On the basis of standard electrode potential values, suggest which of the following reactions would take place

NCERT Exemplar · Q34

Q. On the basis of standard electrode potential values, suggest which of the following reactions would take place? (Consult the book for E⊖ value).

(i) Cu + Zn^2+ → Cu^2+ + Zn

(ii) Mg + Fe^2+ → Mg^2+ + Fe

(iii) Br2 + 2Cl^- → Cl2 + 2Br^-

(iv) Fe + Cd^2+ → Cd + Fe^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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A spontaneous redox reaction occurs only when the species with the more negative (or less positive) reduction potential is oxidised and the species with the more positive reduction potential is reduced. Using standard electrode potentials, reactions (ii) and (iv) are spontaneous.

The key to predicting whether a given redox reaction will take place lies in comparing the standard reduction potentials ( $E^\ominus$ ) of the two half-reactions involved. A reaction is spontaneous (i.e., it will proceed as written) if the cell potential $E^\ominus_{\text{cell}}$ is positive. This cell potential is calculated as:

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

where the cathode is where reduction occurs (gain of electrons) and the anode is where oxidation occurs (loss of electrons). The more positive the reduction potential, the stronger the tendency to be reduced. So, the species with the higher (more positive) $E^\ominus$ will act as the oxidising agent and get reduced, while the species with the lower (more negative) $E^\ominus$ will be oxidised.

Let’s examine each reaction using standard values (in volts, at 298 K):

\text{Cu}^{2+}/\text{Cu}: +0.34

1. Reaction (i): $\text{Cu} + \text{Zn}^{2+} \rightarrow \text{Cu}^{2+} + \text{Zn}$

Here, Cu is being oxidised (loses electrons to become $\text{Cu}^{2+}$ ) and $\text{Zn}^{2+}$ is being reduced (gains electrons to become Zn).

Oxidation half-reaction (anode): $\text{Cu} \rightarrow \text{Cu}^{2+} + 2e^-$ ; $E^\ominus_{\text{ox}} = -0.34$ V (reverse of reduction).
Reduction half-reaction (cathode): $\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}$ ; $E^\ominus_{\text{red}} = -0.76$ V.

E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}} = (-0.76) - (+0.34) = -1.10 \text{ V}

The cell potential is negative, so this reaction is non-spontaneous. In fact, the reverse reaction (Zn displacing Cu) is spontaneous — that’s the familiar zinc-copper cell.

Watch out

A common mistake is to forget that the anode potential must be taken as the oxidation potential (reverse sign of the reduction potential). Always use $E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}}$ directly with the reduction potentials of the half-cells as written.

2. Reaction (ii): $\text{Mg} + \text{Fe}^{2+} \rightarrow \text{Mg}^{2+} + \text{Fe}$

Mg is oxidised to $\text{Mg}^{2+}$ , and $\text{Fe}^{2+}$ is reduced to Fe.

Oxidation (anode): $\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^-$ ; $E^\ominus_{\text{red}} = -2.37$ V.
Reduction (cathode): $\text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}$ ; $E^\ominus_{\text{red}} = -0.44$ V.

E^\ominus_{\text{cell}} = (-0.44) - (-2.37) = +1.93 \text{ V}

A large positive value — this reaction is spontaneous. Magnesium, being a very strong reducing agent (very negative $E^\ominus$ ), readily displaces iron from its salt solution.

3. Reaction (iii): $\text{Br}_2 + 2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{Br}^-$

Here, $\text{Cl}^-$ is oxidised to $\text{Cl}_2$ , and $\text{Br}_2$ is reduced to $\text{Br}^-$ .

Oxidation (anode): $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-$ ; $E^\ominus_{\text{red}} = +1.36$ V.
Reduction (cathode): $\text{Br}_2 + 2e^- \rightarrow 2\text{Br}^-$ ; $E^\ominus_{\text{red}} = +1.09$ V.

$E^\ominus_{\text{cell}} = (+1.09) - (+1.36) = -0.27 \text{ V}$ …

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Q.On the basis of standard electrode potential values, suggest which of the following reactions would take place? (Consult the book for E⊖ value).

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