JEE MainChemistry

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Redox Reactions and Electrochemistry: Chemistry | JEE Main

The standard electrode potential of

\text{M}^+/\text{M}

in aqueous solution does not depend on

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AIonisation of a solid metal atomBSublimation of a solid metalCIonisation of a gaseous metal atomDHydration of a gaseous metal ion

Step-by-step solutionView

Correct answer

Standard reduction potential depends on enthalpy of sublimation, ionisation enthalpy of gaseous atoms, and hydration enthalpy of gaseous ions, not direct ionisation of a solid metal atom.

Option analysis

Why each option works or fails

A · Ionisation of a solid metal atom

None. This is the correct choice because ionisation is defined only for isolated gaseous atoms, not solid metal atoms. Recognise that in the Born-Haber thermochemical cycle for electrode potentials, the metal must first sublime into gaseous atoms before ionisation occurs.

B · Sublimation of a solid metal

Believing that phase change from solid to gas is not part of the standard electrode potential thermodynamic cycle. Include the enthalpy of sublimation ( $\text{M(s)} \rightarrow \text{M(g)}$ ) as the necessary first step in the cycle.

C · Ionisation of a gaseous metal atom

Confusing the question's 'does not depend on' negative constraint and selecting a factor that genuinely determines the potential. Carefully read negative stems; ionisation of gaseous metal atoms ( $\text{M(g)} \rightarrow \text{M}^+\text{(g)} + e^-$ ) is an essential component of the cycle.

D · Hydration of a gaseous metal ion

Assuming that solvent interactions do not contribute to standard reduction potential in aqueous solution. Remember that standard electrode potentials are measured in aqueous solution, where hydration enthalpy ( $\text{M}^+\text{(g)} + \text{aq} \rightarrow \text{M}^+\text{(aq)}$ ) plays a major role.

Reviewed route

Solution

StepWorking

01concept

The overall standard electrode potential of $M^+(aq)/M(s)$ corresponds to the process $M(s) \rightarrow M^+(aq) + e^-$ . According to the thermodynamic cycle (Born-Haber type), this total enthalpy change is the sum of three terms: (1) Enthalpy of sublimation: $M(s) \rightarrow M(g)$ , (2) Ionisation enthalpy: $M(g) \rightarrow M^+(g) + e^-$ , and (3) Enthalpy of hydration: $M^+(g) + aq \rightarrow M^+(aq)$ . Direct ionisation of a solid metal atom ( $M(s) \rightarrow M^+(s) + e^-$ ) is not part of this cycle.

02option_verdict

Ionisation enthalpy is defined strictly for isolated gaseous atoms ( $M(g) \rightarrow M^+(g) + e^-$ ), not solid metal atoms directly. Therefore, the standard electrode potential does not depend on 'Ionisation of a solid metal atom'.

03option_verdict

Sublimation of solid metal ( $M(s) \rightarrow M(g)$ ) is the first mandatory step in the thermodynamic cycle, so standard potential does depend on it.

04option_verdict

Ionisation of a gaseous metal atom ( $M(g) \rightarrow M^+(g) + e^-$ ) is the standard ionisation enthalpy step involved in the cycle.

05option_verdict

Hydration of the gaseous metal ion ( $M^+(g) + aq \rightarrow M^+(aq)$ ) releases energy and directly determines the standard potential.

✓discriminator

Ionisation enthalpy is fundamentally defined only for isolated gaseous species; $M(s)$ must sublimate to $M(g)$ before it can be ionised.

Hints that build this answer step by step

Which thermodynamic cycle relates the electrode process $\text{M}^+\text{(aq)} + e^- \rightarrow \text{M(s)}$ to fundamental enthalpy terms?
Born-Haber cycle involving sublimation, ionisation of gaseous atoms, and hydration of gaseous ions
Based on standard definitions in thermochemistry, which process is non-physical or not a distinct step in the cycle?
Ionisation of a solid metal atom directly into ions

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why can't we ionise a solid metal directly in the thermodynamic cycle?

Ionisation enthalpy by definition requires overcoming interatomic metallic bonding first (sublimation to isolated gas atoms) before stripping an electron from an isolated atom.

Answer

Standard reduction potential depends on enthalpy of sublimation, ionisation enthalpy of gaseous atoms, and hydration enthalpy of gaseous ions, not direct ionisation of a solid metal atom.

Why each option works or fails

A · correct: Ionisation of a solid metal atom - None. This is the correct choice because ionisation is defined only for isolated gaseous atoms, not solid metal atoms. Recognise that in the Born-Haber thermochemical cycle for electrode potentials, the metal must first sublime into gaseous atoms before ionisation occurs.
B: Sublimation of a solid metal - Believing that phase change from solid to gas is not part of the standard electrode potential thermodynamic cycle. Include the enthalpy of sublimation ( $\text{M(s)} \rightarrow \text{M(g)}$ ) as the necessary first step in the cycle.
C: Ionisation of a gaseous metal atom - Confusing the question's 'does not depend on' negative constraint and selecting a factor that genuinely determines the potential. Carefully read negative stems; ionisation of gaseous metal atoms ( $\text{M(g)} \rightarrow \text{M}^+\text{(g)} + e^-$ ) is an essential component of the cycle.
D: Hydration of a gaseous metal ion - Assuming that solvent interactions do not contribute to standard reduction potential in aqueous solution. Remember that standard electrode potentials are measured in aqueous solution, where hydration enthalpy ( $\text{M}^+\text{(g)} + \text{aq} \rightarrow \text{M}^+\text{(aq)}$ ) plays a major role.

Step-by-step solution

concept: The overall standard electrode potential of $M^+(aq)/M(s)$ corresponds to the process $M(s) \rightarrow M^+(aq) + e^-$ . According to the thermodynamic cycle (Born-Haber type), this total enthalpy change is the sum of three terms: (1) Enthalpy of sublimation: $M(s) \rightarrow M(g)$ , (2) Ionisation enthalpy: $M(g) \rightarrow M^+(g) + e^-$ , and (3) Enthalpy of hydration: $M^+(g) + aq \rightarrow M^+(aq)$ . Direct ionisation of a solid metal atom ( $M(s) \rightarrow M^+(s) + e^-$ ) is not part of this cycle.
option_verdict: Ionisation enthalpy is defined strictly for isolated gaseous atoms ( $M(g) \rightarrow M^+(g) + e^-$ ), not solid metal atoms directly. Therefore, the standard electrode potential does not depend on 'Ionisation of a solid metal atom'.
option_verdict: Sublimation of solid metal ( $M(s) \rightarrow M(g)$ ) is the first mandatory step in the thermodynamic cycle, so standard potential does depend on it.
option_verdict: Ionisation of a gaseous metal atom ( $M(g) \rightarrow M^+(g) + e^-$ ) is the standard ionisation enthalpy step involved in the cycle.
option_verdict: Hydration of the gaseous metal ion ( $M^+(g) + aq \rightarrow M^+(aq)$ ) releases energy and directly determines the standard potential.
discriminator: Ionisation enthalpy is fundamentally defined only for isolated gaseous species; $M(s)$ must sublimate to $M(g)$ before it can be ionised.

Redox Reactions and Electrochemistry: Chemistry | JEE Main

What feels right?

Standard reduction potential depends on enthalpy of sublimation, ionisation enthalpy of gaseous atoms, and hydration enthalpy of gaseous ions, not direct ionisation of a solid metal atom.

Why each option works or fails

Solution

01concept

02option_verdict

03option_verdict

04option_verdict

05option_verdict

✓discriminator

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