Electrochemistry

Electrochemistry

6.1 Redox processes: electron transfer and changes in oxidation number (oxidation state)

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Learning Outcomes

• calculate oxidation numbers of elements in compounds and ions

• describe and explain redox processes in terms of electron transfer and changes in oxidation number

• use changes in oxidation numbers to help balance chemical equations

RedOx

A redox reaction is a reaction in which oxidation and reduction takes place. Oxidation is the loss of electrons, or increase in oxidation number. Reduction is the gain of electrons, or decrease in oxidation number.  

What are the rules for assigning oxidation states?

• Uncombined elements have an oxidation state of 0.

• In neutral compounds, the sum of oxidation states is 0.

• The oxidation state of common basic ions is equal to their charge.

• For highly electronegative species, the more electronegative element is negative. 

What are some exceptions to the rule for oxidation states?

• Hydrogen in metal hydrides has an oxidation state of -1 (instead of +1).

• Oxygen in peroxides has an oxidation state of -1 (instead of -2).

Multiple Select

Tick(s) the correct theoretical maximum oxidation state of the following elements

• P = +5

• Pb = +4

• S = +6

• Cr = +7

Fill in the Blank

Calculate the oxidation state of sulfur in H₂SO₄

Answer

• Oxidation state of oxygen = -2. 4 oxygen atoms so -2 x 4 = -8

• Oxidation state of hydrogen = +1. 2 hydrogen atoms so +1 x 2 = +2

• Overall charge on the compound is 0 so the sum of oxidation states must equal 0.

• X is the oxidation state of sulfur: +2 + X + -8 = 0, X - 6 = 0, X = 6

• Oxidation state of S in H₂SO₄ is +6.

Multiple Select

What is oxidation?

• Loss of electrons

• Gain of electrons

• increase in oxidation number

• decrease in oxidation number

Multiple Select

What is reduction?

• increase in oxidation number

• Gain of electrons

• Loss of electrons

• decrease in oxidation number

Multiple Choice

What is the oxidation state of each element in H₃PO₄?

• P = +3, H = +1, O = -2

• P = +5, H = +1, O = -2

• P = +5, H = +1, O = -4

Multiple Select

Tick(s) if the changes involve oxidation

• C₂O₄^2- —> CO₂

• H₂O₂ —> H₂O

• Cr₂O₇^2- —> CrO₄^2-

• H₂O₂ —> O₂

Multiple Select

Tick(s) the correct name for the formulae

• PbO₂ lead(II) oxide

• SnCl₂ tin(II) chloride

• BrF₅ bromine(V) fluoride

Balancing equations with oxidation states

If an oxidation state increases by one unit, one electron is lost from that substance. If an oxidation state decreases by one unit, one electron has been gained. In a reaction, if the oxidation state of one substance decreases, this must be balanced by an increase in the oxidation state of something else.

BALANCING REDOX HALF EQUATIONS

• Work out formulae of the species before and after the change; balance if required

• Work out oxidation state of the element before and after the change

• Add electrons to one side of the equation so that the oxidation states balance

• If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges

• If equation still doesn’t balance, add sufficient water molecules to one side

MnO₄¯ being reduced to Mn²⁺ in acidic solution

• Work out formulae of the species before and after the change; balance if required

• Step 1         MnO₄¯   ———>    Mn²⁺

• Work out oxidation state of the element before and after the change

• Step 1         MnO₄¯   ———>    Mn²⁺

  Step 2              +7        +2

• 
  

MnO₄¯ being reduced to Mn²⁺ in acidic solution

• Add electrons to one side of the equation so that the oxidation states balance

• Step 3           MnO4¯ +  5e¯    ———>   Mn2+

• The oxidation states on either side are different (REDUCTION)

  To balance; add 5 negative charges to the LHS    [+7 + (5 x -1) = +2]

  You must ADD 5 ELECTRONS to the LHS of the equation

MnO₄¯ being reduced to Mn²⁺ in acidic solution

• If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H⁺ ions to one of the sides to balance the charges

• Step 4 MnO₄¯ +  5e¯ +  8H⁺  ———>   Mn²⁺

• Total charges on either side are not equal; 

  LHS  = -1 and -5 = -6, RHS = +2

  Balance them by adding 8 positive charges to the LHS  

  [ -6 + (8 x +1) = +2 ]

  You must ADD 8 PROTONS (H+ ions) to the LHS of the equation

MnO₄¯ being reduced to Mn²⁺ in acidic solution

• If equation still doesn’t balance, add sufficient water molecules to one side

• Step 5  MnO4¯ +  5e¯ +  8H+  ——>   Mn2+   +   4H2O    

• Everything balances apart from oxygen and hydrogen 

  O    LHS = 4  RHS = 0

  H    LHS = 8  RHS = 0

  You must ADD 4 WATER MOLECULES to the RHS; the equation is now balanced

Cr₂O₇^2- being reduced to Cr³⁺ in acidic solution

• Step 1  Cr₂O₇^2-  ———>  2Cr³⁺  both sides now have 2 Cr

• Step 2     2 @ +6, 2 @ +3   both Cr’s are reduced

• Step 3     Cr₂O₇^2- +  6e¯   ——>   2Cr³⁺    each Cr needs 3 electrons

• Step 4   Cr₂O₇^2- +  6e¯ +  14H⁺     ——>   2Cr³⁺

• Step 5     Cr₂O₇^2- +  6e¯ +  14H⁺     ——>   2Cr³⁺ +   7H₂O

Multiple Select

Tick(s) any balanced half equation(s).

• NO₃^- + 4H⁺ + 3e^- —> NO + 2H₂O

• NO₃^- + 2H⁺ + e^- —> NO + H₂O

• SO₄^2- + 4H⁺ + 2e^- —> SO₂ + 2H₂O

• H₂O₂ —> O₂ + 2H⁺ + 2e^-

COMBINING HALF EQUATIONS

• A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows...

• Step 1: Write out the two half equations

• Step 2: Multiply the equations so that the number of electrons in each is the same

• Step 3: Add the two equations and cancel out the electrons on either side

• Step 4: If necessary, cancel any other species which appear on both sides

The reaction between manganate(VII) and iron(II)

The reaction between manganate(VII) and iron(II)

• Step 1:

  Fe²⁺  ——>    Fe³⁺  +   e¯                       Oxidation

  MnO₄¯ +  5e¯ +  8H⁺    ——>   Mn²⁺  + 4H₂O      Reduction

• Step 2:           

  5Fe²⁺    ——>   5Fe³⁺ + 5e¯                              multiplied by 5

  MnO₄¯ +  5e¯ +  8H⁺    ——>   Mn²⁺  + 4H₂O  multiplied by 1

• Step 3:

  MnO₄¯ +  5e¯ +  8H⁺+ 5Fe²⁺ ——>    Mn²⁺  + 4H₂O + 5Fe³⁺ + 5e¯

• Step 4: MnO₄¯ +  8H⁺ + 5Fe²⁺——>   Mn²⁺  +  4H₂O  +  5Fe³⁺

• 
  

Multiple Choice

Construct balanced redox equation for the reaction between Cr₂O₇^2- and Fe²⁺

• Cr₂O₇^2- + 14H⁺ + 6Fe²⁺ ——> 2Cr³⁺ + 6Fe²⁺ + 7H₂O

• Cr₂O₇^2- + 14H⁺ + 3Fe²⁺ ——> 2Cr³⁺ + 3Fe²⁺ + 7H₂O

Multiple Choice

Construct balanced redox equation for the reaction between S₂O₃^2- and I₂

• S₂O₃^2- + I₂ —-> S₄O₆^2- + I^¯

• 2S₂O₃^2- + I₂ —-> S₄O₆^2- + 2I^¯

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