Q.Assertion: Cu cannot liberate hydrogen from acids.
Reason: Because it has positive electrode potential.

The key idea is Standard Reduction Potential: a metal can displace hydrogen from acids only if its reduction potential is less than that of hydrogen ($E^\circ = 0.00\ \text{V}$).

1. For copper: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$, $E^\circ = +0.34\ \text{V}$. This is positive (greater than 0 V).
2. A positive reduction potential means Cu is less reactive than H⁺ — it cannot reduce H⁺ to H₂ gas. So Cu does not liberate H₂ from acids. …

When a metal reacts with an acid to liberate hydrogen gas, it's a classic example of a redox reaction. The metal gets oxidized (loses electrons), and the hydrogen ions ($\text{H}^+$) from the acid get reduced (gain electrons) to form hydrogen gas ($\text{H}_2$). The ability of a metal to do this depends on its tendency to lose electrons compared to hydrogen. This tendency is quantified by its standard electrode potential.

The standard electrode potential ($E^\circ$) measures the tendency of a chemical species to be reduced (gain electrons) under standard conditions. A more positive $E^\circ$ value indicates a greater tendency for reduction, while a more negative $E^\circ$ value indicates a greater tendency for oxidation (and thus, a stronger reducing agent). The standard hydrogen electrode (SHE) is the reference point, with its standard reduction potential defined as $0 \text{ V}$.

For a metal to liberate hydrogen from an acid, the metal must be a stronger reducing agent than hydrogen. In other words, the metal must be more easily oxidized than hydrogen.

1. Understand the Redox Process:

   When a metal ($\text{M}$) reacts with an acid ($\text{H}^+$), the overall reaction can be represented as:

   $\text{M}(s) + n\text{H}^+(aq) \rightarrow \text{M}^{n+}(aq) + \frac{n}{2}\text{H}_2(g)$

   This reaction consists of two half-reactions:

   • Oxidation (at anode): $\text{M}(s) \rightarrow \text{M}^{n+}(aq) + n\text{e}^-$
   • Reduction (at cathode): $2\text{H}^+(aq) + 2\text{e}^- \rightarrow \text{H}_2(g)$

2. Relate to Standard Electrode Potentials:

   For a spontaneous reaction, the standard cell potential ($E^\circ_{\text{cell}}$) must be positive.

   $E^\circ_{\text{cell}} = E^\circ_{\text{reduction}}(\text{cathode}) - E^\circ_{\text{reduction}}(\text{anode})$

   In this case, $\text{H}^+$ is reduced at the cathode, and the metal $\text{M}$ is oxidized at the anode.

   The standard reduction potential for hydrogen is $E^\circ(\text{H}^+/\text{H}_2) = 0 \text{ V}$.

   So, for the reaction to be spontaneous:

   $E^\circ_{\text{cell}} = E^\circ(\text{H}^+/\text{H}_2) - E^\circ(\text{M}^{n+}/\text{M}) > 0$

   $0 - E^\circ(\text{M}^{n+}/\text{M}) > 0$

   $-E^\circ(\text{M}^{n+}/\text{M}) > 0$

   This implies that $E^\circ(\text{M}^{n+}/\text{M})$ must be negative.

   Important

   A metal can liberate hydrogen from acids if and only if its standard reduction potential ($E^\circ_{\text{M}^{n+}/\text{M}}$) is negative. This means the metal is more easily oxidized than hydrogen.

3. Apply to Copper:

   The standard reduction potential for copper is $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34 \text{ V}$.

   Since this value is positive ($+0.34 \text{ V} > 0$), copper does not meet the condition for liberating hydrogen from acids. Copper has a greater tendency to be reduced than hydrogen, meaning it is a weaker reducing agent than hydrogen. Therefore, copper cannot reduce $\text{H}^+$ ions to $\text{H}_2$ gas. …

Method: Standard Reduction Potential Analysis

Concept: The ability of a metal to displace hydrogen from acids depends on its standard reduction potential ($E^\circ$).

Steps:

1. Recall the standard hydrogen electrode (SHE) The reference half-reaction is:

2. Compare the metal's reduction potential

   • Metals with negative $E^\circ$ (e.g., Zn, Fe, Mg) can reduce $H^+$ to $H_2$ — they liberate hydrogen.
   • Metals with positive $E^\circ$ (e.g., Cu, Ag, Au) cannot reduce $H^+$ — they do not liberate hydrogen from acids.

3. Check copper's value

   For copper:

Since $+0.34\ \text{V} > 0.00\ \text{V}$, Cu is less reactive than hydrogen.

4. Interpret the assertion and reason …

Common Mistakes Students Make on This Question

Mistake 1: Confusing "Positive Electrode Potential" with "Positive Reduction Potential"

Many students think that because Cu has a positive standard reduction potential ($E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\ \text{V}$), it can reduce H⁺ ions. This is backwards.

Why it's wrong:

• A positive reduction potential means Cu²⁺ prefers to be reduced (gain electrons) — i.e., Cu²⁺ is a better oxidising agent than H⁺.
• To liberate H₂ from acid, Cu must lose electrons (be oxidised). That requires a negative reduction potential (or positive oxidation potential).
• Cu has a positive reduction potential, so it does not readily oxidise to Cu²⁺ — hence it cannot displace H⁺.

How to avoid:

Always check the direction of the reaction. For hydrogen liberation:

This is the reverse of the reduction half-reaction. Use the relation:

So for Cu: $E^\circ_{\text{oxidation}} = -0.34\ \text{V}$ (negative → non-spontaneous).

Mistake 2: Thinking "Positive Potential" Always Means "Reactive"

Some students memorise that metals with negative reduction potentials (like Zn, Mg) liberate H₂, and those with positive potentials (like Cu, Ag) do not — but they don't understand why.

Why it's wrong:

They may choose option (ii) — "Both true but reason is not correct explanation" — because they think the reason is incomplete or unrelated.

How to avoid:

Understand that the reason is the correct explanation:

• Cu has $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\ \text{V}$
• H⁺ has $E^\circ_{\text{H}^+/\text{H}_2} = 0.00\ \text{V}$
• For Cu to reduce H⁺, the cell potential would be:

• Negative $E^\circ_{\text{cell}}$ → non-spontaneous → Cu cannot liberate H₂.

So the positive electrode potential of Cu directly explains why it cannot displace hydrogen.

Mistake 3: Choosing Option (iii) — "Assertion False, Reason True"

Some students think Cu can liberate H₂ from acids (e.g., with concentrated HNO₃ or H₂SO₄), so they mark assertion as false.

Why it's wrong:

• With oxidising acids like HNO₃, the H⁺ is not reduced — instead, the nitrate ion ($\text{NO}_3^-$) is reduced.
• The assertion specifically says "from acids" — in standard chemistry context, this means dilute non-oxidising acids (HCl, dilute H₂SO₄).
• Cu does not liberate H₂ from dilute HCl or dilute H₂SO₄.

How to avoid:

Remember the activity series:

• Metals above H₂ (negative $E^\circ$) displace H₂.
• Metals below H₂ (positive $E^\circ$) do not. Cu is below H₂ → assertion is true.

Mistake 4: Misinterpreting "Positive Electrode Potential" as "Positive Oxidation Potential" …