Use the data given in Q.8 ($E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33\ V$; $E^\circ_{Cl_2/Cl^-} = 1.36\ V$; $E^\circ_{MnO_4^-/Mn^{2+}} = 1.51\ V$; $E^\circ_{Cr^{3+}/Cr} = -0.74\ V$) and find out the most stable ion in its reduced form. (i) $Cl^-$ (ii) $Cr^{3+}$ (iii) $Cr$ (iv) $Mn^{2+}$

Use the data given in Q.8 (E^°Cr_2O_7²⁻/Cr³⁺ = 1.33V; E^°Cl_2/Cl^- = 1.36V; E^°MnO_4^-/Mn²⁺ = 1.51V; E^°Cr³⁺/C

NCERT Exemplar · Q11

E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33\ V

(i)

Cl^-

(ii)

Cr^{3+}

(iii)

Cr

(iv)

Mn^{2+}

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Concept understanding — Standard Electrode Potentials

Standard Electrode Potentials: A Number for "How Badly It Wants Electrons"

Dip a zinc rod into a zinc-salt solution and a tiny tug-of-war begins at the surface:

metal atoms tend to dissolve as ions (leaving electrons behind on the rod) while ions

from the solution tend to deposit as metal (consuming electrons). The rod ends up with

a characteristic electrical potential relative to the solution — the electrode potential. It is a direct measure of the tendency of that redox couple to gain or

lose electrons.

The Core Idea

Different couples pull electrons with very different strengths. Copper's ion grabs

them readily; zinc's barely wants them. Put a number on each couple and you can

predict, before mixing anything, who will oxidise whom.

Two conventions make the numbers comparable:

Standard conditions. Every species at unit concentration (1 M), any gas at 1 atm, temperature 298 K. The potential measured then is the standard electrode potential, written $E^\ominus$ .
A common zero. Potentials can only be measured as differences, so one electrode is defined as the reference: the standard hydrogen electrode (SHE), $2\text{H}^+ + 2e^- \rightarrow \text{H}_2$ , is fixed at exactly 0.00 V. Every $E^\ominus$ is the voltage of a couple measured against it.

By convention the values are tabulated for the reduction direction:

\text{Oxidised form} + ne^- \rightarrow \text{Reduced form} \qquad E^\ominus \text{ (in volts, at 298 K)}

Reading the Table

The standard-potential table (Table 7.1 in the Class 11 chapter) runs from

$\text{F}_2/\text{F}^-$ at $+2.87$ V down to $\text{Li}^+/\text{Li}$ at $-3.05$ V.

Two rules unlock it:

More positive $E^\ominus$ → stronger oxidising agent (the oxidised form is hungrier for electrons). F₂ tops the table; that is why fluorine oxidises almost everything.
More negative $E^\ominus$ → stronger reducing agent (the reduced form gives electrons up most easily). Li, K, Ca, Na at the bottom are the great electron donors. A negative $E^\ominus$ means the couple is a stronger reducing agent than the H⁺/H₂ couple; a positive one, weaker.

Predicting Whether a Reaction Goes

For any proposed redox reaction, the species being reduced acts as the cathode couple

and the species being oxidised as the anode couple:

E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}}

Worked feel: can Fe³⁺ oxidise iodide? $E^\ominus(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77$ V is above $E^\ominus(\text{I}_2/\text{I}^-) = +0.54$ V, so

$E^\ominus_{\text{cell}} = +0.23$ V — yes. Can silver metal reduce Fe³⁺?

$0.77 - 0.80 = -0.03$ V — no.

This is also the logic of the activity series: a metal displaces, from solution, …

Why this formula?

Galvanic Corrosion: Why the Key Formulas Hold

Galvanic corrosion occurs when two dissimilar metals are electrically connected in the presence of an electrolyte. The key formula that governs this is the mixed potential theory, which leads to the galvanic current and corrosion rate expressions.

Let's build the reasoning step-by-step.

1. The Core Idea: Two Electrodes, One Circuit

When metals M₁ (more active, e.g., zinc) and M₂ (more noble, e.g., copper) are connected:

M₁ acts as the anode — it oxidizes (corrodes):

\text{M}_1 \rightarrow \text{M}_1^{n+} + n e^-

M₂ acts as the cathode — it reduces something (e.g., oxygen or H⁺):

\text{O}_2 + 2\text{H}_2\text{O} + 4e^- \rightarrow 4\text{OH}^- \quad \text{(in neutral/alkaline)}

2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \quad \text{(in acidic)}

The two metals are electrically connected (via a wire or direct contact), and the electrolyte completes the circuit. Electrons flow from M₁ to M₂.

2. The Mixed Potential: Why It Exists

Each metal, when alone in the electrolyte, has its own open-circuit potential (OCP) — the equilibrium potential for its half-reaction. For M₁, it's $E_{\text{corr},1}$ ; for M₂, it's $E_{\text{corr},2}$ .

When connected, the system cannot stay at two different potentials. The entire metal couple must reach a single potential — the mixed potential $E_{\text{mix}}$ .

$E_{\text{mix}}$ lies between $E_{\text{corr},1}$ and $E_{\text{corr},2}$ .
At $E_{\text{mix}}$ , the total anodic current from M₁ equals the total cathodic current from M₂ (charge conservation):

I_{\text{anode}} = I_{\text{cathode}}

This is the fundamental equation of galvanic corrosion.

3. Deriving the Galvanic Current

Assume each electrode follows Butler-Volmer kinetics (for activation-controlled reactions). For the anode (M₁), the anodic current density $i_a$ at potential $E$ is:

i_a = i_{0,1} \exp\left( \frac{\alpha_a F (E - E_{0,1})}{RT} \right)

For the cathode (M₂), the cathodic current density $i_c$ is:

i_c = i_{0,2} \exp\left( -\frac{\alpha_c F (E - E_{0,2})}{RT} \right)

Where:

$i_{0,1}, i_{0,2}$ = exchange current densities
$\alpha_a, \alpha_c$ = transfer coefficients (typically ~0.5)
$F$ = Faraday constant
$R$ = gas constant
$T$ = temperature
$E_{0,1}, E_{0,2}$ = standard reduction potentials

At the mixed potential $E_{\text{mix}}$ :

I_{\text{galvanic}} = A_1 \cdot i_a(E_{\text{mix}}) = A_2 \cdot i_c(E_{\text{mix}})

Where $A_1$ and $A_2$ are the surface areas of the anode and cathode.

Why this holds: The net current from the anode must exactly balance the net current consumed at the cathode — otherwise, charge would accumulate, which is impossible in a steady-state circuit.

4. The Corrosion Rate Formula

The corrosion rate (mass loss per time) of the anode is given by Faraday's law:

\text{Corrosion rate} = \frac{I_{\text{galvanic}} \cdot M}{n \cdot F \cdot \rho}

Where:

$M$ = molar mass of the anode metal
$n$ = number of electrons transferred per atom
$\rho$ = density of the metal
$F$ = Faraday constant (96,485 C/mol)

Why this holds: Each mole of metal oxidized releases $n$ moles of electrons. The total charge passed $Q = I_{\text{galvanic}} \cdot t$ corresponds to moles of metal lost:

\text{moles lost} = \frac{Q}{nF} = \frac{I_{\text{galvanic}} \cdot t}{nF}

Multiply by $M/\rho$ to get volume or thickness loss.

5. The Area Effect: Why It Matters

From the mixed potential equation:

A_1 \cdot i_a(E_{\text{mix}}) = A_2 \cdot i_c(E_{\text{mix}})

If the cathode area $A_2$ is large relative to the anode area $A_1$ , then $i_a(E_{\text{mix}})$ must be large to balance the current. This means:

Small anode + large cathode → severe galvanic corrosion (high current density on the anode). …

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The most stable reduced species is the one with the least negative (or most positive) reduction potential for its half‑reaction, because that indicates the strongest tendency to stay reduced. Comparing the given potentials shows that $Mn^{2+}$ is the most stable reduced ion.

The question asks: among $Cl^-$ , $Cr^{3+}$ , $Cr$ (metal), and $Mn^{2+}$ , which is the most stable in its reduced form?

“Stable in reduced form” means the species has little tendency to get oxidised back — it prefers to stay as it is. In electrochemistry, that tendency is measured by the standard reduction potential $E^\circ$ of the corresponding half‑reaction.

E^\circ_{\text{red}} \text{ (more positive)} \implies \text{stronger oxidising agent (oxidised form)}

But careful: we want the reduced form to be stable. That means the reduced form should have a low tendency to get oxidised. The tendency to get oxidised is the reverse of the reduction potential. So we look at the reduction potential of the oxidised form — if that potential is very positive, the oxidised form is a strong oxidiser and the reduced form is weak (unstable as a reductant). Conversely, if the reduction potential is very negative, the reduced form is a strong reductant (easily oxidised, hence unstable in its reduced state).

Therefore, the most stable reduced form corresponds to the most positive reduction potential of the oxidised species. Let’s list the given data properly.

For $Cl^-$ : The half‑reaction is

Cl_2 + 2e^- \rightarrow 2Cl^- \quad E^\circ = +1.36\ \text{V}

The reduced form is $Cl^-$ . The potential is quite positive, so $Cl_2$ is a strong oxidiser and $Cl^-$ is a weak reductant — fairly stable.

For $Cr^{3+}$ : There are two relevant half‑reactions:

Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \quad E^\circ = +1.33\ \text{V}

and

Cr^{3+} + 3e^- \rightarrow Cr \quad E^\circ = -0.74\ \text{V}

The reduced form $Cr^{3+}$ appears on the right of the first reaction and on the left of the second. To judge its stability, we need the potential for the half‑reaction that produces $Cr^{3+}$ from a higher oxidation state — that’s the $+1.33\ \text{V}$ one. But also, $Cr^{3+}$ can be further reduced to $Cr$ metal at $-0.74\ \text{V}$ , which tells us $Cr^{3+}$ is a weak oxidiser (hard to reduce). However, the question asks about stability of the reduced form — here $Cr^{3+}$ is the reduced form relative to dichromate. Its $E^\circ = +1.33\ \text{V}$ is slightly less positive than $Cl_2/Cl^-$ .

For $Cr$ (metal): The half‑reaction is $Cr^{3+} + 3e^- \rightarrow Cr \quad E^\circ = -0.74\ \text{V}$ …

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Common Mistakes & How to Avoid Them

Mistake 1: Confusing "most stable reduced form" with "strongest reducing agent"

The error: Students often think the species with the most negative $E^\circ$ is the most stable reduced form. They pick Cr ( $-0.74\ V$ ) because it's the most negative.

Why it's wrong:

A more negative $E^\circ$ means the reduced form is a stronger reducing agent (easily gets oxidised) — meaning it is less stable in its reduced state.
A more positive $E^\circ$ means the reduced form is harder to oxidise — hence more stable.

How to avoid:

Remember: Higher $E^\circ$ → Reduced form is more stable (it "wants" to stay reduced).
Think of it like a tug-of-war: a high positive $E^\circ$ means the oxidised form pulls electrons strongly, so the reduced form is content and stable.

Mistake 2: Comparing potentials of half-cells with different numbers of electrons

The error: Students directly compare $E^\circ$ values without noticing that some half-reactions involve different $n$ (number of electrons). For example:

$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ ( $n=6$ )
$Cl_2 + 2e^- \rightarrow 2Cl^-$ ( $n=2$ )
$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ ( $n=5$ )
$Cr^{3+} + 3e^- \rightarrow Cr$ ( $n=3$ )

Why it's wrong:

$E^\circ$ is an intensive property — it does not depend on $n$ . So comparing $E^\circ$ directly is actually correct here.
However, some students mistakenly try to "scale" $E^\circ$ by multiplying with $n$ (like $\Delta G^\circ = -nFE^\circ$ ). That would be wrong for stability comparison.

How to avoid:

For stability of the reduced form, compare $E^\circ$ values directly — not $\Delta G^\circ$ or $n \times E^\circ$ .
Only use $\Delta G^\circ$ if the question asks for spontaneity or thermodynamic favourability of a specific reaction.

Mistake 3: Forgetting to check which species is the "reduced form"

The error: Students look at the given $E^\circ$ values and pick the oxidised form (like $MnO_4^-$ or $Cr_2O_7^{2-}$ ) instead of the reduced form.

Why it's wrong:

The question asks for the most stable ion in its reduced form.
For each half-cell, the reduced form is:
- $Cl^-$ (from $Cl_2/Cl^-$ )
- $Cr^{3+}$ (from $Cr_2O_7^{2-}/Cr^{3+}$ )
- $Mn^{2+}$ (from $MnO_4^-/Mn^{2+}$ )
- $Cr$ (from $Cr^{3+}/Cr$ )

How to avoid:

Always write the half-reaction in the reduction direction:

\text{Oxidised form} + ne^- \rightarrow \text{Reduced form}

The species on the right is the reduced form — that's what you compare.

Mistake 4: Misinterpreting the sign of $E^\circ$ for $Cr^{3+}/Cr$ …

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

Showing the 12 most recent of 16 on this concept.

CBSE 2026Set 56/3/11 markMCQ
$H_2$
›Reveal solutionSolution
In very dilute aqueous NaCl with inert Pt electrodes, water’s reduction to $H_2$ at the cathode and water’s oxidation to $O_2$ at the anode outcompete the NaCl reactions. So $H_2$ is produced at the cathode and $O_2$ at the anode — making option (C) and (D) correct.

Why standard electrode potentials decide the outcome

Electrolysis is a battle of competing half-reactions. At each electrode, the species that is easier to oxidise (at the anode) or easier to reduce (at the cathode) will react first. “Easier” means having a more positive reduction potential for reduction, or a more negative reduction potential for oxidation (equivalently, a more positive oxidation potential).

For a very dilute aqueous solution of NaCl, the possible species are:

Cathode (reduction): $Na^+$ ions and $H_2O$ molecules.

Anode (oxidation): $Cl^-$ ions and $H_2O$ molecules.

We compare their standard reduction potentials (at 298 K, 1 M concentration, 1 atm pressure). But remember: concentration matters. In very dilute NaCl, $[Cl^-]$ is tiny, which shifts the actual potential of the chlorine half-reaction significantly.

Step-by-step reasoning

1. What happens at the cathode?

Two reduction half-reactions compete:
$\begin{aligned} Na^+ + e^- &\rightarrow Na(s) \quad E^\circ = -2.71\ \text{V} \\[4pt] 2H_2O + 2e^- &\rightarrow H_2(g) + 2OH^- \quad E^\circ = -0.83\ \text{V} \end{aligned}$
The reduction of water to hydrogen gas has a much less negative (i.e., more positive) standard potential. Even though the actual potential for water reduction depends slightly on pH (here neutral to slightly basic), it remains far above $-2.71\ \text{V}$ . So water is reduced preferentially.

Watch out

A common mistake is to think that because $Na^+$ is present, sodium metal will plate out. But sodium’s reduction potential is so negative that water (even in neutral solution) is reduced first. Sodium metal would instantly react with water anyway — it’s never produced in aqueous electrolysis.

Result at cathode: $H_2$ gas is evolved. This matches option (D).

2. What happens at the anode?

Two oxidation half-reactions compete (written as reductions for comparison):

$\begin{aligned} Cl_2(g) + 2e^- &\rightarrow 2Cl^- \quad E^\circ = +1.36\ \text{V} \\[4pt] O_2(g) + 4H^+ + 4e^- &\rightarrow 2H_2O \quad E^\circ = +1.23\ \text{V} \end{aligned}$ …
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CBSE 2026Set 56/2/11 markMCQ
$Zn(s) + Ag_2O(s) + H_2O(l) \rightarrow Zn^{2+}(aq) + 2Ag(s) + 2OH^-(aq)$
›Reveal solutionSolution

Zinc is oxidised and silver is reduced, giving $E^o_{cell} = 0.80 - (-0.76) = 1.56$ V with $n = 2$ . Then $\Delta_r G^o = -nFE^o_{cell} = -301.080$ kJ mol $^{-1}$ , which is option (A).

The standard Gibbs energy of a cell reaction is linked to its standard cell potential by
$\Delta_r G^o = -nFE^o_{cell}$
so we first find $E^o_{cell}$ , then $n$ , and finally $\Delta_r G^o$ .

1. Identify the electrodes. Zinc is oxidised (anode) and silver is reduced (cathode):
$\text{Anode:}\quad Zn \rightarrow Zn^{2+} + 2e^-$ $\text{Cathode:}\quad Ag_2O + H_2O + 2e^- \rightarrow 2Ag + 2OH^-$
2. Standard cell potential. Using the given reduction potentials,
$E^o_{cell} = E^o_{cathode} - E^o_{anode} = 0.80 - (-0.76) = 1.56\ \text{V}$
3. Electrons transferred. Each half-reaction involves 2 electrons, so $n = 2$ . …
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CBSE 2025Set D1 markMCQ
$Q. The electromotive force of the following cell is: Zn | Zn2+ (1M) || Fe2+ (1M) | Fe, given E°Zn2+|Zn = -0.76 V, E°Fe2+|Fe = -0.44 V$
(a) 1.2 V
(b) 0.32 V
(c) -1.2 V
(d) -0.32 V
›Reveal solutionSolution

E(cell) = E(cathode) - E(anode) = -0.44 - (-0.76) = +0.32 V.

In the cell notation Zn | Zn2+ || Fe2+ | Fe, zinc is the anode (oxidation, written left) and iron is the cathode (reduction, written right). The standard cell EMF is:

E(cell) = E(cathode) - E(anode)

E(cell) = E(Fe2+/Fe) - E(Zn2+/Zn)

E(cell) = (-0.44) - (-0.76) …

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CBSE 2025Set A1 mark
$Q. Write True or False: The cell potential is the addition of the electrode potentials (reduction potentials) of the cathode and anode.$
›Reveal solutionSolution

Cell potential is obtained by subtracting the anode's reduction potential from the cathode's, not by adding the two reduction potentials.

The standard cell potential is defined as:

$E^{\circ}_{cell} = E^{\circ}_{cathode(reduction)} - E^{\circ}_{anode(reduction)}$

If both electrode potentials are taken as reduction potentials (as the statement specifies), the correct operation is a subtraction (cathode minus anode), not an addition. The 'addition' phrasing is only valid if the anode's contribution is expressed as an oxidation potential (= −reduction potential): then …

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CBSE 2025Set ANNUAL1 mark
$Q. Answer in one word/sentence: Given the standard electrode potentials, arrange these metals in their increasing order of reducting power: K+/K = -2.93 V, Ag+/Ag = 0.80 V, Hg2+/Hg = 0.79 V, Mg2+/Mg = -2.37 V.$
›Reveal solutionSolution

Reducing power increases as the standard electrode (reduction) potential becomes more negative, so we simply rank the four E° values from most positive to most negative.

Given standard reduction potentials:

$K^+/K = -2.93\ V,\quad Mg^{2+}/Mg = -2.37\ V,\quad Hg^{2+}/Hg = +0.79\ V,\quad Ag^+/Ag = +0.80\ V$

A more negative (or less positive) standard reduction potential means the metal has a greater tendency to lose electrons (be oxidised) — i.e. it is a stronger reducing agent. Conversely, a metal with a highly positive reduction potential prefers to stay reduced (gain electrons), making it a poor reducing agent (like Ag, a "noble" metal).

…

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CBSE 2025Set ANNUAL1 mark
$Q. Write two applications of electrochemical series.$
›Reveal solutionSolution
Electrochemical series: predicts reaction feasibility and metal-displacement reactivity.

The electrochemical series arranges elements/ions in order of their standard reduction potentials ( $E°$ ). Two common applications:

Predicting feasibility of a redox reaction: a reaction is spontaneous if the species with the higher (more positive) reduction potential is reduced while the species with the lower (more negative) reduction potential is oxidized, i.e. $E°_{cell} = E°_{cathode} - E°_{anode} > 0$ . …
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CBSE 2024Set 56/3/11 markMCQ
$Cl^-$
›Reveal solutionSolution
In aqueous NaCl electrolysis, the cathode is where reduction occurs. The competing reductions are $Na^+$ and $H_2O$ ; water has a much less negative reduction potential, so it is reduced instead of sodium. The correct answer is (D) Reduction of $H_2O$ .

The key to this question lies in understanding Standard Electrode Potentials — the numerical measure of a species’ tendency to gain electrons (be reduced). In electrolysis, the cathode is the negative electrode where reduction happens. When you have an aqueous solution, you must consider all possible reducible species, not just the obvious cation from the salt.

For aqueous NaCl, the solution contains:

$Na^+$ ions (from the salt)

$H_2O$ molecules (the solvent)

$Cl^-$ ions (from the salt — but these are oxidised at the anode, not reduced at the cathode)

At the cathode, two reduction reactions compete:

Reduction of $Na^+$ :

$Na^+ + e^- \rightarrow Na(s)$

Standard reduction potential: $E^\circ = -2.71\ \text{V}$

Reduction of water:

$2H_2O + 2e^- \rightarrow H_2(g) + 2OH^-$

Standard reduction potential: $E^\circ = -0.83\ \text{V}$

Watch out

A common mistake is to assume that because $Na^+$ is the cation, it must be reduced at the cathode. But the more positive (or less negative) the reduction potential, the easier the reduction. Here, water’s potential ( $-0.83\ \text{V}$ ) is far less negative than sodium’s ( $-2.71\ \text{V}$ ), meaning water is much more readily reduced.

Now, let’s work through the reasoning step by step.

Identify the cathode process.

The cathode is the electrode where reduction occurs — gain of electrons. So we look for which species can accept electrons.

List all reducible species in the solution.

In aqueous NaCl: $Na^+$ ions and $H_2O$ molecules. (The $Cl^-$ ions are already in their lowest oxidation state for a halide; they cannot be reduced further under these conditions — they are oxidised at the anode.)

Compare their reduction potentials.

$Na^+ + e^- \rightarrow Na$ : $E^\circ = -2.71\ \text{V}$

$2H_2O + 2e^- \rightarrow H_2 + 2OH^-$ : $E^\circ = -0.83\ \text{V}$

The more positive (or less negative) the potential, the stronger the oxidising agent — i.e., the more likely it is to be reduced. Since $-0.83 > -2.71$ , water is a much stronger oxidising agent than $Na^+$ in this system. …
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CBSE 2024Set ANNUAL1 mark
$Q. In which electrode of a Galvanic cell, oxidation reaction takes place?$
›Reveal solutionSolution

The anode of a galvanic cell is where oxidation (electron loss) occurs.

A Galvanic (voltaic) cell converts the chemical energy of a spontaneous redox reaction into electrical energy, splitting the reaction into two half-cells. The electrode at which oxidation (loss of electrons) takes place is called the anode; in a galvanic cell this is the negative electrode. For example, in the Daniell cell, $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$ o …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Emf of a cell with Nickel and Copper electrode will be (Given E0 Ni+2/Ni = -0.25 V, E0 Cu2+/Cu = +0.34 V)$
(a) -0.59 V
(b) +0.59 V
(c) +0.09 V
(d) -0.09 V
›Reveal solutionSolution

The electrode with the higher (more positive) standard reduction potential acts as the cathode; the cell EMF is Ecathode - Eanode.

Given: E-degree(Ni2+/Ni) = -0.25 V, E-degree(Cu2+/Cu) = +0.34 V.

Since Cu2+/Cu has the higher reduction potential, copper is reduced (cathode) and nickel is oxidised (anode):

Anode (oxidation): Ni -> Ni2+ + 2e-

Cathode (reduction): Cu2+ + 2e- -> Cu

…

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CBSE 2023Set 56/1/11 markMCQ
$\Delta G$
›Reveal solutionSolution
A spontaneous reaction releases free energy ( $\Delta G < 0$ ) and generates a positive cell potential ( $E^\circ_{\text{cell}} > 0$ ); the answer is (C).

The connection between thermodynamics and electrochemistry rests on a beautiful relationship: the Gibbs free energy change tells us whether a reaction will proceed on its own, while the standard cell potential measures the driving force behind electron flow. For a reaction to be spontaneous, it must release free energy to do useful work—including pushing electrons through a circuit.

The fundamental bridge between these quantities is:
$\Delta G^\circ = -nFE^\circ_{\text{cell}}$
where $n$ is the number of moles of electrons transferred, $F$ is Faraday's constant ( $96{,}485 \, \text{C/mol}$ ), and $E^\circ_{\text{cell}}$ is the standard cell potential.

The negative sign in this equation is the key. It tells us that a positive cell potential (electrons flowing spontaneously from anode to cathode, releasing energy) corresponds to a negative Gibbs free energy change (energy released, reaction spontaneous). Think of it this way: when a battery drives current through a device, it's doing work on the surroundings, which means the battery's chemical reaction is losing free energy—hence $\Delta G < 0$ .

Now let's apply this to the question:

What does spontaneity require thermodynamically?

A spontaneous process proceeds without external intervention and releases free energy. The criterion is $\Delta G < 0$ (negative). This is the defining condition—if $\Delta G$ were positive, we'd need to supply energy to make the reaction go.

What does the equation tell us about $E^\circ_{\text{cell}}$ ?

Rearranging: $E^\circ_{\text{cell}} = -\dfrac{\Delta G^\circ}{nF}$ . Since $n$ and $F$ are always positive, and we've established that $\Delta G^\circ < 0$ for a spontaneous reaction, the negative sign in front flips the inequality: $E^\circ_{\text{cell}} > 0$ (positive).

Physical interpretation

A positive standard cell potential means the cathode (reduction site) has a higher reduction potential than the anode (oxidation site). Electrons naturally flow "downhill" in potential, from lower to higher reduction potential, generating voltage. This is exactly what happens in a galvanic (voltaic) cell—the spontaneous reaction produces electrical energy. …
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CBSE 2023Set 56/3/11 markMCQ
$Q. Assertion (A): Electrolysis of aqueous solution of NaCl gives chlorine gas at anode instead of oxygen gas. Reason (R): Formation of oxygen gas at anode requires overpotential. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.$
›Reveal solutionSolution
In the electrolysis of aqueous NaCl, chlorine is produced at the anode instead of oxygen because the overpotential for oxygen evolution makes the actual potential needed for oxygen formation higher than that for chlorine, even though the standard potential for oxygen is lower. Both Assertion and Reason are true, and the Reason correctly explains the Assertion — so the answer is (A).

Why this question is about real-world electrochemistry

Standard electrode potentials tell you which reaction is thermodynamically favoured. But electrolysis happens under kinetic conditions. The key twist here: oxygen evolution at an inert anode (like platinum or graphite) has a large overpotential — an extra voltage needed to overcome the activation barrier. Chlorine evolution, on the other hand, has a much smaller overpotential. So the reaction that actually occurs at the anode is not the one with the lower standard potential, but the one that requires the lower actual voltage (standard potential + overpotential).

Let’s see the numbers.

1. What are the possible anode reactions?

In aqueous NaCl, the solution contains these ions:

$\text{Na}^+$ , $\text{Cl}^-$ , $\text{H}^+$ (from water), and $\text{OH}^-$ (from water).

At the anode, oxidation happens. The two candidates are:

Oxidation of chloride ions:

$2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^- \quad E^\circ = +1.36\ \text{V}$

Oxidation of water (to oxygen):

$2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^- \quad E^\circ = +1.23\ \text{V}$
Note

Standard potentials are given as reduction potentials. For oxidation, we reverse the sign. But when comparing which oxidation is easier, we compare the actual potentials needed — the more negative the oxidation potential (or the lower the reduction potential), the easier it is to oxidise. Here, water oxidation has $E^\circ = +1.23\ \text{V}$ (reduction), so its oxidation potential is $-1.23\ \text{V}$ . Chloride oxidation has $E^\circ = +1.36\ \text{V}$ (reduction), so its oxidation potential is $-1.36\ \text{V}$ . Since $-1.23 > -1.36$ , water oxidation is thermodynamically easier — it should occur first.

So why doesn’t it?

2. The role of overpotential

Overpotential ( $\eta$ ) is the extra voltage beyond the thermodynamic value required to drive a reaction at a noticeable rate. For oxygen evolution on common anode materials (Pt, graphite), $\eta$ is substantial — typically around $0.4$ – $0.6\ \text{V}$ . For chlorine evolution on the same materials, $\eta$ is very small (often $<0.1\ \text{V}$ ).

So the actual potential needed for each reaction is:

For oxygen:

$E_{\text{actual}}(\text{O}_2) = 1.23\ \text{V} + \eta_{\text{O}_2} \approx 1.23 + 0.5 = 1.73\ \text{V}$

For chlorine:

$E_{\text{actual}}(\text{Cl}_2) = 1.36\ \text{V} + \eta_{\text{Cl}_2} \approx 1.36 + 0.05 = 1.41\ \text{V}$
Now compare: chlorine requires a lower actual voltage (1.41 V) than oxygen (1.73 V). So chlorine is produced preferentially. …
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CBSE 2023Set ANNUAL1 markMCQ
$Q. When concentration of Zn2+ and Cu2+ ions is unity (1 mol dm-3), then electrical potential of Daniell cell will be -$
(a) 0.00 V
(b) 1.10 V
(c) 1.35 V
(d) 2.00 V
›Reveal solutionSolution

A Daniell cell is Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s); its EMF is the difference of the two standard reduction potentials, and at unit concentration this IS the standard cell potential.

The Daniell cell has the cell reaction Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s), with Zn as the anode (oxidation) and Cu as the cathode (reduction).

Standard reduction potentials: E-standard(Cu2+/Cu) = +0.34 V, E-standard(Zn2+/Zn) = -0.76 V.

…

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←Previous questionQ10 — Using the data given in Q.8 (E^°Cr_2O_7²⁻/Cr³⁺ = 1.33V; E^°Cl_2/Cl^- =…→Next questionQ12 — Use the data of Q.8 (E^°Cr_2O_7²⁻/Cr³⁺ = 1.33V; E^°Cl_2/Cl^- = 1.36V; …

Q.Use the data given in Q.8 ( $E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33\ V$ ; $E^\circ_{Cl_2/Cl^-} = 1.36\ V$ ; $E^\circ_{MnO_4^-/Mn^{2+}} = 1.51\ V$ ; $E^\circ_{Cr^{3+}/Cr} = -0.74\ V$ ) and find out the most stable ion in its reduced form.

Concept understanding — Standard Electrode Potentials

Standard Electrode Potentials: A Number for "How Badly It Wants Electrons"

The Core Idea

Reading the Table

Predicting Whether a Reaction Goes

Why this formula?

Galvanic Corrosion: Why the Key Formulas Hold

1. The Core Idea: Two Electrodes, One Circuit

2. The Mixed Potential: Why It Exists

3. Deriving the Galvanic Current

4. The Corrosion Rate Formula

5. The Area Effect: Why It Matters

Method: Comparing Standard Reduction Potentials ( $E^\circ$ )

Steps

Common Mistakes & How to Avoid Them

Mistake 1: Confusing "most stable reduced form" with "strongest reducing agent"

Mistake 2: Comparing potentials of half-cells with different numbers of electrons

Mistake 3: Forgetting to check which species is the "reduced form"

Mistake 4: Misinterpreting the sign of $E^\circ$ for $Cr^{3+}/Cr$ …

Why standard electrode potentials decide the outcome

Step-by-step reasoning

1. What happens at the cathode?

2. What happens at the anode?

Why this question is about real-world electrochemistry

1. What are the possible anode reactions?

2. The role of overpotential

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