Assertion: Cu²⁺ iodide is not known. Reason: Cu²⁺ oxidises I^- to iodine. (i) Both assertion and reason are tr

NCERT Exemplar · Q67

Cu^{2+}

(i) Both assertion and reason are true, and reason is the correct explanation of the assertion.

(ii) Both assertion and reason are true but reason is not the correct explanation of assertion.

(iii) Assertion is not true but reason is true.

(iv) Both assertion and reason are false.

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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 \rightarrow cC + dD

The cell potential $E$ under non-standard conditions is given by:

E = E^\circ - \frac{RT}{nF} \ln Q

Where:

$E$ = cell potential under the given conditions (in volts)
$E^\circ$ = 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 = $\frac{[C]^c [D]^d}{[A]^a [B]^b}$ (using concentrations for dilute solutions)

At 25°C (298 K), the equation simplifies to a very practical form:

E = E^\circ - \frac{0.0591}{n} \log_{10} Q

The 0.0591 comes from $\frac{2.303RT}{F}$ at 298 K. Notice it uses $\log_{10}$ (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:

\text{Zn(s)} \mid \text{Zn}^{2+}(\text{aq}) \parallel \text{Cu}^{2+}(\text{aq}) \mid \text{Cu(s)}

The single vertical line $\mid$ represents a phase boundary (solid electrode | solution). The double line $\parallel$ 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): $\text{Zn(s)} \rightarrow \text{Zn}^{2+}(\text{aq}) + 2e^-$
Cathode (reduction): $\text{Cu}^{2+}(\text{aq}) + 2e^- \rightarrow \text{Cu(s)}$

Overall: $\text{Zn(s)} + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + \text{Cu(s)}$

Here $n = 2$ (two electrons transferred). The reaction quotient is:

Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}

So the Nernst equation becomes:

E = E^\circ - \frac{0.0591}{2} \log_{10} \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}

Note

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 $[\text{Zn}^{2+}] = 0.1$ M and $[\text{Cu}^{2+}] = 1.0$ M at 25°C. $E^\circ$ 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 ( $E^\circ$ ) 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: $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$

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:

2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \quad E^\circ = 0.00 \text{ 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:

E^\circ_{\text{cell}} = E^\circ_{\text{Cu}} - E^\circ_{\text{Zn}} = (+0.34) - (-0.76) = +1.10 \text{ 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 $E^\circ$ ), the anode is where oxidation happens (lower $E^\circ$ ).

3. The Nernst Equation: Why $E = E^\circ - \frac{RT}{nF} \ln Q$

This is the thermodynamic derivation — the real "why."

From Gibbs free energy:

\Delta G = \Delta G^\circ + RT \ln Q

For an electrochemical cell:

\Delta G = -nFE \quad \text{and} \quad \Delta G^\circ = -nFE^\circ

Substituting:

-nFE = -nFE^\circ + RT \ln Q

Rearranging:

E = E^\circ - \frac{RT}{nF} \ln Q

Why this makes physical sense:

$RT \ln Q$ 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."

--- …

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Both the assertion and the reason are true, and the reason is the correct explanation: $CuI_2$ is not known precisely because $Cu^{2+}$ oxidises $I^-$ to $I_2$ (being itself reduced to $Cu^+$ , which separates as insoluble $CuI$ ). The correct option is (i).

Assertion — is it true?

Copper(II) iodide, $CuI_2$ , cannot be isolated as a stable compound. So the assertion is TRUE.

Reason — is it true?

When $Cu^{2+}$ meets $I^-$ , the following redox reaction occurs:

2Cu^{2+} + 4I^- \longrightarrow 2CuI\downarrow + I_2

Here $Cu^{2+}$ is reduced to $Cu^+$ while $I^-$ is oxidised to $I_2$ . So $Cu^{2+}$ does oxidise $I^-$ to iodine — the reason is TRUE.

Although $E^\circ(Cu^{2+}/Cu^+) = +0.15\ \text{V}$ is lower than $E^\circ(I_2/I^-) = +0.54\ \text{V}$ , the reaction is driven forward by the very low solubility of $CuI$ , which removes $Cu^+$ from solution and makes the overall process spontaneous. …

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Common Mistakes & How to Avoid Them

Mistake 1: Thinking $Cu^{2+}$ iodide is known

Why it happens: Students see $CuI_2$ written in some textbooks or recall copper(II) halides like $CuCl_2$ and $CuBr_2$ exist.
The truth: $CuI_2$ is unstable because $Cu^{2+}$ oxidises $I^-$ to $I_2$ , getting reduced to $Cu^+$ itself. The reaction is:

2Cu^{2+} + 4I^- \rightarrow 2CuI \downarrow + I_2

The product is copper(I) iodide ( $CuI$ ), not copper(II) iodide.

How to avoid: Remember the standard reduction potentials:
- $E^\circ(Cu^{2+}/Cu^+) = +0.15\ \text{V}$
- $E^\circ(I_2/I^-) = +0.54\ \text{V}$
Since $E^\circ(I_2/I^-) > E^\circ(Cu^{2+}/Cu^+)$ , $I^-$ can reduce $Cu^{2+}$ to $Cu^+$ . So $CuI_2$ cannot exist.

Mistake 2: Confusing the reason with the explanation

Why it happens: Students see both statements are true and pick option (i) without checking if the reason correctly explains the assertion.
The truth: The reason is the correct explanation. The assertion says " $Cu^{2+}$ iodide is not known" — the reason tells you why: because $Cu^{2+}$ oxidises $I^-$ to iodine.
How to avoid: For assertion-reason questions, always ask: "Does the reason directly cause the assertion to be true?" Here, yes — the redox reaction prevents $CuI_2$ from forming.

Mistake 3: Thinking the assertion is false

Why it happens: Some students think "not known" means "never prepared" and argue that $CuI_2$ might exist under special conditions.
The truth: Under normal conditions, $CuI_2$ is thermodynamically unstable. The reaction is spontaneous:

2Cu^{2+} + 4I^- \rightarrow 2CuI + I_2 \quad (\Delta G < 0)

How to avoid: In exam context, "not known" means "does not exist under standard conditions." Trust the redox potential logic.

Mistake 4: Forgetting the role of $CuI$ precipitation …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2025Set 2025-A1 markMCQ
$E_{M^{3+} / M^{2+}}^o$
›Reveal solutionSolution
The ease of oxidation from +2 to +3 is measured by the standard reduction potential $E^\circ$ for $M^{3+}/M^{2+}$ ; the more negative (or less positive) this value, the easier it is to remove an electron. Cr has the lowest (most negative) value at $-0.41\,\text{V}$ , so it is easiest to oxidize. The correct option is (C).

Concept & Intuition

The question asks: For which metal is the change in oxidation state from +2 to +3 easiest?

That means: which metal loses an electron most readily?

The given $E^\circ$ values are for the reduction half-reaction:
$M^{3+} + e^- \rightarrow M^{2+}$
A more negative reduction potential means the reverse reaction (oxidation: $M^{2+} \rightarrow M^{3+} + e^-$ ) is more spontaneous. So the metal with the lowest (most negative) $E^\circ$ is the easiest to oxidize from +2 to +3.

Step-by-step reasoning

Identify the relevant process We want: $M^{2+} \rightarrow M^{3+} + e^-$ (oxidation). The given $E^\circ$ values are for the reduction: $M^{3+} + e^- \rightarrow M^{2+}$ . The oxidation potential is the negative of the reduction potential:

$E^\circ_{\text{ox}} = -E^\circ_{\text{red}}$
A larger oxidation potential means easier oxidation.

List the given reduction potentials

Cr: $-0.41\,\text{V}$

Mn: $+1.57\,\text{V}$

Fe: $+0.77\,\text{V}$

Co: $+1.97\,\text{V}$

Convert to oxidation potentials

Cr: $E^\circ_{\text{ox}} = -(-0.41) = +0.41\,\text{V}$

Mn: $E^\circ_{\text{ox}} = -(+1.57) = -1.57\,\text{V}$

Fe: $E^\circ_{\text{ox}} = -(+0.77) = -0.77\,\text{V}$

Co: $E^\circ_{\text{ox}} = -(+1.97) = -1.97\,\text{V}$

Compare ease of oxidation

The most positive oxidation potential indicates the easiest loss of an electron. …
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KCET 2024Set B-21 markMCQ
$Q. Which one of the following properties is generally not applicable to ionic hydrides? (A) Non-volatile (B) Non-conducting in solid state (C) Crystalline (D) Volatile$
›Reveal solutionSolution
Ionic hydrides are held together by strong electrostatic lattice forces, so they are high-melting crystalline solids — the one property they never show is volatility.

Step 1 — What an ionic hydride is

Hydrogen forms ionic (saline) hydrides with the strongly electropositive s-block metals (group 1 and the heavier group 2): $NaH$ , $KH$ , $CaH_2$ , $BaH_2$ … Here hydrogen actually gains an electron to become the hydride ion $H^-$ , and the solid is an ionic lattice:
$2Na + H_2 \longrightarrow 2\,Na^{+}H^{-}$
Step 2 — Derive the properties from the bonding

Because the lattice is held by strong electrostatic (ion–ion) forces:

Crystalline — ions pack into a regular 3-D array (NaH and KH take the rock-salt structure). ✓ (option C is true)

Non-volatile / high melting — a great deal of energy is needed to overcome the lattice enthalpy, so these solids have high melting points and negligible vapour pressure. ✓ (option A is true) …
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Q.Assertion: $Cu^{2+}$ iodide is not known.
Reason: $Cu^{2+}$ oxidises $I^-$ to iodine.

Concept understanding — Standard Reduction Potential

Cell Representation and the Nernst Equation: From Intuition to Precision

The Intuition First

The Precise Statement

Cell Representation: How We Write It

Applying the Nernst Equation to a Cell Representation

A Worked Example

Why this formula?

Standard Reduction Potential: Why the Formula Holds

1. The Core Idea: A Half-Cell's "Tendency to Gain Electrons"

2. The Formula: $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$

Why subtraction, not addition?

The reasoning step-by-step:

3. The Nernst Equation: Why $E = E^\circ - \frac{RT}{nF} \ln Q$

From Gibbs free energy:

Substituting:

Rearranging:

Why this makes physical sense:

Assertion — is it true?

Reason — is it true?

Method: Electrochemical Feasibility — Why the Naive $E^\circ$ Comparison Is Misleading

Step 1: Write the relevant half-reactions with their $E^\circ$ values

Step 2: A naive comparison suggests the reaction should NOT occur

Step 3: Account for the precipitation of CuI

Step 4: Connect to the assertion …

Common Mistakes & How to Avoid Them

Mistake 1: Thinking $Cu^{2+}$ iodide is known

Mistake 2: Confusing the reason with the explanation

Mistake 3: Thinking the assertion is false

Mistake 4: Forgetting the role of $CuI$ precipitation …

Step 1 — What an ionic hydride is

Step 2 — Derive the properties from the bonding

Q.Assertion: Cu2+Cu^{2+}Cu2+ iodide is not known. Reason: Cu2+Cu^{2+}Cu2+ oxidises I−I^-I− to iodine.

Assertion — is it true?

Reason — is it true?

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Q.Assertion: $Cu^{2+}$ iodide is not known.
Reason: $Cu^{2+}$ oxidises $I^-$ to iodine.

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