The more positive the value of E⊖, the greater is the tendency of the species to get reduced. Using the standa

NCERT Exemplar · Q2

Q. The more positive the value of E⊖, the greater is the tendency of the species to get reduced. Using the standard electrode potential of redox couples given below find out which of the following is the strongest oxidising agent. E⊖ values: Fe^3+/Fe^2+ = +0.77; I2(s)/I^- = +0.54; Cu^2+/Cu = +0.34; Ag^+/Ag = +0.80V

(i) Fe^3+

(ii) I2(s)

(iii) Cu^2+

(iv) Ag^+

Chhattisgarh CgbseMCQ· 1mImportance★★★★★

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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,

the ions of any metal whose couple sits above it (less negative / more positive) in

the table — Zn displaces Cu²⁺, Cu displaces Ag⁺, never the reverse.

Common Mistakes

Flipping the sign when a half reaction is reversed and then double-counting. Use $E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}}$ with BOTH values as tabulated (reduction) potentials — the subtraction already handles the reversal.
Multiplying $E^\ominus$ by stoichiometric coefficients. Potentials are intensive: doubling a half reaction does not double its $E^\ominus$ .
Reading "negative" as "impossible". A negative $E^\ominus$ only ranks the couple as a stronger reducing agent than H⁺/H₂ — zinc's $-0.76$ V is exactly why zinc is so good at reducing other ions.

Exam Relevance

Standard-potential questions are staples: pick the strongest oxidising/reducing agent

from given $E^\ominus$ values, decide whether a pair reacts (CBSE Class 11 Exercise

7.26; Exemplar Q2–Q4, Q16, Q34), order metals by reducing power, or justify a

displacement series. In Class 12 the same idea grows into full electrochemistry — the

Nernst equation extends $E^\ominus$ to non-standard concentrations, and cell EMF

connects to thermodynamics via $\Delta G = -nFE$ .

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).
Large anode + small cathode → mild corrosion.

Why this holds: The current density on the anode is $i_a = I_{\text{galvanic}} / A_1$ . For a fixed $I_{\text{galvanic}}$ , smaller $A_1$ gives higher $i_a$ , which accelerates corrosion.

6. The Driving Force: Potential Difference

The driving force for galvanic corrosion is the difference in open-circuit potentials:

\Delta E = E_{\text{corr},2} - E_{\text{corr},1}

A larger $\Delta E$ generally leads to a larger $I_{\text{galvanic}}$ , but the exact relationship depends on the polarization behavior (Tafel slopes) of both electrodes.

Why this holds: The mixed potential $E_{\text{mix}}$ is determined by the intersection of the anodic and cathodic polarization curves. A larger separation between the two curves shifts the intersection to a higher current.

Summary of Key Takeaways

Concept	Formula	Why It Holds
Mixed potential	$I_{\text{anode}} = I_{\text{cathode}}$	Charge conservation in a closed circuit
Galvanic current	$A_1 i_a(E_{\text{mix}}) = A_2 i_c(E_{\text{mix}})$	Butler-Volmer kinetics + area balance
Corrosion rate	$\frac{I_{\text{galvanic}} M}{n F \rho}$	Faraday's law of electrolysis
Area effect	Small anode → high $i_a$	Current density inversely proportional to area
Driving force	$\Delta E = E_{\text{corr},2} - E_{\text{corr},1}$	Larger potential difference → larger current (generally)

Exam tip: Always start with the mixed potential condition — it's the foundation. Then apply Faraday's law for the rate. Never forget the area ratio — it's the most common trick in exam problems.

The standard electrode potential ( $E^\ominus$ ) directly indicates the tendency of a species to get reduced. A more positive $E^\ominus$ means a greater tendency for reduction, making the species a stronger oxidising agent. Comparing the given $E^\ominus$ values, $\text{Ag}^+$ has the highest positive value, making it the strongest oxidising agent. The correct option is (iv).

In electrochemistry, the standard electrode potential ( $E^\ominus$ ) is a crucial measure that tells us about the inherent tendency of a species to gain electrons (get reduced) or lose electrons (get oxidised) under standard conditions.

An oxidising agent is a chemical species that causes another species to be oxidised, while itself getting reduced. Therefore, to find the strongest oxidising agent, we need to identify the species that has the greatest tendency to get reduced.

The problem statement itself provides the key principle: "The more positive the value of $E^\ominus$ , the greater is the tendency of the species to get reduced." This directly links a high positive $E^\ominus$ value to strong oxidising power.

E^\ominus

Let's apply this principle to the given data:

Identify the potential oxidising agents and their reduction potentials:

We are given standard electrode potentials for various redox couples. For each couple, the species on the left side of the reduction half-reaction (the one that gains electrons) is the oxidising agent.
- For $\text{Fe}^{3+}/\text{Fe}^{2+}$ , the reduction half-reaction is $\text{Fe}^{3+} + \text{e}^- \rightarrow \text{Fe}^{2+}$ .
  
  The oxidising agent is $\text{Fe}^{3+}$ , and its $E^\ominus = +0.77 \text{ V}$ .
- For $\text{I}_2(\text{s})/\text{I}^-$ , the reduction half-reaction is $\text{I}_2(\text{s}) + 2\text{e}^- \rightarrow 2\text{I}^-$ .
  
  The oxidising agent is $\text{I}_2(\text{s})$ , and its $E^\ominus = +0.54 \text{ V}$ .
- For $\text{Cu}^{2+}/\text{Cu}$ , the reduction half-reaction is $\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}$ .
  
  The oxidising agent is $\text{Cu}^{2+}$ , and its $E^\ominus = +0.34 \text{ V}$ .
- For $\text{Ag}^{+}/\text{Ag}$ , the reduction half-reaction is $\text{Ag}^{+} + \text{e}^- \rightarrow \text{Ag}$ .
  
  The oxidising agent is $\text{Ag}^{+}$ , and its $E^\ominus = +0.80 \text{ V}$ .
Compare the standard electrode potential ( $E^\ominus$ ) values:

We list the $E^\ominus$ values for the identified oxidising agents:
- $\text{Fe}^{3+}$ : $+0.77 \text{ V}$
- $\text{I}_2(\text{s})$ : $+0.54 \text{ V}$
- $\text{Cu}^{2+}$ : $+0.34 \text{ V}$
- $\text{Ag}^{+}$ : $+0.80 \text{ V}$
Determine the strongest oxidising agent:

According to the principle, the species with the most positive $E^\ominus$ value will be the strongest oxidising agent.

Comparing the values, $+0.80 \text{ V}$ is the highest positive value among them. This value corresponds to the $\text{Ag}^{+}/\text{Ag}$ redox couple, where $\text{Ag}^{+}$ is the oxidising agent.

Therefore, $\text{Ag}^{+}$ has the greatest tendency to get reduced and is the strongest oxidising agent among the given options.

✓Final answer

\boxed{\text{Ag}^+}

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