Faraday's Laws of Electrolysis

CHEElectrolysis & Faraday's Laws

Faraday's laws relate the mass of substance liberated at an electrode to the quantity of electricity passed. These are the workhorse formulae for electrolysis numericals on deposited metals and gases liberated, including series-connected cells.

Master charge $Q = It$ and the role of $n$ (electrons per ion).

First law (mass deposited)

m = ZIt = \dfrac{M I t}{nF}

m

= mass,

Z

= electrochemical equivalent,

I

= current (A),

t

= time (s),

M

= molar mass,

n

= electrons per ion,

F = 96500\,\text{C mol}^{-1}

Second law (cells in series)

\dfrac{m_1}{m_2} = \dfrac{E_1}{E_2} = \dfrac{M_1/n_1}{M_2/n_2}

equal charge deposits masses in the ratio of equivalent weights

E = M/n

Moles of electrons / charge

Q = It \qquad \text{moles of } e^- = \dfrac{It}{F} \qquad \text{moles deposited} = \dfrac{It}{nF}

1\,F = 96500\,\text{C}

= charge of one mole of electrons.

Gas volume at STP

V = \left(\dfrac{It}{nF}\right)\times 22400\ \text{mL}

n

= electrons per molecule of gas;

22400\,\text{mL} = 22.4\,\text{L}

per mole at STP.

First law: mass deposited $m \propto Q$ , i.e. $m = Z \times I \times t$ , where $Z$ is the electrochemical equivalent and $Q = It$ (current in amperes $\times$ time in seconds).
Second law: for the same quantity of electricity through different electrolytes (cells in series), masses deposited are proportional to their equivalent weights: $\dfrac{m_1}{m_2} = \dfrac{E_1}{E_2}$ .
Key constant: $1\,F = 96500\,\text{C mol}^{-1}$ = charge on one mole of electrons; equivalent weight $E = \dfrac{M}{n}$ where $n$ = electrons gained/lost per ion.
Electrochemical equivalent $Z = \dfrac{E}{F} = \dfrac{M}{nF}$ is the mass deposited by $1\,\text{C}$ ( $1\,\text{A}$ for $1\,\text{s}$ ); units $\text{g C}^{-1}$ .
Working formula: moles deposited $= \dfrac{It}{nF}$ , so mass $= \dfrac{M \cdot I \cdot t}{n F}$ . Always convert time to seconds (e.g. $30$ min $= 1800\,\text{s}$ ).
Charge per ion sets $n$ : $Cu^{2+} + 2e^- \rightarrow Cu$ ( $n=2$ ); $Ag^+ + e^- \rightarrow Ag$ ( $n=1$ ); $Al^{3+} + 3e^- \rightarrow Al$ ( $n=3$ ). Using the wrong $n$ is the most common error.
Gas volumes at STP: use moles $\times 22400\,\text{mL}$ (i.e. $22.4\,\text{L mol}^{-1}$ ). At the anode $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$ ( $n=4$ per $O_2$ ); $2Cl^- \rightarrow Cl_2 + 2e^-$ ( $n=2$ per $Cl_2$ ).
Series-cell shortcut: equal charge means equal moles of electrons, so equal numbers of equivalents are deposited in each cell — equate equivalents (moles $\times n$ ), not moles.
Worked check (Cu, $2\,\text{A}$ , $30\,\text{min}$ ): $Q = 2\times1800 = 3600\,\text{C}$ ; $m = \dfrac{63.5\times3600}{2\times96500} \approx 1.18\,\text{g}$ of Cu.
Worked check (Ag, $0.5\,\text{A}$ to deposit $1.08\,\text{g}$ ): $t = \dfrac{mnF}{MI} = \dfrac{1.08\times1\times96500}{108\times0.5} \approx 1930\,\text{s}$ .
For the SAME charge, more electrons per ion ( $n$ ) means LESS metal deposited per gram-equivalent — e.g. Al ( $n=3$ ) deposits less mass than Ag ( $n=1$ ) for the same coulombs.
Total charge can also come from $Q = n_e F$ where $n_e$ is moles of electrons; use this when the problem gives moles of product directly.

Where the marks go

Using the wrong $n$ (valency / electrons per ion) — the single most common slip; check the half-reaction, not the formula.
Forgetting to convert minutes or hours to seconds before applying $Q = It$ .
Using molar mass instead of equivalent weight (or vice versa) inconsistently — $m = \dfrac{MIt}{nF}$ already contains $M/n$ , so do not divide by $n$ twice.
For gases, dividing by the wrong $n$ : $O_2$ needs $4e^-$ per molecule and $Cl_2$ needs $2e^-$ — using $n=2$ for $O_2$ doubles the volume.
In series cells, equating moles deposited instead of equivalents — equal charge gives equal equivalents, not equal moles.

How the board asks it

Numerical $m = \dfrac{MIt}{nF}$
A current of $2\,\text{A}$ is passed through molten $CuCl_2$ for $30$ minutes. Calculate the mass of copper deposited at the cathode. $(Cu = 63.5,\ 1\,F = 96500\,\text{C})$
Numericalsecond law / cells in series
The same quantity of electricity that deposits $1.08\,\text{g}$ of silver from $AgNO_3$ is passed through a $CuSO_4$ solution connected in series. Calculate the mass of copper deposited. $(Ag = 108,\ Cu = 63.5)$
Numericalgas volume at STP
On electrolysis of acidified water, a charge of $9650\,\text{C}$ is passed. Calculate the volume of $O_2$ liberated at the anode at STP. $(2H_2O \rightarrow O_2 + 4H^+ + 4e^-)$
Numerical $t = \dfrac{mnF}{MI}$
Calculate the time for which a current of $1.5\,\text{A}$ must be passed to deposit $1.27\,\text{g}$ of copper from a $CuSO_4$ solution. $(Cu = 63.5,\ 1\,F = 96500\,\text{C})$
Define / state $1\,F = 96500\,\text{C mol}^{-1}$
State Faraday's first and second laws of electrolysis. Define the term electrochemical equivalent $(Z)$ and give its SI unit.
Give reasonsrole of $n$ for the same charge
Account for the fact that when the same quantity of electricity is passed through solutions of $AgNO_3$ and $AlCl_3$ , the mass of silver deposited is greater than the mass of aluminium deposited.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.