Match the items of Column I and Column II on the basis of data given below: E^°F_2/F^- = 2.87V, E^°Li^+/Li = -

NCERT Exemplar · Q58

E^\circ_{F_2/F^-} = 2.87\ V

(i)

F_2

(ii)

Li

(iii)

Au^{3+}

(iv)

Br^-

(v)

Au

(vi)

Li^+

(vii)

F^-

Column II:

(a) metal is the strongest reducing agent

(b) metal ion which is the weakest oxidising agent

(d) unreactive metal

(e) anion that can be oxidised by

Au^{3+}

(f) anion which is the weakest reducing agent

(g) metal ion which is an oxidising agent

Uttar Pradesh UpmspShort· 3mImportance★★★★★

88% · 101/115 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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). …

Start your 14-day free trial to unlock the full solution →

The standard electrode potentials tell us the relative strengths of oxidising and reducing agents. By comparing the given $E^\circ$ values, we match each species in Column I to its correct description in Column II. The final pairings are: (i)→(c), (ii)→(a), (iii)→(g), (iv)→(e), (v)→(d), (vi)→(b), (vii)→(f).

The key to this problem is understanding what standard electrode potential ( $E^\circ$ ) actually measures. For a half-reaction written as a reduction:

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

A more positive $E^\circ$ means the oxidised form has a stronger tendency to get reduced — it is a better oxidising agent. Conversely, a more negative $E^\circ$ means the reduced form has a stronger tendency to get oxidised — it is a better reducing agent.

So the scale is simple: high $E^\circ$ → good oxidising agent; low (negative) $E^\circ$ → good reducing agent. Let's arrange the given values in order:

Half-reaction	$E^\circ$ (V)
$\text{Li}^+ + e^- \rightarrow \text{Li}$	$-3.5$
$\text{F}_2 + 2e^- \rightarrow 2\text{F}^-$	$+2.87$
$\text{Au}^{3+} + 3e^- \rightarrow \text{Au}$	$+1.4$
$\text{Br}_2 + 2e^- \rightarrow 2\text{Br}^-$	$+1.09$

Now, the reduced forms (Li, F $^-$ , Au, Br $^-$ ) are the ones that get oxidised — their reducing power is opposite to the $E^\circ$ of their oxidised form. So Li (from Li $^+$ with $E^\circ = -3.5$ V) is an extremely strong reducing agent, while F $^-$ (from F $_2$ with $E^\circ = +2.87$ V) is an extremely weak reducing agent.

Let's go through each match step by step.

F $_2$ (i) → (c) non-metal which is the best oxidising agent

F $_2$ has the highest $E^\circ$ ( $+2.87$ V) among all given species. That means it has the greatest tendency to get reduced (gain electrons), making it the strongest oxidising agent. It is a non-metal, so (c) fits perfectly.
Li (ii) → (a) metal is the strongest reducing agent

Li is the reduced form of Li $^+$ / $E^\circ = -3.5$ V. The more negative the $E^\circ$ , the stronger the reducing agent. Li is a metal, and its $E^\circ$ is the most negative here, so it is the strongest reducing agent.
Au $^{3+}$ (iii) → (g) metal ion which is an oxidising agent

Au $^{3+}$ has $E^\circ = +1.4$ V, which is positive. So it can act as an oxidising agent (it can get reduced to Au). It is a metal ion, so (g) is correct. Note that it is not the strongest oxidising agent (that's F $_2$ ), but it is indeed an oxidising agent.
Br $^-$ (iv) → (e) anion that can be oxidised by Au $^{3+}$

To check this, we need to see if Au $^{3+}$ can oxidise Br $^-$ to Br $_2$ . The relevant half-reactions:

\text{Au}^{3+} + 3e^- \rightarrow \text{Au} \quad E^\circ = +1.4\ \text{V}

\text{Br}_2 + 2e^- \rightarrow 2\text{Br}^- \quad E^\circ = +1.09\ \text{V}

For Au $^{3+}$ to oxidise Br $^-$ , the overall reaction must be spontaneous. The cell potential for:

2\text{Au}^{3+} + 6\text{Br}^- \rightarrow 2\text{Au} + 3\text{Br}_2

is $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 1.4 - 1.09 = +0.31\ \text{V} > 0$ . So yes, it is spontaneous. Br $^-$ can indeed be oxidised by Au $^{3+}$ .

Au (v) → (d) unreactive metal …

Start your 14-day free trial to unlock the full solution →

Here’s a clear solution method for matching based on Standard Electrode Potentials.

Method Name: The E° Hierarchy Method (More Positive = Better Oxidising Agent)

Step 1: Arrange the given $E^\circ$ values in decreasing order

The more positive the reduction potential, the stronger the oxidising agent (it wants to get reduced itself).

The more negative the reduction potential, the stronger the reducing agent (it wants to get oxidised).

From the data:

$F_2/F^-$ : $+2.87\ V$ (most positive)
$Au^{3+}/Au$ : $+1.40\ V$
$Br_2/Br^-$ : $+1.09\ V$
$Li^+/Li$ : $-3.50\ V$ (most negative)

Step 2: Identify the extremes

Best oxidising agent = species with the most positive $E^\circ$ → $F_2$ (strongest oxidiser)
Best reducing agent = species with the most negative $E^\circ$ → $Li$ (strongest reducer)

Step 3: Match each item logically

Column I	Reasoning	Column II
(i) $F_2$	Highest $E^\circ$ → best oxidising agent	(c) non metal which is the best oxidising agent
(ii) $Li$	Most negative $E^\circ$ → strongest reducing agent	(a) metal is the strongest reducing agent
(iii) $Au^{3+}$	Positive $E^\circ$ ( $+1.40\ V$ ) → readily reduced, so it acts as an oxidising agent	(g) metal ion which is an oxidising agent
(iv) $Br^-$	Its oxidised form $Br_2$ has a lower $E^\circ$ ( $+1.09\ V$ ) than $Au^{3+}/Au$ ( $+1.40\ V$ ), so $Br^-$ can be oxidised by $Au^{3+}$	(e) anion that can be oxidised by $Au^{3+}$
(v) $Au$	Positive $E^\circ$ but not extremely high; unreactive (noble metal)	(d) unreactive metal

Start your 14-day free trial to unlock the full solution →

Here are the common mistakes students make with this Standard Electrode Potential matching problem, and how to avoid each.

Mistake 1: Confusing "Strongest Reducing Agent" with "Highest $E^\circ$ "

The error:

Students see $F_2$ has the highest $E^\circ$ (+2.87 V) and think it is the strongest reducing agent.

Why it’s wrong:

A high positive $E^\circ$ means the species is easily reduced — it is a strong oxidising agent, not a reducing agent.

How to avoid:

Remember the mnemonic:

High $E^\circ$ → likes to gain electrons → oxidising agent (gets reduced).
Low (most negative) $E^\circ$ → likes to lose electrons → reducing agent (gets oxidised).

So for this data:

Strongest reducing agent = Li (most negative $E^\circ = -3.5\ V$ ) → matches (a).
Best oxidising agent = $F_2$ (most positive $E^\circ = +2.87\ V$ ) → matches (c).

Mistake 2: Thinking "Weakest Oxidising Agent" means "Most Negative $E^\circ$ "

The error:

Students pick $Li^+$ because its $E^\circ$ is very negative.

Why it’s wrong:

$Li^+$ is indeed a very weak oxidising agent (it does not want to gain electrons), but the question asks for the weakest oxidising agent among metal ions.

$Li^+$ has $E^\circ = -3.5\ V$ (very negative) → very weak oxidiser.
$Au^{3+}$ has $E^\circ = +1.4\ V$ (positive) → stronger oxidiser.

How to avoid:

For oxidising power of ions:

More positive $E^\circ$ → stronger oxidising agent.
More negative $E^\circ$ → weaker oxidising agent. So $Li^+$ is the weakest oxidising agent → matches (b).

Mistake 3: Forgetting that $F^-$ is the conjugate of $F_2$

The error:

Students match $F^-$ with "weakest reducing agent" because they think it is inert.

Why it’s wrong:

$F^-$ is indeed a very weak reducing agent — but the reasoning must be based on $E^\circ$ :

$F_2$ has a very high $E^\circ$ (+2.87 V), so its reduced form $F^-$ has very little tendency to be oxidised back.
Hence $F^-$ is the weakest reducing agent among anions.

How to avoid:

Use the reverse logic:

High $E^\circ$ for the oxidised form → its reduced form is a weak reducing agent.
Low $E^\circ$ for the oxidised form → its reduced form is a strong reducing agent.

So $F^-$ → (f).

Mistake 4: Misinterpreting "Anion that can be oxidised by $Au^{3+}$ "

The error:

Students think any anion with a more negative $E^\circ$ than $Au^{3+}/Au$ can be oxidised.

Why it’s wrong:

For $Au^{3+}$ to oxidise an anion $X^-$ , the reaction must be spontaneous:

$Au^{3+}$ is reduced ( $E^\circ = +1.4\ V$ ).
The anion $X^-$ must be oxidised to $X_2$ — this requires $E^\circ_{X_2/X^-} < +1.4\ V$ .

Check the data:

$Br^-$ : $E^\circ_{Br_2/Br^-} = 1.09\ V$ (less than 1.4 V) → yes, $Au^{3+}$ can oxidise $Br^-$ .
$F^-$ : $E^\circ_{F_2/F^-} = 2.87\ V$ (greater than 1.4 V) → no.

How to avoid:

Compare $E^\circ$ values directly:

Oxidising agent (here $Au^{3+}$ ) has higher $E^\circ$ than the reducing agent (here $Br^-$ ) for spontaneity.
So $Br^-$ → (e).

Mistake 5: Confusing "Unreactive Metal" with "Noble Metal" …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ57 — Match the items of Column I and Column II. Column I: (i) Lechlanche ce…→Next questionQ59 — Assertion: Cu is less reactive than hydrogen. Reason: E^°Cu²⁺/Cu is ne…

🎓Unlock everything free for 14 days

Unlock everything free for 14 days