Q.The increasing order of reduction of alkyl halides with zinc and dilute HCl is

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):

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

or

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):

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:

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

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 $E_{\text{mix}}$:

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:

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:

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

5. The Area Effect: Why It Matters

From the mixed potential equation:

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:

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

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.