Topic 7: Oxidation and reduction (redox reactions) (physical chem) Flashcards
define oxidation
loss of electrons
define reduction
gain of electrons
When does a redox reaction occur
when oxidation and reduction occur simultaneously (one species looses electrons which are then donated and gained by the other species)
What is an oxidation number
Gives the oxidation state of an element or ionic substance
-e.g hydrogen = +1
-oxygen = -2
Group 7 = -1
Group 1 = +1
Oxidising agent
-accepts electrons from the species that is being oxidised so it gains electrons and is reduced (oxidation no. becomes more negative)
Reducing agent
-donates electrons to the species being reduced
-looses electrons and is oxidised (positive oxidation no.)
half equations
-used to show separate oxidation and reduction reactions that occur
state in terms of electrons the meaning of the term oxidising agent
electron accepter
equation for oxidation of iron
Fe2+ –> Fe3+ + e-
when is oxidation number the same as charge
only for monotomic elements
Cl2 (aq) + 2Br (aq) –> 2Cl- (aq) + Br2 (aq) –> what is being oxidised and what is being reduced
Cl2 = reduced (neutral to negative)
Br = oxidised (negative to neutral)
negative to neutral
oxidisation
reduction vs oxidation
-Elements can either be oxidised or reduced (transfer of electrons between elements and the system)
-reduction = gain of electrons
-oxidation = loss of electrons
Al3+ + 3e- –> Al what is happening in this equation
aluminum is reduced (gains electrons)
2Cl- –> Cl2 + 2e- what is happening here
chlorine is oxidsed (looses electron)
what side of the equation has oxidised and what side is reduction
oxidised = right hand side
reduction = left hand side
steps to calculating half equations
1) molecule –> compound or compound –> molecule
2) Balance atoms
3) See charges to tell whether it is reduced or oxidised
4) Add electrons
5) Balance equation
solid vs aqueous
Solid = no charge
Aqueous = charge
Zn + CuSO4 –> ZnSO4 + Cu
write the two half equations
Zn –> Zn2+ + 2e- (oxidised)
Cu2+ + 2e- –> Cu (reduction)
molecule to compound
reduction
calculate the oxidation number in Fe2O3
Fe = +3
O = -2
state in terms of electrons why iodine is classified as an oxidising agent in this reaction
-electron acceptor
-iodine gains electrons and becomes reduced
what is oxidised and reduced in this equation using oxidation numbers
H2S + Cl2 –> S + 2HCl
S has been oxidised from -2 in H2S to 0 in S
Cl has been reduced from 0 in Cl2 to -1 in 2HCl
define oxidation number
-Oxidation number shows us how many electrons are gained or lost by an element when atoms turn into ions
-Oxidation number –> the charge that an ion has or the charge it would have if the bonding electrons are completely transferred
more positive vs negative oxidation number
The more electrons removed (positive oxidation number) = oxidation
The more electrons added (negative oxidation number) = reduction
rules for oxidation numbers
-uncombined element = zero
-ions of just one atom = charge of the ion
-neutral compound = zero
-sum of oxidation numbers for an ion = charge of ion
-Flourine in its compound = -1
-Group 1 metals = +1
-H = +1 except in metal hydrides which would be –1
-Oxygen = -2 expect when bonded to flourine or H2O2
oxidising agent
oxidises another element and gets reduced itself by receiving electrons from the element becoming oxidised. (element reduced)
reducing agent
Reducing agent –> reduces another element and gets oxidised by transferring its electrons to the elements being reduced (element oxidised)
reducing and oxidation number in equation
Mg + CuSO4 –> MgSO4 + Cu
-> Mg is oxidised (0 –> +2)
–> Cu is reduced (+2 –> 0)
–> Cu in CuSO4 is oxidising agent
—> Mg is the reducing agent
equation for Cr3+ –> Cr2O7
2Cr3+ + 7H2O –> Cr2O7^2- + 14H+ + 6e-
disproportionation
-Disproportionation is where an element is simultaneously oxidised and reduced in the same reaction
-e.g Cl2 can be both oxidised and reduced in two separate compounds so has experienced disproportionation
balancing redox equations
-chemical equation –> ionic equation –> half equation –> redox
1) Write half equations for reduction and oxidation
2) Add H2O to balance oxygen
3) Add H+ to balance hydrogen atoms
4) Add electrons to balance the charges (must be the same on both sides e.g -1 on both sides)
combining half equations
1) Write both half equations balanced
2) Multiply so number of electrons on both equations are equal
3) Add half equations together and cancel out electrons, H+ ions and common species
Explain in terms of oxidation states why magnesium is the reducing agent
Mg changes oxidation state from 0 to +2 so electrons are lost
oxidation
ion –> molecule + electron
A solution of sodium chlorate was added to a colourless solution of potassium iodidie. Suggest what happens and why
-solution goes brown
-bc of iodine
-iodine is oxidised
oxidiation vs reduction
oxidation = increase in oxidation number
reduction = decrease in oxidation number
reducing agents are
electron donors
(species that is oxidised)
write a half equation for the reduction of chlorate ions to form chlorine in acidic conditions
2ClO- + 2e- + 4H+ —> Cl2 + 2H2O
true or false - oxidation state must equal charge of compound
e.g SO4^2- –> charge of s is 6 instead of 8
true
why is the oxidation state of chromium is zero in Cr(PF3)6
PF3 is neutral and the complex is neutral, so chromium must have an oxidation state of zero
