4.4.3.5 - Electrode half equations

4.4.3.5 - Electrode half equations

Learn to represent cathode reductions and anode oxidations with balanced half equations. Work from the reacting ions to the element formed, then conserve atoms and charge using electrons. The method applies to molten compounds, aluminium extraction and aqueous electrolysis, including hydrogen and oxygen from ions supplied by water.

Reduction at the cathode (Higher)

Ions gain electrons at the negative cathode, so the cathode reaction is reduction.

The negative electrode supplies electrons to positively charged ions at its surface. A copper ion with charge 2+2+ needs two electrons to become a neutral copper atom:

CuX2++2 eX−→Cu\ce{Cu^2+ + 2e- -> Cu}

The left-hand charge is +2−2=0+2-2=0, equal to the neutral product's charge. The nucleus remains a copper nucleus; gaining electrons changes its charge, not its element identity.

For a second worked example, aluminium has ion charge 3+3+, so choose three electrons:

AlX3++3 eX−→Al.\ce{Al^3+ + 3e- -> Al}.

The left charge is +3−3=0+3-3=0, matching the right. Molten sodium chloride instead needs one electron for each NaX+\ce{Na+}: NaX++eX−→Na\ce{Na+ + e- -> Na}. Use the actual ion charge. A coefficient of two before AlX3+\ce{Al^3+} describes two ions and requires six electrons overall, rather than changing either ion's charge.

Choose the correct product before writing its half equation. In aqueous sodium chloride, hydrogen forms instead of sodium because sodium is more reactive than hydrogen. A correctly balanced sodium half equation would therefore represent the wrong cathode product for that aqueous solution. Balancing verifies conservation; it does not replace the electrolysis prediction rules.

Oxidation at the anode (Higher)

Negative ions lose electrons at the positive anode, so the anode reaction is oxidation.

A chloride ion becomes a neutral chlorine atom by losing an electron, but elemental chlorine occurs as ClX2\ce{Cl2}. Two chloride ions are therefore required:

2 ClX−→ClX2+2 eX−\ce{2Cl- -> Cl2 + 2e-}

There are two chlorine atoms on each side. Total charge is −2 on the left and −2 from the electrons on the right. The product is chlorine, not chloride: the name distinguishes the neutral element from the reacting ion.

Bromide and iodide follow the same pattern: 2 BrX−→BrX2+2 eX−\ce{2Br- -> Br2 + 2e-} and 2 IX−→IX2+2 eX−\ce{2I- -> I2 + 2e-}. The element formula determines the required atom count, while ion charges determine the electron count.

In a molten oxide, oxide ions form oxygen:

2 OX2−→OX2+4 eX−\ce{2O^2- -> O2 + 4e-}

Two oxide ions contain two oxygen atoms and carry total charge −4. The two-atom oxygen molecule is neutral, so four electrons must appear on the right. Using only two electrons would balance the oxygen atoms but not charge. This is the electrode equation for oxygen formation from oxide in aluminium extraction; the subsequent reaction between oxygen and the carbon anode is a separate reaction.

Hydrogen and oxygen from aqueous electrolytes (Higher)

Ions from water can provide electrode products when the dissolved compound's ions remain in solution.

At the cathode of an aqueous solution containing a metal more reactive than hydrogen, hydrogen ions gain electrons and form hydrogen gas:

2 HX++2 eX−→HX2\ce{2H+ + 2e- -> H2}

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