U3AOS1 Topic 5: Redox Basics

What is redox?

Redox reaction involves the transfer of electrons 

When breaking the word down, red - stands for reduction and ox - stands for oxidation

Reduction cannot occur without oxidation occurring simultaneously and vice versa

1. Reduction is the gain of an electron
2. Oxidation is the loss of an electron

A helpful acronym to remember is OIL RIG:

Oxidation Is Loss Reduction Is Gain

Now where does the electron gained during reduction come from? It can’t just magically appear

It comes from the species undergoing oxidation

Hence, both reduction and oxidation must occur together

Remember how depending on the group that the element sits in, the valence electron number can be determined (exclusive of transition metals from groups 3 to 12)

Eg. Sodium (Na) is in group 1 therefore has 1 valence electron, Fluorine (F) is in group 17 therefore has 7 valence electrons

All elements ‘try’ to have a full valence shell of 8 electrons as per the octet rule. Thus, gaining stability therefore halogens (group 17) gain an electron via reduction to become an anion whilst alkali metals (group 1) lose an electron via oxidation to become a cation

Half equations

Half equations only represent either oxidation or reduction, not both at the same time.

Oxidation:

Ag(s) →Ag³⁺(aq) + 3e^-               Check if e^-  is on right side of equation to verify oxidation

2Cl^-(aq) →Cl₂(g) + 2e^-

Reduction:

Na⁺(aq) + e^- →Na(s)                  Check if e^- is on left side of equation to verify reduction

Br₂(l) +2e^- → 2Br^-(aq)

Normal tip: Ensure the charges are balanced on both sides, as you can see the net charge on left hand side is 0 and on the right it's also 0 (+3-3=0)

Oxidant (oxidising agent)/reductant (reducing agent)

Just like how a teacher provides the ‘teaching’, oxidant provides the oxidation and reductant provides the reduction

Essentially oxidant is called oxidant as it causes the other species to oxidise

But as we discussed earlier Redox is a pair of reactions therefore the oxidant species itself is actually reduced

Likewise, reductant is called reductant as it causes the other species to reduce

Reductant itself is oxidised

Oxidation numbers:

Oxidation numbers is basically the theoretical charge if the molecule is thought of as having ionic bonds

Here is a list of rules to remember:

·Diatomic molecules or isolated elements (Eg. Na or N₂) -> 0

·Ions -> current charge eg. Na⁺ has +1

·Ionic compounds -> current charge eg. NaCl is Na⁺ and Cl^- which is +1 and -1 respectively

·Oxygen -> generally -2 however -1 when with peroxides (eg. H₂O₂)

·Hydrogen -> generally +1 in covalent molecules however -1 with ionic compounds

·Oxidation number must sum to total polyatomic ion charge

Here is an example

To find the oxidation number of Cr in Cr₂O₇^2- we pretend as if Cr is the unknown for a simple algebra equation

2 Cr + 7(-2) = -2

As we have 2 Cr, each O is -2 oxidation number from the rule above and the total charge is -2 as Cr₂O₇^2-

Now solve for Cr:

2 Cr -14 = -2

2 Cr = +12

Cr = +6

Electrochemical series

·Supplied on page 2 of the databook

·Oxidants are always found on the LHS

·Reductants are always found on the RHS

·Strong oxidants are top left (weak bottom left)

·Strong reductants are bottom right (weak top right)

Always circle or take note by writing a mini electrochemical series on page of the reactants present in the reaction (not the products) as only the reactants can react

Always strongest oxidant present reacting with strongest reductant present. This is true for any redox reaction throughout the course.

There are four ions which you need to take note of:

·Fe²⁺

·Sn²⁺

·Pb²⁺

·Mn²⁺

These each appear twice in the electrochemical series on both oxidant and reductant side. So if you cannot find it then you are probably looking on the wrong side.

Spontaneity:

Spontaneous reaction just means that the reaction is naturally occurring without the need for input of energy

For a spontaneous redox reaction, a negative gradient must be formed when joining the strongest oxidant and strongest reductant reactants with an imaginary line.

Tip: Negative gradient looks like a hill. Think of a ball on top of the hill on the LHS. You don’t need to input energy to keep it rolling down the hill

For non-spontaneous reaction a positive gradient must be formed when joining the strongest oxidant and the strongest reductant reactants with an imaginary line

Tip: Positive gradient looks like a hill. Think of a ball on the bottom of the hill on the LHS. You need to keep inputting energy to roll the ball up the hill.

Electrochemical series limitations:

·Constructed at SLC 1.0 M - therefore cannot predict reactions out of these conditions

·Cannot predict rate of reaction – therefore if a reaction cannot be observed, it may still be occurring (just very slowly)

KOHES

Usually for the simple redox reactions, we can find the equations in the databook and simply copy it in reverse for oxidation and forwards for reduction

However, most times VCAA asks questions involving chemicals not found in the databook. This is where KOHES comes in

KOHES is an acronym for balancing HALF EQUATIONS only

K – Balance key element

O – Balance oxygen by adding water (same amount of water as oxygens needed)

H – Balance hydrogens by adding H⁺

E – Balance overall charge on LHS and RHS by adding e^- on the side with highest charge

S – Include states

Note: KOHES is defaulted to acidic conditions as H⁺ is acidic

If question asks for alkaline conditions

You still use KOHES however you add an extra step of neutralising H⁺ by adding OH^- (acid and base reaction)

Here is an example

Dichromate can react to form chromium ions

Acidic conditions:

1.Write out both species on each side and put arrow in the middle (make sure to leave space to fill in with KOHES later)

Cr₂O₇^2-                         -> Cr³⁺

Note: Whenever you encounter an unknown molecule especially a polyatomic anion (eg. dichromate), make sure to check page 7 of the databook

2.Balance K (key elements) which in this case is chromium as we have 2 on LHS and 1 on RHS

Cr₂O₇^2-                         -> 2Cr³⁺

3.Balance O (oxygens) by placing H₂O. We can see that the LHS has 7 O whilst RHS has 0 O. Therefore we must add 7 H₂O to RHS

Cr₂O₇^2-                         -> 2Cr³⁺ + 7H₂O

4.Balance H (hydrogens) by adding H⁺. We can see that RHS has 14 H from 7 * 2. Therefore 14 H⁺ needs to be added to LHS

Cr₂O₇^2- + 14H⁺             -> 2Cr³⁺ + 7H₂O

5.Balance charge with electrons (E).

LHS: -2 + 14 = +12

RHS: 2 * 3 = +6

As we can see there is a 6 difference in charge between LHS and RHS thus we add 6 electrons. Always add electrons to highest charge side therefore LHS

Cr₂O₇^2- + 14H⁺ + 6e^- -> 2Cr³⁺ + 7H₂O

6.Now we need to do the final step which most people forget to do. Add the states! Whenever it is an ion, assume aqueous unless otherwise stated in context.

Cr₂O₇^2-(aq) + 14H⁺(aq) + 6e^- -> 2Cr³⁺(aq) + 7H₂O(l)

Alkaline/basic conditions

1.You simply use the answer from acidic condition and continue from there. So using same example:

Cr₂O₇^2-(aq) + 14H⁺(aq) + 6e^- -> 2Cr³⁺(aq) + 7H₂O(l)

2.Now alkaline condition cannot have H⁺ therefore we count how many there are (14) and add same amount of OH^- to neutralise

Cr₂O₇^2-(aq) + 14H⁺(aq) + 14OH^- + 6e^- -> 2Cr³⁺(aq) + 7H₂O(l)

We can clearly see that the LHS and RHS is both imbalanced in atoms and charge therefore we must the exact same amount of OH^- to RHS too

Cr₂O₇^2-(aq) + 14H⁺(aq) + 14OH^- + 6e^- -> 2Cr³⁺(aq) + 7H₂O(l) + 14OH^-

3.The H⁺ and OH^- can combine to form water as it neutralizes via this reaction:

OH^- + H⁺ ->H₂O

Cr₂O₇^2-(aq) + 14H₂O+ 6e^- -> 2Cr³⁺(aq) + 7H₂O(l) + 14OH^-

4.We can see H₂O appears on both LHS and RHS therefore we can simplify by subtracting 7 H₂O on each side

Cr₂O₇^2-(aq) + 7H₂O+ 6e^- -> 2Cr³⁺(aq) + 14OH^-

5.Last step is to make sure to add all the states

Cr₂O₇^2-(aq) + 7H₂O(l)+ 6e^- -> 2Cr³⁺(aq) + 14OH^-(aq)

Note: something interesting to notice is that the number of electrons required is the same. This makes sense because it is still basically the same reaction. Only the electrolyte has changed (from H⁺ to OH^-) and this is what we expect. If you get different electron numbers after double checking if both atoms and charge is balanced on LHS and RHS then a mistake is probably made!