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Electrochemistry - Electroplating

Grade 12IGCSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Electroplating is the process of coating the surface of a conducting object with a layer of metal using an electrolytic cell.

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The Cathode (negative electrode) is the object to be plated. Reduction occurs here: Mn+(aq)+ne−→M(s)M^{n+}(aq) + ne^- \rightarrow M(s).

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The Anode (positive electrode) is made of the pure metal that is to be deposited. Oxidation occurs here: M(s)→Mn+(aq)+ne−M(s) \rightarrow M^{n+}(aq) + ne^-.

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The Electrolyte must be an aqueous solution of a soluble salt of the plating metal (e.g., AgNO3AgNO_3 for silver plating or CuSO4CuSO_4 for copper plating).

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The concentration of metal ions in the electrolyte remains constant because the rate of metal dissolution at the anode is equal to the rate of metal deposition at the cathode.

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Purposes of electroplating include prevention of corrosion (e.g., chrome plating on steel) and improving aesthetic appearance (e.g., silver plating on jewelry).

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According to Faraday's Laws, the mass of metal deposited is directly proportional to the quantity of electricity (QQ) passed through the cell.

📐Formulae

Q=I×tQ = I \times t

n(e−)=QFn(e^-) = \frac{Q}{F}

F≈96500 C mol−1F \approx 96500 \text{ C mol}^{-1}

m=I⋅t⋅Mrz⋅Fm = \frac{I \cdot t \cdot M_r}{z \cdot F}

Moles of metal=Moles of electronsvalency (z)\text{Moles of metal} = \frac{\text{Moles of electrons}}{\text{valency } (z)}

💡Examples

Problem 1:

A spoon is to be electroplated with silver (AgAg) using a current of 0.5 A0.5\text{ A} for 2020 minutes. Calculate the mass of silver deposited on the spoon. (Given: Ar(Ag)=108A_r(Ag) = 108, F=96500 C mol−1F = 96500\text{ C mol}^{-1})

Solution:

  1. Calculate total charge: Q=I×t=0.5 A×(20×60) s=600 CQ = I \times t = 0.5\text{ A} \times (20 \times 60)\text{ s} = 600\text{ C}.
  2. Calculate moles of electrons: n(e−)=QF=60096500≈0.00622 moln(e^-) = \frac{Q}{F} = \frac{600}{96500} \approx 0.00622\text{ mol}.
  3. Use the half-equation: Ag+(aq)+e−→Ag(s)Ag^+(aq) + e^- \rightarrow Ag(s). Since 11 mole of e−e^- deposits 11 mole of AgAg, n(Ag)=0.00622 moln(Ag) = 0.00622\text{ mol}.
  4. Calculate mass: m=n×Ar=0.00622×108≈0.672 gm = n \times A_r = 0.00622 \times 108 \approx 0.672\text{ g}.

Explanation:

The quantity of electricity is found first, then converted to moles of electrons using Faraday's constant. The stoichiometry of the silver reduction (1:11:1 ratio) allows us to find the moles of silver, which is then converted to mass using the relative atomic mass.

Problem 2:

State the half-equations occurring at the anode and cathode during the electroplating of an iron nail with copper (CuCu) using a CuSO4CuSO_4 electrolyte.

Solution:

At the Anode (Positive): Cu(s)→Cu2+(aq)+2e−Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-. At the Cathode (Negative): Cu2+(aq)+2e−→Cu(s)Cu^{2+}(aq) + 2e^- \rightarrow Cu(s).

Explanation:

At the anode, the copper atoms lose electrons to form ions, which enter the solution. At the cathode, those copper ions gain electrons to form solid copper metal, which coats the iron nail.

Electroplating - Revision Notes & Key Formulas | IGCSE Grade 12 Chemistry