Products of Electrolysis & Reasoning

CHEElectrolysis & Faraday's Laws

This subtopic explains which species are actually discharged at each electrode during electrolysis, using preferential discharge and the difference between molten and aqueous electrolytes. Classic cases are molten vs aqueous $NaCl$ , dilute vs concentrated $H_2SO_4$ , and electrorefining of copper.

Molten NaCl

Na^+ + e^- \rightarrow Na \quad (\text{cathode}) \qquad 2Cl^- \rightarrow Cl_2 + 2e^- \quad (\text{anode})

no water present, so the metal Na is obtained.

Aqueous NaCl (brine)

2H_2O + 2e^- \rightarrow H_2 + 2OH^- \quad (\text{cathode}) \qquad 2Cl^- \rightarrow Cl_2 + 2e^- \quad (\text{anode})

water is reduced in preference to

Na^+

;

NaOH

remains in solution.

Dilute

H_2SO_4

(electrolysis of water)

2H^+ + 2e^- \rightarrow H_2 \quad (\text{cathode}) \qquad 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \quad (\text{anode})

SO_4^{2-}

has a high discharge potential and is not oxidised.

Electrorefining of copper

Cu \rightarrow Cu^{2+} + 2e^- \quad (\text{impure anode}) \qquad Cu^{2+} + 2e^- \rightarrow Cu \quad (\text{pure cathode})

in aqueous

CuSO_4

; noble impurities collect as anode mud.

Preferential discharge: at the cathode the cation with the higher (more positive) reduction potential is discharged; at the anode the species more easily oxidised (lower discharge potential) is liberated.
Molten $NaCl$ : cathode $Na^+ + e^- \rightarrow Na$ ; anode $2Cl^- \rightarrow Cl_2 + 2e^-$ — no water, so the metal is obtained.
Aqueous $NaCl$ (brine): cathode gives $H_2$ ( $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$ ) because water is reduced more easily than $Na^+$ is; anode gives $Cl_2$ (high $Cl^-$ concentration and overpotential favour it over $O_2$ ), leaving $NaOH$ in solution.
Dilute $H_2SO_4$ : cathode $2H^+ + 2e^- \rightarrow H_2$ ; anode $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$ — effectively electrolysis of water; $SO_4^{2-}$ is not discharged because water is oxidised more readily.
Concentrated $H_2SO_4$ gives the same gases ( $H_2$ at cathode, $O_2$ at anode) because $SO_4^{2-}$ has a very high discharge potential and is not oxidised.
Overvoltage (overpotential) explains why $O_2$ evolution is kinetically hindered, so $Cl_2$ is preferred at the anode in concentrated chloride solutions despite $E^\circ$ values.
The discharge order at the cathode roughly follows reduction potential: ions like $Cu^{2+}, Ag^+, H^+$ are discharged in preference to active-metal ions ( $Na^+, K^+, Ca^{2+}, Al^{3+}$ ), which stay in solution in aqueous media.
At the anode the ease of discharge roughly follows $OH^-/Cl^-/Br^-/I^- >$ oxoanions like $SO_4^{2-}, NO_3^-$ (which are usually not discharged; water/ $OH^-$ is oxidised to $O_2$ instead).
Nature of the electrode matters: with an attackable (active) anode such as copper, the anode itself dissolves ( $Cu \rightarrow Cu^{2+} + 2e^-$ ) instead of an anion being discharged — this is the basis of electrorefining and electroplating.
Electrorefining of copper uses aqueous $CuSO_4$ with an impure-Cu anode and pure-Cu cathode: $Cu \rightarrow Cu^{2+} + 2e^-$ at the anode and $Cu^{2+} + 2e^- \rightarrow Cu$ at the cathode; nobler impurities (Ag, Au, Pt) settle as anode mud while $Cu$ is purified.
More reactive impurities than copper (Fe, Zn) dissolve into solution as ions but are not deposited at the cathode (their ions stay in solution), so the cathode collects pure copper.
Concentration of the electrolyte can change products: dilute NaCl tends to give $O_2$ at the anode, but concentrated brine gives $Cl_2$ because high $Cl^-$ concentration plus overvoltage favours chloride discharge.

Where the marks go

Claiming aqueous $NaCl$ gives $Na$ at the cathode — water is reduced to $H_2$ instead; sodium metal forms only from MOLTEN NaCl.
Writing $SO_4^{2-}$ being discharged at the anode in $H_2SO_4$ electrolysis — it is not; water is oxidised to $O_2$ .
Ignoring overvoltage: by $E^\circ$ alone $O_2$ should appear at the anode of brine, but overpotential makes $Cl_2$ the actual product.
Forgetting that an active (copper) anode dissolves rather than evolving a gas — key to electrorefining and electroplating.
Stating that reactive-metal impurities (Fe, Zn) deposit on the cathode during copper refining — they remain in solution as ions; only copper is deposited.

How the board asks it

Give reasonspreferential discharge of water over active-metal cations
Give reasons: When aqueous $NaCl$ solution is electrolysed, $H_2$ is liberated at the cathode and not $Na$ , even though $Na^+$ ions are present in large amounts.
Predict the productmolten vs aqueous electrolytes
Name the products obtained at the cathode and the anode, with balanced electrode reactions, when (i) molten $NaCl$ and (ii) concentrated aqueous $NaCl$ are electrolysed using inert electrodes.
Predict the productnature of the electrode (inert vs active anode)
State the products liberated at the anode and the cathode when an aqueous solution of $CuSO_4$ is electrolysed using (i) platinum electrodes and (ii) copper electrodes.
Give reasonsactive anode dissolution in electrorefining
Account for the following: During the electrorefining of copper using an impure copper anode in $CuSO_4$ solution, the anode dissolves while pure copper is deposited at the cathode, and impurities such as $Ag$ and $Au$ collect as anode mud.
Give reasonsovervoltage favouring $Cl_2$ over $O_2$
Explain why, on electrolysing concentrated brine with inert electrodes, $Cl_2$ is evolved at the anode rather than $O_2$ , even though water is oxidised more readily than $Cl^-$ on the basis of $E^\circ$ values alone.
Distinguishwater electrolysis vs metal deposition
How will you distinguish, in terms of the products formed at each electrode, between the electrolysis of dilute $H_2SO_4$ and the electrolysis of aqueous $CuSO_4$ , both using platinum electrodes?

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.