Chemsitry: Redox Equations

Question 1

When is a redox equation balanced?

Answer 1

-numbers of atoms of each element on both sides are the same
-total charge same on both sides

Question 2

What steps do you take in order to balance an unbalanced redox equation?

Answer 2

-write oxidation states above each element
-identify species oxidised and reduced
-determine amount of electrons gained/lost
-balance number of electrons gained/lost
-balance remaining atoms

Question 3

Using the unbalanced equation, MnO4- + Fe2+ + H+ -> Mn2+ + Fe3+ + H2O, show the steps involved to balance this redox equation?

Answer 3

1. oxidation states above each species
   Mn: +7 O4: -2 Fe: +2 H: +1 -> Mn: +2 Fe: +3 H: +1 O: -2
2. identify species oxidised and reduced
   Mn: +7 -> Mn:+2 reduction, gain 5e-
   Fe: +2 -> +3 oxidation, loss 1e-
3. balance number of electrons transferred
   MnO4- + 5e- -> Mn2+
   Fe2+ -> Fe3+ + e-
   therefore multiple by 5
   MnO4- + 5Fe2+ + H+ -> Mn2+ + 5Fe3+ + H2O
4. balance all other elements involved
   MnO4- + 5Fe2+ + 8H+ -> Mn2+ + 5Fe3+ + 4H2O

Question 4

What happen in a disproportionation reaction?

Answer 4

atoms of same element both reduced and oxidised

Question 5

If an element has been oxidised, on what side are the electrons added in a half equation?

Answer 5

side of products

Question 6

If an element has been reduced, what side are the electrons added to?

Answer 6

side with reactants

Question 7

How do you find the number of electrons transferred from oxidation states?

Answer 7

the difference between oxidation states

Question 8

If the oxidation state increases, has it been oxidised or reduced?

Answer 8

oxidised

Question 9

If the oxidation state decreases, has the element been oxidised or reduced?

Answer 9

reduced

Question 10

Using the example, 3Cu + 8H+ + 2NO3 -> 3Cu2+ + 2NO + H2O, show how you would find the half equations

Answer 10

Cu: 0 H+: +1 N: +5 O: -2 -> Cu: 2+ N: +2 O: -2 H:+1 O: -2
3Cu -> 3Cu2+ Cu: increase by +2 3Cu: 2x3 = 6 6 electrons transferred
3Cu -> 3Cu2+ + 6e-
2NO3- -> 2NO N: increase by +3 2N: 3x2=6 6 electrons transferred
2NO3- + 6e- -> 2NO
however oxygen molecules not accounted for, add H+ to reactants and H2O to products
2NO3- + 8H+ + 6e- -> 2NO + 4H2O