Q.$E^\ominus$ of Cu is $+0.34\ V$ while that of Zn is $-0.76\ V$. Explain.

The key idea is stability of oxidation states — specifically, how the electronic configuration of an ion influences its tendency to gain or lose electrons.

Reasoning:

1. Copper has the configuration $[Ar]\,3d^{10}4s^1$. The $\ce{Cu+}$ ion ($3d^{10}$) has a completely filled d-subshell, which is highly stable. However, $\ce{Cu^2+}$ ($3d^9$) is also common because the extra hydration energy of $\ce{Cu^2+}$ compensates for the loss of a filled shell. The net result is that $\ce{Cu^2+}$ is more stable in aqueous solution than $\ce{Cu+}$, giving Cu a positive reduction potential.

2. Zinc has the configuration $[Ar]\,3d^{10}4s^2$. The $\ce{Zn^2+}$ ion ($3d^{10}$) achieves a stable, completely filled d-subshell. This makes Zn lose electrons very readily, giving it a strong tendency to oxidise — hence a highly negative reduction potential. …

The question asks why copper has a positive standard reduction potential ($+0.34\ \text{V}$) while zinc has a negative one ($-0.76\ \text{V}$). This is not just a number to memorise — it reflects deep differences in how these metals behave at the atomic level.

The Concept: What $E^\ominus$ Actually Measures

Standard electrode potential measures the tendency of a metal ion to get reduced (gain electrons) relative to the standard hydrogen electrode. A positive $E^\ominus$ means the ion is easily reduced — the metal is noble. A negative $E^\ominus$ means the metal prefers to oxidise (lose electrons) — it is reactive.

But why? The reduction process $M^{2+}(aq) + 2e^- \rightarrow M(s)$ involves several hidden steps. To understand the sign, we need to break the process into its components.

Step-by-Step Reasoning

1. The Born-Haber cycle for reduction

When we write $M^{2+}(aq) + 2e^- \rightarrow M(s)$, we are really combining three energy changes:

• Ionization: $M(g) \rightarrow M^{2+}(g) + 2e^-$ (requires energy — endothermic)
• Hydration: $M^{2+}(g) \rightarrow M^{2+}(aq)$ (releases energy — exothermic)
• Sublimation: $M(s) \rightarrow M(g)$ (requires energy — endothermic)

The overall enthalpy change for reduction is the reverse of the oxidation process. So:

$\Delta H_{\text{reduction}} = -[\Delta H_{\text{sub}} + \text{IE}_1 + \text{IE}_2 + \Delta H_{\text{hyd}}]$

Where $\Delta H_{\text{sub}}$ is sublimation enthalpy, IE are ionization enthalpies, and $\Delta H_{\text{hyd}}$ is hydration enthalpy.

A more negative value inside the bracket means a more positive $E^\ominus$ (easier reduction).

2. Compare Zn and Cu — the key numbers

3. Why is Cu's sum larger?

Two factors dominate:

• Sublimation enthalpy: Cu has a much higher $\Delta H_{\text{sub}}$ (340 vs 130 kJ/mol). Copper atoms are held more tightly in the metallic lattice due to stronger metallic bonding (from d-electrons contributing to bonding). This makes it harder to turn solid Cu into gaseous atoms — which favours the reverse (reduction) direction.

• Ionization enthalpy: Cu has slightly higher IE₁ + IE₂ (2700 vs 2640 kJ/mol). This is because Cu has a $3d^{10}$ configuration after losing one electron, making the second ionization harder. Zn, with $4s^2$ configuration, loses both s-electrons more easily. …

Method: Standard Electrode Potential & Stability Analysis

This question is solved using the Standard Reduction Potential Comparison Method, which links electrode potential values to the relative stability of oxidation states.

Step 1: Recall the meaning of $E^\ominus$

• More positive $E^\ominus$ → the species is easily reduced (acts as a strong oxidising agent).
• More negative $E^\ominus$ → the species is easily oxidised (acts as a strong reducing agent).

Step 2: Write the half-reactions

For copper:

$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$

$E^\ominus = +0.34\ \text{V}$

For zinc:

$\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}$

$E^\ominus = -0.76\ \text{V}$

Step 3: Compare the values

• Cu has a positive $E^\ominus$ → $\text{Cu}^{2+}$ is stable in solution; it does not easily lose electrons to become Cu metal.

  → The +2 oxidation state of copper is stable under standard conditions.

• Zn has a negative $E^\ominus$ → $\text{Zn}$ metal readily loses electrons to form $\text{Zn}^{2+}$.

  → The +2 oxidation state of zinc is even more stable (Zn metal is a strong reducing agent).

Step 4: Final explanation

| Metal | $E^\ominus$ (V) | Tendency to oxidise | Stability of +2 state | …

Here are the common mistakes students make when explaining why the standard electrode potential ($E^\ominus$) of Cu is $+0.34\ V$ and Zn is $-0.76\ V$, along with how to avoid each.

Mistake 1: Confusing "Stability" with "Reactivity" in the Wrong Direction

• The Mistake: Students often say "Cu is more stable, so it has a higher $E^\ominus$" without clarifying what is stable. They might incorrectly think a positive $E^\ominus$ means the metal is harder to oxidize (which is true) but then fail to link it to the electronic configuration. Worse, some think a negative $E^\ominus$ means the ion is stable.
• Why it’s wrong: $E^\ominus$ measures the tendency to get reduced (gain electrons). A positive $E^\ominus$ (Cu) means $\text{Cu}^{2+}$ is easily reduced — so the metal (Cu) is stable, not the ion. A negative $E^\ominus$ (Zn) means $\text{Zn}^{2+}$ is hard to reduce — so the ion ($\text{Zn}^{2+}$) is stable, not the metal.
• How to avoid: Always state the species being discussed.
  • Cu: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$; $E^\ominus = +0.34\ V$ → Cu metal is stable (resists oxidation).
  • Zn: $\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}$; $E^\ominus = -0.76\ V$ → $\text{Zn}^{2+}$ ion is stable (resists reduction).

Mistake 2: Ignoring the Role of Hydration Enthalpy

• The Mistake: Students explain the difference using only ionization enthalpy (IE) or sublimation enthalpy, forgetting that the hydration enthalpy of the ion is a major factor. They might say "Zn has higher IE, so it should be harder to oxidize" — which contradicts the observed $E^\ominus$.

• Why it’s wrong: The overall $E^\ominus$ depends on the sum of three enthalpies:

  • Sublimation (solid → gas)
  • Ionization (atom → ion)
  • Hydration (ion → aqueous ion)

  For Zn, the hydration enthalpy of $\text{Zn}^{2+}$ is very high (due to its small size and $3d^{10}$ configuration), which compensates for its high IE. For Cu, the hydration enthalpy is lower, so despite Cu having a lower IE, the overall process is less favorable.

• How to avoid: Always mention the hydration enthalpy explicitly. Write the Born-Haber cycle for the oxidation process:

The enthalpy change is:

For Zn, $\Delta H_{hyd}$ is very negative, making the overall $\Delta H$ less positive (more favorable oxidation) → more negative $E^\ominus$.

Mistake 3: Forgetting the "Stability of Half-Filled/Full d-Orbitals" Nuance

• The Mistake: Students blindly say "Zn has a full $3d^{10}$ configuration, so it is stable" — but they apply this to the metal instead of the ion. They then get confused why Zn has a negative $E^\ominus$.

• Why it’s wrong: The stability of a full $d^{10}$ subshell applies to $\text{Zn}^{2+}$ (which has $3d^{10}$), not to Zn metal (which has $3d^{10}4s^2$). The $\text{Zn}^{2+}$ ion is stable because it has a filled d-subshell, making it hard to reduce back to Zn. This is exactly why $E^\ominus$ is negative — the ion is stable.

• How to avoid: Draw the electronic configurations:

  • Zn: $[\text{Ar}] 3d^{10} 4s^2$ → loses two $4s$ electrons to form $\text{Zn}^{2+}$ ($3d^{10}$).
  • Cu: $[\text{Ar}] 3d^{10} 4s^1$ → loses one $4s$ electron to form $\text{Cu}^+$ ($3d^{10}$), but $\text{Cu}^{2+}$ ($3d^9$) is more common in aqueous solution due to higher hydration enthalpy.

  The stability of $\text{Zn}^{2+}$ (full d-subshell) makes it reluctant to accept electrons → negative $E^\ominus$.

Mistake 4: Misapplying the "Stability of Cu⁺ vs Cu²⁺"

• The Mistake: Students say "Cu has a positive $E^\ominus$ because $\text{Cu}^+$ is stable" — but the standard electrode potential given is for $\text{Cu}^{2+}/\text{Cu}$, not $\text{Cu}^+/\text{Cu}$.
• Why it’s wrong: The $E^\ominus$ value $+0.34\ V$ is for the half-reaction:

This does not directly involve $\text{Cu}^+$. In fact, $\text{Cu}^+$ is unstable in water (disproportionates to $\text{Cu}^{2+}$ and Cu). The positive $E^\ominus$ arises because $\text{Cu}^{2+}$ has a high hydration enthalpy (due to its smaller size and higher charge) and the metal has a high sublimation enthalpy.

• How to avoid: Always check which half-reaction is being discussed. For $\text{Cu}^{2+}/\text{Cu}$, focus on the d⁹ configuration of $\text{Cu}^{2+}$ and its high hydration enthalpy, not on $\text{Cu}^+$.

Mistake 5: Treating $E^\ominus$ as an Absolute Measure of Stability …