TOPIC 3 - REDOX REACTIONS I + II

a reaction that involves both oxidation and reduction

• loss is e- and Hydrogen but gain in Oxygen - gain in e- and hydrogen but loss in oxygem

Species that oxidises another species and gets reduced itself e.g. F2 in 2NaCl as it oxidises Cl- to Cl2 and accpets e- from Cl- and redduces to F-

Species that reduces another species and gets oxidised itself

An ionic equation used to describe either the loss or gain of electrons during a redox reaction. Use this to FULL IONIC EQ. • LOSS e- write at RIGHT • GAIN e- write at LEFT

Redox reactions which can be used to compare the relative strengths of metals as reducing agents and non-metals as oxidising agents.

Ions which are present in solution but take no part in the reaction.

when an element is simultaneously oxidised and reduced in a reaction.

Charge that an ion has or an ion would have if its species were fully ionic.

. oxid. no. of all elements in a neutral compound/ uncombined element = 0 . sum of oxid. no of all elements in an ion = charge of the ion. . in compound the MORE electronegative element is given a NEGATIVE OXID NO. . oxid. no of F is ALWAYS -1 .oxid. no. of H is +1 EXCEPT when combined with less electronegative elements it is -1 and in metal HYDRIDES. . oxid. no. of oxygen = -2 and when combined with F it is +2 and with peroxides = -1 (Na2O2, H2O2, Li2O2) .

reduced

oxidised

Includes the oxidation number

The number of electrons in either half equation is the same. This can be done by multiplying t either one of both of the whole equations to reach a common denominator of electrons. E.g. if one half equation had 2 electrons, and the other 5, you would multiply the whole of the first by 5, and the whole of the second by 2, to reach 10 electrons on both sides.

• balance atoms apart from oxygen and hydrogen. • balance oxygen by adding H20 molecules on other side. • balance hydrogen by H+ ions to other side • balance charges by adding e- e.g. SO4 2- —-> H2S 8e- + 10H^+ + SO4 2- —-> H2S + 4H2O

• ADDITIONAL STEP to get rid of H+ ions you add OH- ions to both sides. e.g. Ti^+ —-> Ti2O3 6OH- + 3H20 + 2Ti^+ —-> Ti2O3 + 6H+ + 6OH- 6OH- + 3H20 + 2Ti^+ —-> Ti2O3 + 6H20 6OH- + 2Ti^+ —-> Ti2O3 + 3H20 6OH- + 2Ti^+ —-> Ti2O3 + 3H20 + 6e-

is the charge of the ion or the charge it would have if it were fully ionic

• 298 K • 100 kPa for gasses • 1.00 mol dm^-3 concentration for ions

• Porous platinum foil immersed in 1 moldm^–3 HCl, - This is enclosed in a tube containing hydrogen gas at standard pressure (100kPa).

• The position of equilibrium can change if we alter these conditions • In order to make a fair comparison with other metals we make the conditions the same

• Hydrogen gas can adsorb onto the platinum surface - This allows for a Hydrogen ion solution to quickly be established