19.1 - Redox equations

19.1.1: Balance redox equations in acid solution. Half equations and oxidation numbers may be used. H+ (aq) and H₂O should be used where necessary to balance half-equations.

Balancing redox equations

Redox equations are constructed from half equations showing the reduction or oxidation of the species involved.

By convention electrode potential half equations are written as reductions - electrons are added to the species being reduced on the left hand side of the equation:

Zn²⁺_(aq) + 2e Zn(s)

A reduction half equation can only be combined with an oxidation half equation. Using the electrode potentail series species on the left hand side will only react with a species on the right hand side that is more negative (or less positive). The equation E = E(red) - E(ox) must give a positive value greater than 0.3 for spontaneous reaction.

Consider the reaction between zinc metal and copper ions. (electrode potential of zinc is -0.76 and copper is 0.34).

The Eº for the reaction:

Eº = E(red) - E(ox) = 0.34 - (-0.76) = 1.10V

This tells us that the reaction is feasible and spontaneous.

Writing down the half equations in the correct form (reduction for copper and oxidation for zinc)

Zn(s) Zn2+(aq) + 2e

Cu2+(aq) + 2e Cu(s)

We check to make sure that the electrons are balanced. In this case they are, and the equations can be added together directly

Zn(s) Zn2+(aq) + 2e

Cu2+(aq) + 2e Cu(s)

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

If the number of electrons on both sides are different then the half equations must be multiplied through by appropriate quantities to balance the number of electrons on both sides.

Example Balance the redox reaction between manganate VII ions and iron 2+ ions - is the reaction feasible? Fe2+(aq) Fe3+(aq) + e-   Eº = 0.77 MnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O(l)   Eº = 1.49 We can see that the electrons are not balanced as there are 5 electrons on the left hand side and only one on the right hand side. The first equation must be multiplied through by 5 to equalise the electrons. 5Fe2+(aq) 5Fe3+(aq) + 5e-   Eº = 0.77 MnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O(l)   Eº = 1.49 Notice that this does NOT affect the electrode potential - this is just the difference in potential between the two states and is never altered regardless of whether the equation is written as a reduction or an oxidation or whether it is multiplied by any factor. The equations can now be added 5Fe2+(aq) 5Fe3+(aq) + 5e-   Eº = 0.77 MnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O(l)   Eº = 1.49 5Fe2+(aq) + MnO4-(aq) + 8H+(aq) 5Fe3+(aq)+ Mn2+(aq) + 4H2O(l) Is the reaction feasible? Applying: E = E(red) - E(ox) The reduced state is the MnO4- ion and the oxidised state is the Fe2+ Therefore E = E(red) - E(ox) E = 1.49 - 0.77 = 0.72V This result is positive and greater than 0.3 therefore the reaction is spontaneous

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