TOPIC 3 - REDOX REACTIONS I + II

Question 1

Redox reactions

Answer 1

a reaction that involves both oxidation and reduction

Question 2

Oxidation

Reduction

Answer 2

• loss is e- and Hydrogen but gain in Oxygen

- gain in e- and hydrogen but loss in oxygem

Question 3

Oxidising agent

Answer 3

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-

Question 4

Reducing agent

Answer 4

Species that reduces another species and gets oxidised itself

Question 5

Half equation

Answer 5

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

Question 6

displacement reactions

Answer 6

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

Question 7

spectator ions

Answer 7

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

Question 8

Disproportionation reactions

Answer 8

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

Question 9

Oxidation number

Answer 9

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

Question 10

RULES OF OXIDATION NO.

Answer 10

. 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)
.

Question 11

If the oxidation number of a compound decreases, it has been…

Answer 11

reduced

Question 12

If the oxidation number increases in the compound has been…

Answer 12

oxidised

Question 13

Systematic names

Answer 13

Includes the oxidation number

Question 14

What do you need to make sure is the same when writing full equations from ionic half-equations?

Answer 14

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.

Question 15

ACIDIC/ NEUTRALISATION EQ CONDITIONS

Answer 15

• 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

Question 16

BASIC/ ALKALINE EQ CONDITIONS

Answer 16

• 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-

Question 17

oxidation state

Answer 17

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

Question 18

Standard electrode potential Conditions

Answer 18

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

Question 19

SHE (Standard Hydrogen Electrode) Structure

Answer 19

• Porous platinum foil immersed in 1 moldm^–3 HCl,

- This is enclosed in a tube containing hydrogen gas at standard pressure (100kPa).

Question 20

Why is using standard conditions important?

Answer 20

• 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

Question 21

Why is Platinum foil used?

Answer 21

• Hydrogen gas can adsorb onto the platinum surface

- This allows for a Hydrogen ion solution to quickly be established