$ and write $+0.591 \, \text{V}$.

Why it's wrong: The Nernst equation gives $E = -0.0591 \times \text{pH}$. At high pH (low $[H^+]$), the reduction potential becomes more negative — the electrode is less likely to gain electrons.

How to avoid: Always write the Nernst equation in full before simplifying. The negative sign comes from $\log(1/[H^+]) = -\log[H^+]$.

✗ Mistake 2: Using the Wrong Form of the Nernst Equation

What students do: They use $E = E^\circ - \frac{0.0591}{n} \log Q$ but write $Q = \frac{[H^+]^2}{P_{H_2}}$ (inverted).

Why it's wrong: For the reduction half-reaction $2H^+ + 2e^- \rightarrow H_2$, the reaction quotient is:

Products over reactants (excluding solids and pure liquids), with gases in atm.

How to avoid: Memorise the pattern: products (gases, aqueous) / reactants (aqueous, gases). For reduction, products are on the right side of the half-reaction.

✗ Mistake 3: Confusing pH with $[H^+]$

What students do: They substitute pH = 10 directly into the equation without converting.

Why it's wrong: The Nernst equation uses $[H^+]$ in mol/L, not pH. You must use:

How to avoid: Write the conversion step explicitly: $\text{pH} = 10 \Rightarrow [H^+] = 10^{-10} \, \text{M}$.

✗ Mistake 4: Using $n = 1$ Instead of $n = 2$

What students do: They write $E = -0.0591 \times \text{pH}$ but think it's because $n=1$.

Why it's wrong: The half-reaction involves 2 electrons ($2e^-$), so $n=2$. The simplification works because:

The factor of 2 from $\log(1/[H^+]^2) = 2\log(1/[H^+])$ cancels with $n=2$.

How to avoid: Always state $n$ from the balanced half-reaction before simplifying.

✗ Mistake 5: Forgetting Standard Conditions for $P_{H_2}$

What students do: They assume $P_{H_2}$ is not 1 atm and try to include it.

Why it's wrong: The problem states "hydrogen electrode" — by convention, this means $P_{H_2} = 1 \, \text{atm}$ unless specified otherwise.

How to avoid: Remember: standard hydrogen electrode (SHE) always uses $P_{H_2} = 1 \, \text{atm}$ and $[H^+] = 1 \, \text{M}$ for $E^\circ = 0 \, \text{V}$.

✓ Quick Checklist to Avoid All Mistakes

Final answer: $\boxed{-0.591 \, \text{V}}$