Q.Explain why $Cu^+$ ion is not stable in aqueous solutions?

The key idea is Standard Reduction Potential and the tendency of $Cu^+$ to undergo disproportionation.

In aqueous solution, $Cu^+$ is unstable because it spontaneously disproportionates into $Cu^{2+}$ and metallic $Cu$. This happens because the standard reduction potential for $Cu^{2+}/Cu^+$ is more positive than that for $Cu^+/Cu$, making the overall reaction thermodynamically favourable.

The two relevant half-reactions are:

1. $Cu^{2+} + e^- \rightarrow Cu^+$ ($E^\circ = +0.153\ \text{V}$)
2. $Cu^+ + e^- \rightarrow Cu$ ($E^\circ = +0.521\ \text{V}$)

For the disproportionation reaction $2Cu^+ \rightarrow Cu^{2+} + Cu$, the cell potential is:

$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.521 - 0.153 = +0.368\ \text{V}$ …

Why This Happens: The Concept of Standard Reduction Potential

The stability of an ion in aqueous solution depends on its tendency to either gain or lose electrons under standard conditions. This tendency is measured by the standard reduction potential ($E^\circ$). A more positive $E^\circ$ means the species is more easily reduced (it acts as a stronger oxidising agent). Conversely, a more negative $E^\circ$ means the species is more easily oxidised (it acts as a stronger reducing agent).

For copper, we have two key half-reactions:

1. $Cu^{2+}(aq) + e^- \rightarrow Cu^+(aq)$ with $E^\circ = +0.153 \, \text{V}$
2. $Cu^+(aq) + e^- \rightarrow Cu(s)$ with $E^\circ = +0.521 \, \text{V}$

Notice that the second half-reaction has a more positive reduction potential than the first — this is the crucial observation, and it leads somewhere that trips many students up.

But here's the twist: if $Cu^+$ is a good oxidising agent, it can oxidise something else. What can it oxidise? Another $Cu^+$ ion! This is the essence of disproportionation — a single species simultaneously oxidises and reduces itself.

Step-by-Step Reasoning

1. Identify the possible reaction. For $Cu^+$ to be stable, it must not spontaneously convert into other species. The most likely reaction in water is disproportionation:

We need to check if this reaction is thermodynamically favourable.

2. Break the reaction into two half-reactions.

- Oxidation: $Cu^+(aq) \rightarrow Cu^{2+}(aq) + e^-$ (reverse of reduction 1)

$E^\circ_{\text{ox}} = -0.153 \, \text{V}$ (sign reversed)

- Reduction: $Cu^+(aq) + e^- \rightarrow Cu(s)$ (reduction 2)

$E^\circ_{\text{red}} = +0.521 \, \text{V}$

3. Calculate the standard cell potential. For the overall disproportionation, the cell potential is:

> [!FORMULA]
> For a disproportionation reaction: $E^\circ_{\text{cell}} = E^\circ_{\text{reduction of the species}} - E^\circ_{\text{reduction of the product}}$  
> Here: $E^\circ_{\text{cell}} = E^\circ_{Cu^+/Cu} - E^\circ_{Cu^{2+}/Cu^+} = 0.521 - 0.153 = +0.368 \, \text{V}$

4. Interpret the sign.

A positive $E^\circ_{\text{cell}}$ means the reaction is spontaneous under standard conditions (since $\Delta G^\circ = -nFE^\circ_{\text{cell}}$). Therefore, $Cu^+$ ions in water will spontaneously convert into $Cu^{2+}$ and metallic copper.

5. Check the equilibrium constant. …

Method: Using Standard Reduction Potentials to Predict Disproportionation

Method Name: Disproportionation Analysis via Standard Electrode Potentials

Concept: A species is unstable in aqueous solution if it can spontaneously undergo disproportionation — where the same element simultaneously oxidises and reduces itself. This is predicted by comparing the standard reduction potentials ($E^\circ$) of the two relevant half-reactions.

Steps to Solve

Step 1: Write the two possible half-reactions involving $\ce{Cu+}$

For copper(I) ion in water, the relevant equilibria are:

• Reduction to copper metal:

  $\ce{Cu+ + e- -> Cu}$

  $E^\circ = +0.52\ \text{V}$

• Oxidation to copper(II) ion:

  $\ce{Cu+ -> Cu^{2+} + e-}$

  (This is the reverse of reduction: $\ce{Cu^{2+} + e- -> Cu+}$)

  $E^\circ = +0.16\ \text{V}$

Step 2: Combine the two half-reactions to get the disproportionation reaction

Add the reduction half-reaction and the oxidation half-reaction:

• Reduction: $\ce{Cu+ + e- -> Cu}$
• Oxidation: $\ce{Cu+ -> Cu^{2+} + e-}$

Overall:

$\ce{2Cu+ -> Cu + Cu^{2+}}$

Step 3: Calculate the cell potential ($E^\circ_{\text{cell}}$) for this reaction

For a spontaneous reaction, $E^\circ_{\text{cell}} > 0$.

• The reduction half-reaction has $E^\circ_{\text{red}} = +0.52\ \text{V}$
• The oxidation half-reaction has $E^\circ_{\text{ox}} = -0.16\ \text{V}$ (reverse sign of the reduction potential for $\ce{Cu^{2+}/Cu+}$) …

Common Mistakes: Instability of $Cu^+$ in Aqueous Solutions

Mistake 1: Confusing $Cu^+$ with $Cu^{2+}$ Stability

The error: Students often assume $Cu^+$ is stable because copper commonly shows +1 oxidation state in solid compounds (like $Cu_2O$, $CuCl$).

Why it's wrong: Stability in solid compounds ≠ stability in aqueous solution. In water, $Cu^+$ undergoes disproportionation — a reaction where the same species simultaneously oxidizes and reduces itself.

How to avoid: Always check the standard reduction potentials for the relevant half-reactions in aqueous medium:

The net disproportionation reaction is:

Calculate $E^\circ_{\text{cell}}$:

Key result: A positive $E^\circ_{\text{cell}}$ means the reaction is spontaneous — $Cu^+$ will spontaneously convert to $Cu^{2+}$ and metallic copper.

Mistake 2: Forgetting to Check the Sign Convention

The error: Students incorrectly subtract reduction potentials or use the wrong half-reaction as cathode/anode.

How to avoid: Follow this systematic approach:

1. Write both half-reactions as reductions (as given in standard tables)
2. Identify which species is being oxidized (loses electrons) — that's the anode
3. Identify which species is being reduced (gains electrons) — that's the cathode
4. Use: $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$

For $Cu^+$ disproportionation:

• $Cu^+ \rightarrow Cu^{2+} + e^-$ (oxidation, anode) — reverse of the first reduction
• $Cu^+ + e^- \rightarrow Cu(s)$ (reduction, cathode)

So: $E^\circ_{\text{cell}} = (+0.521) - (+0.153) = +0.368 \, \text{V}$

Mistake 3: Ignoring the Role of Water

The error: Students think $Cu^+$ simply "decomposes" without understanding water's role.

The truth: Water acts as the medium enabling ion mobility and solvation. The $Cu^{2+}$ ion is more stable in water because:

• $Cu^{2+}$ has higher hydration enthalpy (more negative $\Delta H_{\text{hyd}}$) due to its smaller size and higher charge
• The Jahn-Teller distortion in $Cu^{2+}$ complexes provides additional stabilization

How to avoid: Remember that aqueous stability depends on hydration energy and lattice energy trade-offs. For $Cu^+$, the hydration energy is insufficient to overcome the tendency to disproportionate.

Mistake 4: Memorizing Without Understanding the Trend

The error: Students memorize "$Cu^+$ is unstable" without connecting to the broader periodic trend. …