Q.Explain why Cu+ ion is not stable in aqueous solutions?
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Standard Reduction Potential
Cell Representation and the Nernst Equation: From Intuition to Precision
Imagine you have a Daniell cell — a zinc rod in zinc sulphate solution connected by a salt bridge to a copper rod in copper sulphate solution. You know it produces a voltage. But what happens if you dilute the copper sulphate solution? Or if you change the temperature? The voltage changes. The Nernst equation is the tool that tells you exactly how much it changes.
The Intuition First
A battery works because the two half-cells "want" to react — zinc wants to lose electrons, copper ions want to gain them. This "want" is measured as a tendency, or potential. But the strength of that tendency depends on how crowded the ions are.
Think of it like this: If you have a room full of people who all want to leave (like zinc ions wanting to form), the push to get out is stronger when the room is packed. If the room is nearly empty, the push is weaker. Similarly, for copper ions wanting to enter the metal (gain electrons), the pull is stronger when there are many copper ions around, and weaker when there are few.
The Nernst equation quantifies this: the actual cell potential depends on the concentrations (or activities) of the ions involved.
The Precise Statement
For a general cell reaction:
aA+bB→cC+dD
The cell potential E under non-standard conditions is given by:
E=E∘−nFRTlnQ
Where:
- E = cell potential under the given conditions (in volts)
- E∘ = standard cell potential (when all reactants/products are at 1 M, 1 atm, 25°C)
- R = universal gas constant (8.314 J/mol·K)
- T = temperature in Kelvin
- n = number of moles of electrons transferred in the balanced half-reactions
- F = Faraday constant (96,485 C/mol)
- Q = reaction quotient = [A]a[B]b[C]c[D]d (using concentrations for dilute solutions)
At 25°C (298 K), the equation simplifies to a very practical form:
E=E∘−n0.0591log10Q
The 0.0591 comes from F2.303RT at 298 K. Notice it uses log10 (common log), not natural log.
Cell Representation: How We Write It
In electrochemistry, we represent a cell with a shorthand notation. For the Daniell cell:
Zn(s)∣Zn2+(aq)∥Cu2+(aq)∣Cu(s)
The single vertical line ∣ represents a phase boundary (solid electrode | solution). The double line ∥ represents the salt bridge.
The anode (oxidation) is written on the left, the cathode (reduction) on the right. Electrons flow from left to right in the external circuit.
Applying the Nernst Equation to a Cell Representation
For the Daniell cell, the half-reactions are:
- Anode (oxidation): Zn(s)→Zn2+(aq)+2e−
- Cathode (reduction): Cu2+(aq)+2e−→Cu(s)
Overall: Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)
Here n=2 (two electrons transferred). The reaction quotient is:
Q=[Cu2+][Zn2+]
So the Nernst equation becomes:
E=E∘−20.0591log10[Cu2+][Zn2+]
Solids (Zn, Cu) do not appear in Q because their concentrations are constant (activity = 1).
A Worked Example
Suppose you have a Daniell cell where [Zn2+]=0.1 M and [Cu2+]=1.0 M at 25°C. E∘ for the cell is 1.10 V. …
Why this formula?
Standard Reduction Potential: Why the Formula Holds
Let's build this from first principles — understanding why before memorising what.
1. The Core Idea: A Half-Cell's "Tendency to Gain Electrons"
A standard reduction potential (E∘) measures how strongly a species wants to gain electrons (be reduced) under standard conditions (1 M concentration, 1 atm pressure, 25°C).
But why can't we measure this directly? Because every reduction must be paired with an oxidation — you can't have electrons flowing without a complete circuit.
2. The Formula: Ecell∘=Ecathode∘−Eanode∘
Why subtraction, not addition?
Consider a Daniell cell:
- Zn | Zn²⁺ (1 M) || Cu²⁺ (1 M) | Cu
Experimentally, we measure the cell potential as +1.10 V.
Now, we define the standard hydrogen electrode (SHE) as exactly 0.00 V:
2H++2e−→H2E∘=0.00 V
The reasoning step-by-step:
- We can only measure differences — like measuring height difference between two points.
- If we connect the SHE to the copper half-cell, we measure +0.34 V (Cu²⁺ is reduced).
- If we connect the SHE to the zinc half-cell, we measure −0.76 V (Zn²⁺ is reduced less readily than H⁺).
Now, the cell potential is the difference in their tendencies:
Ecell∘=ECu∘−EZn∘=(+0.34)−(−0.76)=+1.10 V
Key insight: The formula uses subtraction because we're comparing two half-cells against the same reference (SHE). The cathode is where reduction happens (higher E∘), the anode is where oxidation happens (lower E∘).
3. The Nernst Equation: Why E=E∘−nFRTlnQ
This is the thermodynamic derivation — the real "why."
From Gibbs free energy:
ΔG=ΔG∘+RTlnQ
For an electrochemical cell:
ΔG=−nFEandΔG∘=−nFE∘
Substituting:
−nFE=−nFE∘+RTlnQ
Rearranging:
E=E∘−nFRTlnQ
Why this makes physical sense:
- RTlnQ represents the entropy penalty of non-standard concentrations
- nF converts charge to energy (Faraday's constant × number of electrons)
- The minus sign means: as products accumulate (Q increases), the cell potential drops — the reaction is approaching equilibrium
At equilibrium (Q=K), E=0 — the battery is "dead."
--- …
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 Cu2+ and metallic Cu. This happens because the standard reduction potential for Cu2+/Cu+ is more positive than that for Cu+/Cu, making the overall reaction thermodynamically favourable.
The two relevant half-reactions are:
- Cu2++e−→Cu+ (E∘=+0.153 V)
- Cu++e−→Cu (E∘=+0.521 V)
For the disproportionation reaction 2Cu+→Cu2++Cu, the cell potential is:
Ecell∘=Ecathode∘−Eanode∘=0.521−0.153=+0.368 V …
The key idea is that Cu+ in water spontaneously disproportionates into Cu2+ and Cu because the standard reduction potential for Cu2+/Cu+ is more positive than that for Cu+/Cu, making the reaction thermodynamically favourable. The final result: Cu+ is unstable in water and undergoes disproportionation.
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∘). A more positive E∘ means the species is more easily reduced (it acts as a stronger oxidising agent). Conversely, a more negative E∘ means the species is more easily oxidised (it acts as a stronger reducing agent).
For copper, we have two key half-reactions:
- Cu2+(aq)+e−→Cu+(aq) with E∘=+0.153V
- Cu+(aq)+e−→Cu(s) with E∘=+0.521V
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.
A common mistake is to compare the E∘ values directly without considering the direction of the reaction. Remember: E∘ is for reduction. A higher E∘ means the species on the left is more easily reduced. So Cu+ (in the second reaction) is more easily reduced to Cu than Cu2+ is reduced to Cu+. This means Cu+ is a stronger oxidising agent than Cu2+.
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
- 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:
2Cu+(aq)→Cu2+(aq)+Cu(s)
We need to check if this reaction is thermodynamically favourable.
2. Break the reaction into two half-reactions.
- Oxidation: Cu+(aq)→Cu2+(aq)+e− (reverse of reduction 1)
Eox∘=−0.153V (sign reversed)
- Reduction: Cu+(aq)+e−→Cu(s) (reduction 2)
Ered∘=+0.521V
- Calculate the standard cell potential. For the overall disproportionation, the cell potential is:
Ecell∘=Ered∘+Eox∘=0.521V+(−0.153V)=+0.368V
> [!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 Ecell∘ means the reaction is spontaneous under standard conditions (since ΔG∘=−nFEcell∘). Therefore, Cu+ ions in water will spontaneously convert into Cu2+ and metallic copper.
- 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∘) of the two relevant half-reactions.
Steps to Solve
Step 1: Write the two possible half-reactions involving CuX+
For copper(I) ion in water, the relevant equilibria are:
-
Reduction to copper metal:
CuX++eX−Cu
E∘=+0.52 V
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Oxidation to copper(II) ion:
CuX+CuX2++eX−
(This is the reverse of reduction: CuX2++eX−CuX+)
E∘=+0.16 V
Step 2: Combine the two half-reactions to get the disproportionation reaction
Add the reduction half-reaction and the oxidation half-reaction:
- Reduction: CuX++eX−Cu
- Oxidation: CuX+CuX2++eX−
Overall:
2CuX+Cu+CuX2+
Step 3: Calculate the cell potential (Ecell∘) for this reaction
For a spontaneous reaction, Ecell∘>0.
- The reduction half-reaction has Ered∘=+0.52 V
- The oxidation half-reaction has Eox∘=−0.16 V (reverse sign of the reduction potential for CuX2+/CuX+) …
Common Mistakes: Instability of Cu+ in Aqueous Solutions
Mistake 1: Confusing Cu+ with Cu2+ Stability
The error: Students often assume Cu+ is stable because copper commonly shows +1 oxidation state in solid compounds (like Cu2O, 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:
Cu2++e−Cu++e−→Cu+E∘=+0.153V→Cu(s)E∘=+0.521V
The net disproportionation reaction is:
2Cu+→Cu2++Cu(s)
Calculate Ecell∘:
Ecell∘=Ecathode∘−Eanode∘=(+0.521)−(+0.153)=+0.368V
Key result: A positive Ecell∘ means the reaction is spontaneous — Cu+ will spontaneously convert to Cu2+ 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:
- Write both half-reactions as reductions (as given in standard tables)
- Identify which species is being oxidized (loses electrons) — that's the anode
- Identify which species is being reduced (gains electrons) — that's the cathode
- Use: Ecell∘=Ecathode∘−Eanode∘
For Cu+ disproportionation:
- Cu+→Cu2++e− (oxidation, anode) — reverse of the first reduction
- Cu++e−→Cu(s) (reduction, cathode)
So: Ecell∘=(+0.521)−(+0.153)=+0.368V
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 Cu2+ ion is more stable in water because:
- Cu2+ has higher hydration enthalpy (more negative ΔHhyd) due to its smaller size and higher charge
- The Jahn-Teller distortion in Cu2+ 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. …
- TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQQ.Copper matte is a mixture of (A) Oxides of Cu and Fe (B) Carbonates of Cu and Fe (C) Sulphides of Cu and Fe (D) Silicates of Cu and Fe
›Reveal solutionSolution
Copper matte is the intermediate product in copper smelting, consisting mainly of copper and iron sulphides, so the correct answer is (C).
The key concept here is the pyrometallurgical extraction of copper. In the smelting of copper ores (typically chalcopyrite, CuFeS2), the ore is first concentrated by froth flotation, then roasted and smelted in a furnace. The purpose of smelting is to separate the desired metal values from unwanted gangue (rock). The molten product that separates from the slag is called matte.
Why sulphides?
Copper and iron have a strong affinity for sulphur. During smelting, the oxides of these metals (if present) react with remaining sulphur or with added sulphide minerals to form a liquid sulphide phase. This molten sulphide mixture is immiscible with the lighter slag (which contains silicates and oxides). The matte layer sinks to the bottom and is tapped off for further processing (converting to blister copper).
Step-by-step reasoning:
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Identify the process context: The question is about "copper matte," which is a specific term from the metallurgy of copper. It is not the final pure copper, but an intermediate product.
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Recall the composition of typical copper ores: The most common copper ore is chalcopyrite (CuFeS2). It contains copper, iron, and sulphur. Other ores like bornite (Cu5FeS4) also contain these three elements.
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Understand the smelting reaction: When the concentrated ore is smelted with silica (flux) at high temperature, the iron oxide reacts with silica to form slag (iron silicate), while the copper(I) sulphide and remaining iron(II) sulphide form a separate liquid layer — the matte. The overall reaction can be simplified as:
CuFeS2+O2+SiO2→Cu2S⋅FeS (matte)+FeSiO3 (slag)+SO2
The matte is essentially a solution of Cu2S and FeS.
- Eliminate other options:
- (A) Oxides of Cu and Fe: Oxides are not the main components of matte; they are either reduced or slagged off. Copper oxide would be reduced by sulphide or by added coke. …
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- TG EAPCET 2022Set ap-2022-07-30-FN1 markMCQQ.A substance which loses its water of crystallisation [ex: CuSOX4⋅2HX2O] on exposure to the atmosphere is called (A) hygroscopic (B) deliquescent (C) efflorescent (D) isomorphous
›Reveal solutionSolution
The key idea is that a substance that spontaneously loses its water of crystallisation when exposed to air is called efflorescent. The correct answer is (C).
Concept & Intuition
This question tests your understanding of three common terms describing how solids interact with atmospheric moisture.
- Hygroscopic substances absorb moisture from the air but do not necessarily change their physical state (e.g., silica gel).
- Deliquescent substances absorb so much moisture that they dissolve in it, forming a solution (e.g., calcium chloride).
- Efflorescent substances do the opposite: they lose their own water of crystallisation to the air, often crumbling or becoming powdery (e.g., washing soda, NaX2COX3⋅10HX2O).
- Isomorphous refers to substances that have the same crystal structure, not a moisture-related property.
The question explicitly says “loses its water of crystallisation on exposure to the atmosphere” — that is the textbook definition of efflorescence.
Step-by-step reasoning
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Identify the process described
The substance is losing water that is chemically bound in its crystal structure (water of crystallisation) simply by being in open air. This is a spontaneous dehydration driven by a lower vapour pressure of the hydrated crystal compared to the surrounding air.
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Match the process to the correct term
- Efflorescence is exactly this: hydrated salts that give up water to the atmosphere.
- Hygroscopy involves gaining water from air, not losing it.
- Deliquescence is an extreme form of hygroscopy where the substance becomes wet and dissolves.
- Isomorphism is unrelated to water loss.
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Eliminate the other options
- (A) Hygroscopic: Incorrect — these absorb moisture. …
- TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQQ.Lyophilic sols are more stable than lyophobic sols, because (A) The colloidal particles are not solvated (B) There is a strong electrostatic interaction between the colloidal particles (C) The colloidal particles have no change (D) The Brownian movement
›Reveal solutionSolution
Lyophilic sols are more stable than lyophobic sols primarily because their particles are heavily solvated (surrounded by a solvent layer), which prevents coagulation — the correct option is (B) is not correct; the actual reason is solvation, so the answer is (A) is false, (C) is false, (D) is irrelevant; the key is that lyophilic particles are solvated, giving them stability.
Concept & Intuition
Colloidal stability is about keeping particles from clumping together (coagulating). Lyophilic (“solvent-loving”) sols, like gelatin or starch in water, have particles that strongly attract the solvent molecules. This creates a thick, tightly bound solvent layer around each particle — a “solvation shell.” This shell physically prevents particles from getting close enough to stick. In contrast, lyophobic (“solvent-hating”) sols, like gold or silver in water, lack this shell; their stability depends mainly on electrostatic repulsion from surface charges, which is easier to disrupt (e.g., by adding salt). So the primary reason lyophilic sols are more stable is solvation, not electrostatic interactions or Brownian motion.
Step-by-step reasoning
-
Understand the terms
- Lyophilic: particles have a strong affinity for the dispersion medium (e.g., water). They are often macromolecules or highly hydrated.
- Lyophobic: particles have little affinity for the medium; they are typically inorganic or metallic.
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Identify the key stabilizing factor
For lyophilic sols, the dominant stabilizing force is solvation — the solvent molecules form a protective layer around each particle. This layer is often several molecules thick and acts as a physical barrier, preventing particle-particle contact even if electrostatic repulsion is weak.
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Evaluate each option
- (A) “The colloidal particles are not solvated” — This is false. Lyophilic particles are solvated; that’s exactly why they’re stable. So (A) is incorrect.
- (B) “There is a strong electrostatic interaction between the colloidal particles” — Electrostatic repulsion can help, but it’s not the main reason for lyophilic stability. In fact, lyophilic sols often have little or no charge; their stability comes from solvation. So (B) is not the best answer.
- (C) “The colloidal particles have no charge” — Many lyophilic sols are uncharged (e.g., starch), but that doesn’t make them stable; it’s the solvation that matters. Also, some lyophilic sols do carry charge. So (C) is misleading and incorrect.
- (D) “The Brownian movement” — Brownian motion keeps particles suspended but doesn’t explain relative stability between lyophilic and lyophobic sols. Both types exhibit Brownian motion. So (D) is irrelevant.
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Conclude the correct reasoning …
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- TG EAPCET 2021Set ap-2021-08-10-FN1 markMCQQ.Lyophilic sols are more stable than lyophobic sols, because (A) The colloidal particles are not solvated (B) There is a strong electrostatic interaction between the colloidal particles (C) The colloidal particles have no change (D) The Brownian movement
›Reveal solutionSolution
Lyophilic sols are more stable than lyophobic sols primarily because their particles are strongly solvated (coated with a solvent layer), which prevents coagulation — the correct option is (B) is not correct; the actual reason is solvation, so the answer is not among the given choices as stated, but the closest intended answer is (B) if interpreted as "strong interaction" meaning solvation, though the question's options are flawed. The standard answer is (B) in many textbooks, but we must explain carefully.
Watch outMany students pick (A) because they misread "not solvated" — but lyophilic sols are solvated, so (A) is false. Option (C) is also false because lyophilic particles often carry charge. Option (D) is a general property of all colloids, not a distinguishing factor.
Concept and Intuition
The stability of a colloidal solution depends on how well the particles resist coming together (coagulating). Lyophilic ("solvent-loving") sols, like gelatin or starch in water, have particles that are strongly solvated — each particle is surrounded by a thick shell of solvent molecules. This solvation layer acts like a cushion, physically preventing particles from sticking even when they collide. In contrast, lyophobic ("solvent-hating") sols, like gold or silver in water, lack this protective layer; their stability relies mainly on electrostatic repulsion from surface charges, which is more easily disrupted by adding electrolytes.
Thus, the key reason lyophilic sols are more stable is solvation, not electrostatic repulsion or Brownian motion.
Step-by-Step Reasoning
-
Understand the terms
- Lyophilic: particles have a strong affinity for the solvent (e.g., water-loving).
- Lyophobic: particles have little or no affinity for the solvent. In water, lyophilic sols are called hydrophilic, lyophobic are hydrophobic.
-
Identify the main stabilizing factor for each
- Lyophilic sols: stability comes from solvation — solvent molecules form a tightly bound layer around each particle. This layer is often many molecules thick and is thermodynamically favorable.
- Lyophobic sols: stability comes from electrostatic repulsion — particles carry like charges, and the repulsive force prevents aggregation. This is more fragile; adding salt neutralizes charges and causes coagulation.
-
Evaluate each option
- (A) "The colloidal particles are not solvated" — This is false for lyophilic sols; they are solvated. So (A) is incorrect. …
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