Topic 5.1 Redox equilibria

5.1a Oxidation number and stoichiometry
The total increase in oxidation number in a reaction = the total decrease.
For the reaction
3ClO^-(aq) ---> 2Cl^-(aq) + ClO₃^-(aq) the oxidation numbers are
  +1                   -1           +5
Total decrease 2 Cl atoms at +1 to 2 Cl ions at -1 =4
Total increase =1 Cl atom ox. no. +1 to ClO₃^- ion ox. no. +5 =4
Task 5.1a

5.1b Estimating oxidising agents and reducing agent
Titrations involving potassium manganate(VII) ions,  MnO₄^-, are used to estimate concentrations of reducing agents like ethanedioate ions, C₂O₄^2- and iron II ions,Fe²⁺ .
The ionic equations are:
2MnO₄^- + 16H⁺ + 5C₂O₄^2- ---> 2Mn²⁺ + 8H₂O + 10CO₂
MnO₄^- + 5 Fe²⁺ + 8H⁺ ---> 5 Fe³⁺ + Mn²⁺ + 4H₂O
The purple aqueous manganate VII is added from the burette and the end point is signalled by a permanent pink colour in the flask. The reaction with ethanedioate needs a temperature of about 60^oC.  Both reactions requires excess dilute sulphuric acid.   

Titrations involving iodine, I₂ and thiosulphate ions, S₂O₃^2- are used to estimate concentrations of oxidising agents like manganate VII ions, iodate V ions, IO₃^- or chlorine.  In each case a know amount of the oxidising agent reacts with iodide ions to liberate iodine in a conical flask. e.g.
IO₃^- + 5I^-  + 6H⁺---> 3I₂ + 3H₂O
2MnO₄^- + 10I^- + 16H⁺ ---> 2Mn²⁺ + 8H₂O + 5I₂
The iodine in the flask is titrated with standardised aqueous sodium thiosulphate in the burette.  The reaction is:
2S₂O₃^2- + I₂ = 2I^- + S₄O₆^2-
The iodine solution in the flask begins with a yellow /brown colour. Near the end point of the titration, when it becomes very pale, starch should be added turning the solution a dark blue/black colour. The end-point is the appearance of a colourless solution produced by the titration of the blue/black solution.
Task 5.1b

5.1c Standard electrode potential
This is sometimes called the standard reduction potential.  The symbol is E^o.
e.g Pb²⁺(aq)|Pb(s)    E^o = -0.13V  or

Pb²⁺(aq)  + 2e^- -----> Pb(s)    E^o = -0.13V 
This is the potential of a half - cell measured relative to the Standard Hydrogen Electrode under standard conditions of 298K, 1atm and in a solution of concentration 1 mol dm^-3.
For the reaction
Cu²⁺(aq) + Zn(s) = Cu(s) + Zn²⁺(aq)
The convention for representation of cells is
Zn(s) | Zn ⁺²(aq) || Cu⁺²(aq) | Cu(s)
   oxidation                reduction
Reduced substances on the outside, oxidised substances on the inside.
The standard electrode potential of a metal and its solution cannot be measured directly. Only potential differences between a metal and a standard electrode can be measured. A hydrogen electrode, defined as having a potential of 0.00V, is used as the standard.
2H⁺(aq) + 2e^- ---> H₂(g)   E^o = 0.00V 
Task 5.1c Write some simple ionic equations or cell diagrams for the following:
Mg(s) | Mg²⁺(aq) || Cu⁺²(aq) | Cu(s)
Zn(s) | Zn²⁺(aq) || Pb²⁺(aq) | Pb(s)
2Ag⁺(aq) + Cu --> Cu⁺²(aq) + 2Ag(s)
Mg(s) + Zn²⁺(aq) --> Mg²⁺(aq) + Zn(s)

5.1d Spontaneous change in redox reactions
The anti-clockwise rule:
Write the more positive electrode potential on the bottom then follow anticlockwise arrows for direction of spontaneous change:
    <----------
Cu²⁺(aq) +e^-  ---> Cu(s) ;E^o = +0.34V
Ag⁺ (aq) + e^- ---> Ag(s)  ;E^o = +0.80V   (most +ve at bottom)
    ---------->
spontaneous reaction is Ag⁺(aq) + Cu(s) ----> Ag(s) + Cu⁺²(aq)
Copper is a stronger reducing agent than silver as the E^o value of copper is more negative than that of Ag.
where the cell diagram for the above reaction is Cu(s)|Cu²⁺(aq) || Ag⁺(aq)|Ag(s)
Ecell = E^oright - E^oleft  
or for half equations written as above
Ecell = E^obottom - E^otop
If Ecell is positive then the reaction is spontaneous
Standard conditions of 1 molar solutions, 1 atmosphere pressure and 298K must exist for these rules to apply.  In practise the conditions in many reactions  are not standard.  If the E^o values are close (or Ecell is small) then non-standard conditions can change E^o values.  In these cases  a reaction does not always happen as expected.  
Task 5.1d

5.1e Disproportionation and standard electrode potentials
Disproportionation is the oxidation and reduction of the same element in the same reaction. For example in solution Cu⁺(aq) disproportionates.

2Cu⁺(aq) ----> Cu(s) + Cu²⁺(aq)
+1 -reduction-> 0
+1 -oxidation ------->    +2

              <--------------
Cu²⁺(aq) + e- -----> Cu⁺(aq) E^o = +0.15V
Cu⁺(aq) + e- ----> Cu(s) E^o = +0.52V
             ---------------->
Task 5.1e

5.1f Electrode potentials and corrosion

Rusting is a redox reaction involving the following half equations:
Fe²⁺(aq) + 2e^- <=> Fe(s) E^o = -0.44V
4H⁺(aq) + O₂(g) + 4e^- <=> 2H₂O(l) E^o = +1.23V
The cell reaction is
4H⁺(aq) + O₂(g) + 2Fe(s) => 2Fe²⁺(aq) +2H₂O(l)
The cell has a positive emf.
Fe(s)|Fe²⁺(aq)||[4H⁺(aq) + O₂(g)],2H₂O(l)|Pt(s)  Ecell = +1.67V

To prevent corrosion iron is coated (galvanised) with zinc. Zinc has a more negative electrode potential than iron, therefore zinc is oxidised in preference to iron. i.e. zinc corrodes in place of iron therefore the iron is protected.
Zn²⁺(aq) + 2e- <=> Zn(s) E^o = - 0.76V
Fe²⁺(aq) + 2e- <=> Fe(s) E^o = - 0.44V
The cell reaction is Zn(s) + Fe²⁺(aq) <=> Zn²⁺(aq) + Fe(s)
The cell has a positive emf.
Zn(s)|Zn²⁺(aq)||Fe²⁺(aq)|Fe(s) Ecell = +0.32V

Tin protects iron so long as there is no scratch.    Iron has a more negative electrode potential than tin, therefore iron is oxidised in preference to tin.  If there is a scratch the iron in tin cans corrodes faster than untreated iron.
Fe²⁺(aq) + 2e^- <=> Fe(s) E^o = -0.44V
Sn²⁺ (aq) + 2e^- <=> Sn(s) E^o = -0.14V
The cell reaction is
Sn²⁺ (aq) + Fe(s) => Fe²⁺(aq) + Sn(s)
The cell has a positive emf.
Fe(s)|Fe²⁺(aq)||Sn²⁺(aq)|Sn(s) Ecell = +0.30V
Task 5.1f 

5.1g Simple storage cells
A simple cell (a primary cell) has two electrodes (one a more and one a less reactive metal e.g. Zinc and Copper), each dipping into an electrolyte with the electrolytes connected by a salt bridge.
Zn is very reactive so gives up its electrons and goes into solution, becoming a -ve electrode. Cu becomes the +ve electrode.
A primary cell cannot have its reactants regenerated by charging.

A lead-acid battery (e.g. a car battery) is a secondary cell or storage cell which can be recharged. It consists of a lead plate, a lead(IV)oxide in a lead grid plate and a sulphuric acid electrolyte.

DURING DISCHARGING
the reaction at the now negative lead plate is oxidation (so it is the anode)
PbSO₄(s) + 2e^- => Pb(s) + SO₄^2-(aq) ; E^o = - 0.36V
the reaction at the now positive lead (IV) oxide plate is reduction (so it is the cathode)
PbO₂(s) + 4H⁺(aq) + SO₄^2-(aq) + 2e^- => PbSO₄(s) + 2H₂O(l) ; E^o = +1.69V

When both plates are connected, a current flows. When all the Pb and PbO₂ have been converted to PbSO₄ the reaction stops.
The spontaneous reaction (E = +2.05V) for discharging the cell is
Pb(s) + PbO₂(s) + 4H⁺(aq) + 2SO₄^2-(aq) ----> 2PbSO₄(s) + 2H₂O(l)
The cell is recharged by the reverse reaction.
6 cells in series give the familiar 12V car battery

DURING CHARGING
the reaction at the positive lead plate is reduction (so it is the cathode):
PbSO₄(s) + 2e^- => Pb(s) + SO₄^2-(aq)
the reaction at the negative lead(IV)oxide plate is oxidation (so it is the anode)
PbSO₄(s) + 2H₂O(l) => PbO₂(s) + 4H⁺(aq) + SO₄^2-(aq) + 2e^-
The overall reaction for charging is:
2PbSO₄(s) + 2H₂O(l) ----> Pb(s) + PbO₂(s) + 4H⁺(aq) + 2SO₄^2-(aq)
Task 5.2g Draw labelled diagrams of (a) a primary cell using iron and tin, (b) a lead acid secondary cell during discharge, (c) a lead acid cell during charging.

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