Standard electrode potential and the hydrogen electrode. Standard electrode potentials of redox systems with inert electrodes. Use of tables of standard electrode potentials to predict feasibility and direction of reactions, and to calculate e.m.f. of a cell. (Calculations involving concentrations or the Nerst equation excluded.)

Standard Electrode Potential E^θ

When a metal is placed in a solution of its ions an equilibrium is established between the tendency of the metal to lose electrons and pass into solution as ions, and the opposing tendency for the ions in solution to gain electrons and be deposited on the metal.

M (s) Mn+ (aq) + ne-

The metal aquires an electrical charge (usually negative) and a potential difference is set up between the metal and the solution called the electrode potential of the metal.

The size of this electrode potential will depend on the position of this equilibrium and this depends on;

1. the metal concerned

2. the concentration of the metal ion in solution

3. temperature

The temperature and concentration are therefore standardised at 298K (25oC) and 1 molar respectively. Under these conditions the electrode potential is known as the standard electrode potential, E^θ. (For a gas/gas ion equilibrium the gas pressure is standardised as 1 atmosphere).

E^θ is in volts at 298K

The standard electrode potential of a standard half cell is determined by coupling it with a standard hydrogen electrode, which is arbitrarily given a standard electrode potential of 0.00 volts.

By convention standard electrode potentials are written as

oxidised state + ne- reduced state

i.e. as reduction potentials. The voltage produced is for the reaction as written, left to right.

Predicting the Feasibility of Reactions

A table of standard redox potentials can be used to predict the likelihood that a reaction will take place

Positive E^θ values indicate systems that have a tendency to gain electrons (i.e. oxidising agents). Negative E^θ values indicate systems with a tendency to lose electrons (i.e. reducing agents).

The larger the positive value of E^θ the more powerful the oxidising agent.

E^θ

F₂ (g) + 2e^- 2F^- (aq) +2.87 V

Cl₂ (g) + 2e^- 2Cl^- (aq) +1.36 V

Therefore fluorine will oxidise chloride ions to chlorine

F₂ + 2Cl^- 2F^- + Cl₂

The larger the negative E^θ value the more powerful the reducing agent.

Mg²⁺ (aq) + 2e^- Mg (s) - 2.38 V

Zn²⁺ (aq) + 2e^- Zn (s) - 0.76 V

Therefore magnesium will reduce zinc ions to zinc

Mg + Zn²⁺ Mg²⁺ + Zn

This can also be done by calculation

(a) Write the more negative ½ equation as producing electrons.

(b) Write the more positive ½ equation as accepting electrons.

(c) Add the two ½ equations.

(d) If the E^Ø value is positive the reaction will proceed as written, if the E^θ

value is negative the reverse reaction will occur. e.g.

Cu ²⁺ + 2e^- Cu E^θ = +0.34V (source)

2Ag⁺ + 2e^- 2Ag E^θ = +0.80V

(a) Cu Cu²⁺ + 2e^- E^θ = -0.34V

(b) 2Ag⁺ + 2e^- 2Ag E^θ = +0.80V

(c) Adding

2Ag⁺ + Cu Cu²⁺ + 2Ag E^θ = +0.46V

Since the value of E^Ø is positive the reaction will occur as written.

· Will lead displace cobalt from a solution of cobalt nitrate?

Co²⁺ + 2e^- Co E^θ = -0.28V (source)

Pb²⁺ + 2e^- Pb E^θ = -0.13V

(a) Co Co²⁺ + 2e^- E^θ = +0.28V

(b) Pb²⁺ + 2e^- Pb E^θ = -0.13V

(c)Adding

Pb²⁺ + Co Co²⁺ + Pb E^θ = +0.15V

The reaction will occur as written i.e. cobalt will displace lead ions from solution and lead will not displace cobalt ions.

· Reaction of Fe²⁺ (aq) with halogens

2Fe²⁺ + Cl₂ 2Fe³⁺ + 2Cl^-

E^θ

2Fe²⁺ 2Fe³⁺ + 2e^- -0.77volts

Cl₂ + 2e^- 2Cl^- +1.36volts

2Fe²⁺ + Cl₂ 2Fe³⁺ + 2Cl^- +0.59volts

Overall the value of E^θ is +ve so the reaction will take place.

• Predict the feasibility of Fe²⁺ (aq) reacting with Br₂ and I₂.

Questions

1 Which of the following reactions will occur spontaneously?

(a) Fe(s) + Zn²⁺ (aq) Fe²⁺ (aq) + Zn(s)

(b) Fe(s) + Sn²⁺ (aq) Sn(s) + Fe²⁺ (aq)

(c) Sn⁴⁺ (aq) + 2I^- Sn²⁺ (aq) + I₂ (s)

(d) Zn(s) + Mg²⁺ (aq) Mg (s) + Zn²⁺ (aq)

(e) Zn(s) + Sn²⁺ (aq) Zn²⁺ (aq) + Sn(s)

(f) Sn⁴⁺ (aq) + 2Fe²⁺ (aq) Sn²⁺ (aq) + Fe³⁺ (aq)

(g) Cr₂O₇^2-(aq) + 14H⁺ + 6Cl^- 2Cr³⁺ (aq) + 7H₂O (l) + 3Cl₂(g)

(h) 2Ce⁴⁺ (aq) + 2Br ^- (aq) 2Ce³⁺ (aq) + Br₂ (l)

2 Which of the following species are oxidised by manganese(IV) oxide?

Br ^-, Ag, I^-, Cl ^-

3 Which of the following species are reduced by tin(II)?

I₂, Ni²⁺, Cu²⁺, Fe²⁺

4 Some standard electrode potentials are listed below

VO₂⁺ (aq)/ VO²⁺ (aq) + 1.00V

Cd²⁺ (aq)/ Cd(s) - 0.40V

BrO₃^- (aq)/ Br₂(l) + 1.52V

S₄O₆^2- (aq)/ S₂O₃^2- + 0.09V

(a) State which of the species MnO₄^-, Ce⁴⁺, Cr₂O₇^2-, VO₂⁺, and Fe³⁺ are able to liberate chlorine from an acidic solution of sodium chloride.

(b) Write a balanced equation for the reaction between MnO₄^- and VO₂⁺, in an acid solution.

(c) Put the following into order of their power as oxidising agents in acid solution. Cr₂O₇^2-, Cl₂, MnO₄^-, I₂, BrO₃^-, S₄O₆^2-

	Reaction	E^θ / V (298K)
Weakest	Li^+(aq) + e^- Li (s)	-3.04	strongest
	K^+ (aq) + e^- K(s)	-2.92
oxidising	Ba^2+ (aq) + 2e^- Ba (s)	-2.90	reducing
	Sr2+ + 2e- Sr (s)	-2.89
agent	Ca^2+ (aq) + 2e^- Ca (s)	-2.87	agent
	Na^+ (aq) + e^- Na(s)	-2.71
	Mg^2+ (aq) + 2e^- Mg (s)	-2.38
	Al^3+ (aq) + 3e^- Al (s)	-1.66
	Mn^2+ + 2e- Mn (s)	-1.18
	Zn^2+ (aq) + 2e^- Zn (s)	-0.76
	Cr^3+ (aq) + 3e^- Cr (s)	-0.74
	Fe^2+ (aq) + 2e^- Fe(s)	-0.44
	Cr^3+(aq) + e^- Cr^2+ (aq)	-0.41
	Cd^2+ (aq) + 2e^- Cd (s)	-0.40
	Co^2+ (aq) + 2e- Co (s)	-0.28
	Ni^2+ (aq) + 2e^- Ni (s)	-0.25
	Sn^2+ (aq) + 2e^- Sn (s)	- 0.14
	Pb^2+ (aq) + 2e^- Pb (s)	-0.13
	2H^+(aq) + 2e^- H2 (g)	0.00
	Sn^4+(aq) + 2e^- Sn^2+(aq)	+0.15
	AgCl (aq) + e^- Ag (s) + C^l- (aq)	+0.22
	Cu^2+(aq) + 2e^- Cu (s)	+0.34
	I2 (s) + 2e^- 2I^- (aq)	+0.54
	Fe^3+ (aq)+ e^- Fe^2+(aq)	+0.77
	Ag^+ (aq) + e^- Ag (s)	+0.80
	Br2(l) + 2e^- 2Br^- (aq)	+1.07
	MnO2 (s) + 4H^+(aq) + 2e^- Mn^2+(aq) + 2H2O (l)	+1.23