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AP Chemistry Notes

9.11.3 Mass Changes at Electrodes: Electroplating and Dissolution

AP Syllabus focus: ‘Determine mass deposited on or removed from an electrode (as in electroplating) using Faraday’s laws and electron accounting.’

Electrolysis can change electrode mass because electrons drive reduction at one electrode and oxidation at the other. Faraday’s laws connect electrical charge to moles of electrons, letting you predict plating or dissolution quantitatively.

Mass changes at electrodes: the core idea

When a species is reduced onto an electrode, that electrode gains mass; when an electrode material is oxidised into solution, that electrode loses mass. In both cases, the link between electricity and chemical change is electron counting.

Electroplating vs dissolution

Electrode mass changes depend on which half-reaction occurs at that surface.

Electroplating: the process of depositing a solid metal onto an electrode surface by reducing metal ions from solution (or melt) to metal atoms at the cathode.

Electroplating typically involves a metal cation (like Ag+\text{Ag}^+, Cu2+\text{Cu}^{2+}, Ni2+\text{Ni}^{2+}) gaining electrons and becoming a metal solid, increasing the electrode’s mass.

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Electroplating cell diagram showing a silver anode dissolving to form Ag+\text{Ag}^+ in solution while Ag+\text{Ag}^+ is reduced and deposited as Ag(s) onto the spoon cathode. The labeled electron flow through the external circuit highlights that reduction occurs at the cathode (mass gain) and oxidation occurs at the anode (mass loss). Source

Dissolution is the reverse: a metal solid becomes ions, decreasing the electrode’s mass.

Using Faraday’s laws to connect charge to mass

Faraday’s laws state that the amount of chemical change is proportional to the total charge passed. The essential workflow is: current and time → charge → moles of electrons → moles of substance → mass change.

A key point is that the balanced half-reaction tells you how many electrons per mole of metal (or other species) are transferred.

q=It q = It

q q = charge passed, C

I I = current, A (C s1\text{C s}^{-1})

t t = time, s

n(e)=qF n(e^-) = \dfrac{q}{F}

n(e) n(e^-) = moles of electrons transferred, mol

F F = Faraday constant, 9.6485×104 C mol1 e9.6485\times 10^4\ \text{C mol}^{-1}\ e^-

m=ItMzF m = \dfrac{ItM}{zF}

m m = mass deposited or dissolved, g

M M = molar mass of deposited/dissolved species, g mol1\text{g mol}^{-1}

z z = electrons transferred per formula unit (from the half-reaction), mol ee^-/mol substance

This calculation only works when you correctly identify zz from the relevant half-reaction at that electrode.

Writing the half-reaction to get the electron ratio

Mass changes are tied to the electrode reaction:

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Diagram of an electrolytic cell with labeled half-reactions, showing oxidation at the anode and reduction at the cathode driven by an external power source. This kind of schematic is useful for mapping each electrode’s half-reaction to the electron count zz, which is the key link between charge passed and predicted mass change. Source

Metal deposition at the cathode (mass increases)

  • General form: Mz+(aq)+zeM(s)\text{M}^{z+}(aq) + z e^- \rightarrow \text{M}(s)

  • The electrode gains mass because solid metal forms on the surface.

  • The moles of metal deposited equal n(e)z\dfrac{n(e^-)}{z}.

Metal dissolution at the anode (mass decreases)

  • General form: M(s)Mz+(aq)+ze\text{M}(s) \rightarrow \text{M}^{z+}(aq) + z e^-

  • The electrode loses mass because metal atoms leave the solid as ions.

  • The moles of metal dissolved equal n(e)z\dfrac{n(e^-)}{z}.

What the balance of electrons implies for both electrodes

In an electrolytic setup, the same current passes through both electrodes, so the same total moles of electrons are transferred at the anode and cathode. However, mass changes can differ if:

  • different substances react at the two electrodes, or

  • zz differs (e.g., Ag+\text{Ag}^+ needs 1 ee^-, but Cu2+\text{Cu}^{2+} needs 2 ee^-).

Common pitfalls that change the predicted mass

  • Using the overall (molecular) equation instead of the electrode half-reaction to find zz.

  • Forgetting to convert time to seconds when using q=Itq=It.

  • Confusing the sign of mass change:

    • cathode (reduction): typically mass increases for metal-ion plating

    • anode (oxidation): typically mass decreases for metal dissolution

  • Ignoring that the deposited/dissolved species must match the half-reaction (use the correct molar mass MM).

FAQ

If side reactions also consume electrons (e.g., hydrogen evolution), not all charge produces metal.

You can account for this by multiplying the theoretical mass by the efficiency (as a decimal).

Deposited metal can be powdery and fall off, or the surface may not adhere well.

Mechanical loss reduces measured mass without changing the charge passed.

Use the actual cathode half-reaction and the molar mass of the solid formed.

If multiple solids can form, you need additional information (conditions or observations) to decide which reaction dominates.

The calculated mass assumes a single metal with a single oxidation state.

Real dissolution may preferentially remove one component, so the effective $M$ and $z$ can differ from a pure-metal assumption.

Yes. Gas bubbles can cling to the electrode, making the apparent mass higher until they detach.

Careful rinsing and drying procedures are needed to obtain a reliable mass change.

Practice Questions

Q1 (3 marks) A nickel object is electroplated in a Ni2+(aq)\text{Ni}^{2+}(aq) solution using a current of 2.00 A2.00\ \text{A} for 10.0 min10.0\ \text{min}. Calculate the mass of Ni deposited. (F=9.6485×104 C mol1F=9.6485\times 10^4\ \text{C mol}^{-1}, M(Ni)=58.69 g mol1M(\text{Ni})=58.69\ \text{g mol}^{-1})

  • Uses q=Itq=It with t=600 st=600\ \text{s} to find q=1200 Cq=1200\ \text{C} (1)

  • Finds n(e)=q/F=1.24×102 moln(e^-)=q/F=1.24\times 10^{-2}\ \text{mol} (1)

  • Uses Ni2++2eNi\text{Ni}^{2+}+2e^-\rightarrow \text{Ni} so n(Ni)=6.22×103 moln(\text{Ni})=6.22\times 10^{-3}\ \text{mol} and m=0.365 gm=0.365\ \text{g} (1)

Q2 (6 marks) Copper is purified by electrolysis using copper electrodes in CuSO4(aq)\text{CuSO}_4(aq). A current of 0.750 A0.750\ \text{A} passes for 1.50 h1.50\ \text{h}. (a) Calculate the mass change of the cathode.
(b) Calculate the mass change of the anode.
(c) Write the half-equations at both electrodes.
(F=9.6485×104 C mol1F=9.6485\times 10^4\ \text{C mol}^{-1}, M(Cu)=63.55 g mol1M(\text{Cu})=63.55\ \text{g mol}^{-1})

  • Converts time: 1.50 h=5400 s1.50\ \text{h}=5400\ \text{s} and calculates q=It=4050 Cq=It=4050\ \text{C} (1)

  • Finds n(e)=q/F=4.20×102 moln(e^-)=q/F=4.20\times 10^{-2}\ \text{mol} (1)

  • Uses z=2z=2 for copper and finds n(Cu)=2.10×102 moln(\text{Cu})=2.10\times 10^{-2}\ \text{mol} (1)

  • Cathode mass change: +1.33 g+1.33\ \text{g} (allow 1.32–1.34 g) (1)

  • Anode mass change: 1.33 g-1.33\ \text{g} (allow 1.32–1.34 g) (1)

  • Half-equations: cathode Cu2++2eCu(s)\text{Cu}^{2+}+2e^-\rightarrow \text{Cu}(s); anode Cu(s)Cu2++2e\text{Cu}(s)\rightarrow \text{Cu}^{2+}+2e^- (1)

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