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Electroplating: Connecting Electricity and Chemistry

Electroplating uses electrical energy to deposit a metal onto the surface of another object. For example, silver plating occurs through the half-reaction

\[ \mathrm{Ag^+(aq)+e^- \rightarrow Ag(s)} \]

Every silver ion requires one electron to be reduced to metallic silver. Consequently, the amount of silver deposited is directly proportional to the amount of electrical charge that passes through the circuit.

Consider a current of

\[ 1.00\ \text{A} \]

flowing for

\[ 1.00\ \text{s} \]

The total charge delivered is

\[ Q=It=(1.00\ \text{A})(1.00\ \text{s})=1.00\ \text{C} \]

This amount of charge deposits

\[ 1.118\ \text{mg of Ag} \]

Since one mole of silver has a mass of

\[ 107.87\ \text{g} \]

the charge required to deposit one mole of silver is

\[ \frac{107.87\ \text{g}} {1.118\times10^{-3}\ \text{g C}^{-1}} = 9.65\times10^4\ \text{C} \]

or

\[ \boxed{96485\ \text{A}\cdot\text{s}} \]

This quantity is called one Faraday. Since the silver-plating reaction requires one electron per silver atom, one Faraday is the charge required to deposit one mole of silver. More generally, one Faraday is defined as the charge carried by

\[ \boxed{1.00\ \text{mol of electrons}} \]

The Faraday provides the essential connection between electrical measurements and chemical amounts in all electrochemical calculations.

The Faraday

Because electrochemical reactions involve enormous numbers of electrons, it is convenient to measure electrical charge in terms of moles of electrons rather than individual electrons.

The charge carried by one mole of electrons is called a Faraday, denoted by \(F\).

\[ \boxed{ F=96485\ \text{C mol}^{-1} } \]

Thus, one Faraday is defined as the charge carried by exactly

\[ 1.00\ \text{mol of electrons} \]

This constant provides the bridge between electrical measurements and chemical amounts.

Relating Current to Chemical Change

Electric current is the rate at which charge flows through a circuit:

\[ I=\frac{Q}{t} \]

where

\(I\) is the current (A),
\(Q\) is the electrical charge (C), and
\(t\) is the time (s).

Rearranging,

\[ Q=It \]

Once the total charge is known, the number of moles of electrons transferred is

\[ n_{e^-} = \frac{Q}{F} \]

Stoichiometry can then be used to determine how many moles—and ultimately how many grams—of material are deposited or consumed.

From Charge to Mass

Electroplating calculations generally follow the same sequence:

Calculate the total charge using \[ Q=It \]
Convert charge to moles of electrons: \[ n_{e^-}=\frac{Q}{F} \]
Use the balanced half-reaction to determine the number of moles of metal deposited.
Convert moles of metal to mass using the molar mass.

Big picture: Electroplating provides a direct connection between electricity and chemistry. By measuring electrical current and time, we can determine the number of electrons transferred, the amount of substance produced or consumed, and the mass of material deposited during an electrochemical process.

Worked examples

Worked Example: Determining the Charge on a Metal Ion

Gold is electroplated onto a metal surface by passing a current of

\[ 2.00\ \text{A} \]

for

\[ 30.0\ \text{s} \]

A total of

\[ 40.8\ \text{mg} \]

of gold is deposited. Determine the charge on the gold ion in solution.

Solution

First calculate the total charge that passed through the cell:

\[ Q=It \] \[ Q=(2.00\ \text{A})(30.0\ \text{s}) \] \[ Q=60.0\ \text{C} \]

Convert the charge to moles of electrons:

\[ n_{e^-} = \frac{Q}{F} \] \[ n_{e^-} = \frac{60.0}{96485} \] \[ n_{e^-} = 6.22\times10^{-4}\ \text{mol} \]

Next, calculate the number of moles of gold deposited:

\[ n_{\mathrm{Au}} = \frac{0.0408\ \text{g}} {196.97\ \text{g mol}^{-1}} \] \[ n_{\mathrm{Au}} = 2.07\times10^{-4}\ \text{mol} \]

The number of electrons required per gold atom is therefore

\[ \frac{n_{e^-}} {n_{\mathrm{Au}}} = \frac{6.22\times10^{-4}\ mol} {2.07\times10^{-4}\ mol} \] \[ =3.00 \]

Thus, three moles of electrons are required to reduce one mole of gold ions:

\[ \boxed{\mathrm{Au^{3+}+3e^- \rightarrow Au(s)}} \]

Physical interpretation: The electroplating experiment shows that each gold ion must gain three electrons before it can become a neutral gold atom. Therefore, the gold ion in solution has a charge of \(+3\).

Practice

Practice Problem: Copper Wire in Silver Nitrate
A copper wire is placed into a solution of \(\mathrm{AgNO_3}\). The copper displaces silver ions according to the net ionic reaction \[ \mathrm{Cu(s)+2Ag^+(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)} \] If \(0.362\ \text{g}\) of copper are lost from the wire: How many grams of silver should be recovered? How many coulombs of charge were transferred?
Show Answer First calculate the moles of copper that reacted: \[ n_{\mathrm{Cu}} = \frac{0.362\ \text{g}} {63.55\ \text{g mol}^{-1}} = 5.70\times10^{-3}\ \text{mol} \] From the balanced reaction, one mole of copper produces two moles of silver: \[ n_{\mathrm{Ag}} = 2(5.70\times10^{-3}) = 1.14\times10^{-2}\ \text{mol} \] \[ m_{\mathrm{Ag}} = (1.14\times10^{-2}\ \text{mol}) (107.87\ \text{g mol}^{-1}) = 1.23\ \text{g} \] \[ \boxed{m_{\mathrm{Ag}}=1.23\ \text{g}} \] Each copper atom loses two electrons: \[ \mathrm{Cu(s)\rightarrow Cu^{2+}(aq)+2e^-} \] \[ n_{e^-} = 2(5.70\times10^{-3}) = 1.14\times10^{-2}\ \text{mol e}^- \] The charge transferred is \[ Q=n_{e^-}F \] \[ Q= (1.14\times10^{-2}\ \text{mol}) (96485\ \text{C mol}^{-1}) = 1.10\times10^3\ \text{C} \] \[ \boxed{Q=1.10\times10^3\ \text{C}} \] The mass of silver is determined by reaction stoichiometry, while the charge transferred is determined by the number of electrons exchanged in the redox process.

Practice Problem: Copper Wire in Silver Nitrate

A copper wire is placed into a solution of \(\mathrm{AgNO_3}\). The copper displaces silver ions according to the net ionic reaction \[ \mathrm{Cu(s)+2Ag^+(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)} \] If \(0.362\ \text{g}\) of copper are lost from the wire:

How many grams of silver should be recovered?
How many coulombs of charge were transferred?

Show Answer

First calculate the moles of copper that reacted:

\[ n_{\mathrm{Cu}} = \frac{0.362\ \text{g}} {63.55\ \text{g mol}^{-1}} = 5.70\times10^{-3}\ \text{mol} \]

From the balanced reaction, one mole of copper produces two moles of silver:

\[ n_{\mathrm{Ag}} = 2(5.70\times10^{-3}) = 1.14\times10^{-2}\ \text{mol} \] \[ m_{\mathrm{Ag}} = (1.14\times10^{-2}\ \text{mol}) (107.87\ \text{g mol}^{-1}) = 1.23\ \text{g} \] \[ \boxed{m_{\mathrm{Ag}}=1.23\ \text{g}} \]

Each copper atom loses two electrons:

\[ \mathrm{Cu(s)\rightarrow Cu^{2+}(aq)+2e^-} \] \[ n_{e^-} = 2(5.70\times10^{-3}) = 1.14\times10^{-2}\ \text{mol e}^- \]

The charge transferred is

\[ Q=n_{e^-}F \] \[ Q= (1.14\times10^{-2}\ \text{mol}) (96485\ \text{C mol}^{-1}) = 1.10\times10^3\ \text{C} \] \[ \boxed{Q=1.10\times10^3\ \text{C}} \]

The mass of silver is determined by reaction stoichiometry, while the charge transferred is determined by the number of electrons exchanged in the redox process.

Practice Problem: Electroplating Silver
Silver metal is deposited according to the half-reaction \[ \mathrm{Ag^+(aq)+e^- \rightarrow Ag(s)} \] A current of \[ 2.50\ \text{A} \] is passed through \[ 100.0\ \text{mL} \] of \[ 1.00\ \text{M AgNO}_3 \] for \[ 60.0\ \text{s}. \] Assuming the solution volume remains constant, what is the concentration of \(\mathrm{Ag^+}\) remaining after electroplating?
Show Answer First calculate the total charge passed through the cell: \[ Q=It \] \[ Q=(2.50\ \text{A})(60.0\ \text{s}) =150\ \text{C} \] Convert the charge to moles of electrons: \[ n_{e^-} = \frac{Q}{F} = \frac{150}{96485} = 1.55\times10^{-3}\ \text{mol} \] Since one electron reduces one silver ion, \[ n_{\mathrm{Ag^+}} = n_{e^-} = 1.55\times10^{-3}\ \text{mol} \] Initially, the solution contains \[ n_{\mathrm{Ag^+}} = (1.00\ \text{mol L}^{-1}) (0.1000\ \text{L}) = 0.1000\ \text{mol} \] Therefore, the moles of silver ions remaining are \[ 0.1000-0.00155 = 0.09845\ \text{mol} \] Since the solution volume remains \[ 0.1000\ \text{L}, \] the final concentration is \[ [\mathrm{Ag^+}] = \frac{0.09845}{0.1000} = 0.984\ \text{M} \] \[ \boxed{[\mathrm{Ag^+}]=0.984\ \text{M}} \]

Practice Problem: Electroplating Silver

Silver metal is deposited according to the half-reaction \[ \mathrm{Ag^+(aq)+e^- \rightarrow Ag(s)} \] A current of \[ 2.50\ \text{A} \] is passed through \[ 100.0\ \text{mL} \] of \[ 1.00\ \text{M AgNO}_3 \] for \[ 60.0\ \text{s}. \] Assuming the solution volume remains constant, what is the concentration of \(\mathrm{Ag^+}\) remaining after electroplating?

Show Answer

First calculate the total charge passed through the cell:

\[ Q=It \] \[ Q=(2.50\ \text{A})(60.0\ \text{s}) =150\ \text{C} \]

Convert the charge to moles of electrons:

\[ n_{e^-} = \frac{Q}{F} = \frac{150}{96485} = 1.55\times10^{-3}\ \text{mol} \]

Since one electron reduces one silver ion,

\[ n_{\mathrm{Ag^+}} = n_{e^-} = 1.55\times10^{-3}\ \text{mol} \]

Initially, the solution contains

\[ n_{\mathrm{Ag^+}} = (1.00\ \text{mol L}^{-1}) (0.1000\ \text{L}) = 0.1000\ \text{mol} \]

Therefore, the moles of silver ions remaining are

\[ 0.1000-0.00155 = 0.09845\ \text{mol} \]

Since the solution volume remains

\[ 0.1000\ \text{L}, \]

the final concentration is

\[ [\mathrm{Ag^+}] = \frac{0.09845}{0.1000} = 0.984\ \text{M} \] \[ \boxed{[\mathrm{Ag^+}]=0.984\ \text{M}} \]

Key points (one glance)

Electroplating uses electrical energy to drive a non-spontaneous reduction reaction, depositing a metal onto the surface of an object.
The total electrical charge delivered by a current is \[ Q = It \] where \(Q\) is the charge (C), \(I\) is the current (A), and \(t\) is the time (s).
One Faraday is the charge carried by one mole of electrons: \[ F = 96485\ \text{C mol}^{-1} \]
The number of moles of electrons transferred is calculated from \[ n_{e^-} = \frac{Q}{F} \]
The balanced half-reaction relates the number of moles of electrons transferred to the number of moles of material deposited or consumed.
Once the number of moles of a substance is known, its mass can be determined from its molar mass.
Electroplating calculations often follow the sequence:

Current \(\rightarrow\) Charge \(\rightarrow\) Moles of electrons \(\rightarrow\) Moles of substance \(\rightarrow\) Mass or concentration
Measuring the amount of material deposited also allows the stoichiometry of an electrochemical half-reaction to be determined, including the charge on the ions involved.

Big picture: Electroplating provides a direct connection between electrical measurements and chemical change. By relating current and time to the number of moles of electrons transferred, the Faraday constant allows electrochemical reactions to be analyzed using ordinary chemical stoichiometry.