If we place a strip of a metal such as zinc into a beaker containing zinc sulphate (the electrolyte) an equilibrium is set up.
Reactive metals lose electrons (oxidation) more readily than unreactive metals.
By convention, these half equations are always written as a reduction reaction:
Zn2+(aq) + 2e– ⇌ Zn(s)
Zinc is a reactive metal – the position of equilibrium lies to the left. As zinc atoms lose electrons to form ions, electrons accumulate on the surface of the metal strip and this attracts a layer of positively charged ions.
Let’s contrast this with a strip of copper metal immersed in copper sulphate solution.
Cu2+(aq) + 2e– ⇌ Cu(s)
Copper is an unreactive metal – the position of equilibrium lies to the right.
We can’t actually measure the potential difference between an electrode and its electrolyte, but we can measure the difference between two electrode potentials.
It was decided once upon a time that all electrode potentials would be measured against a hydrogen electrode under standard conditions (298K, 100kPa), and the standard hydrogen electrode potential (E⦵) was to be zero volts.
2H+(aq) + 2e– ⇌ H2(g) E⦵ = 0.00V
The platinum wire acts as an inert solid surface (that also enables transfer of electrons) for the reaction between aqueous H+ ions and gaseous hydrogen to take place on. The RSC have a quick video clip of a standard hydrogen electrode set up that is worth watching.
We can compare the potential difference of zinc against hydrogen with a simple experiment that connects the two half cells and measures the electromotive force of the electrochemical cell, Ecell⦵, (maximum voltage) using a high resistance voltmeter.
In each half cell there will be either an oxidation reaction happening (the electrons that are lost flow around the circuit to the other half cell) or a reduction reaction happening (the electrons flowing around the circuit are gained). But how do we know which is the oxidation half cell and which is the reduction half cell?
If we carry out this experiment, the voltmeter reads -0.76V. We know that by convention E⦵ = 0.00V for the 2H+(aq) / H2(g) half cell so the Zn2+(aq) / Zn(s) half cell is negative in comparison.
Zinc is preferentially losing electrons.
The position of equilibrium for the half reaction, Zn2+(aq) + 2e– ⇌ Zn(s), lies to the left and so the Zn2+(aq) / Zn(s) half cell must be the oxidation half cell. These electrons are flowing around the circuit to the 2H+(aq) / H2(g) half cell and if they were not being blocked by the high resistance voltmeter, they would be reducing hydrogen ions to form hydrogen gas. The 2H+(aq) / H2(g) half cell is the reduction half-cell.
A note on terminology ….
- half reactions describe the reaction in a half cell (such as a metal electrode in contact with a solution of its own ions) and are always written as reduction, Zn2+(aq) + 2e– ⇌ Zn(s), regardless of whether that half cell is the oxidation or reduction half cell in practice
- we can use the shorthand Zn2+(aq) / Zn(s) to describe a half cell or half reaction – again always written as reduction
- always include state symbols with everything
- the symbol E⦵ is the standard electrode potential for a half cell, where that half cell has been compared against the 2H+(aq) / H2(g) half cell. The ⦵ symbol tells us that we are working under standard conditions of 298K, 100kPa, all solutions at 1.00 mol dm-3.
- the symbol Ecell⦵ is essentially the emf or voltage for an electrochemical cell (two half cells connected by a high resistance voltmeter and a salt bridge), all under standard conditions. Ecell is the voltage recorded for an electrochemical cell not under standard conditions.
Let’s repeat this experiment with a Cu2+(aq) / Cu(s) half cell replacing zinc.
The voltmeter for the electrochemical cell reads +0.34V. The Cu2+(aq) / Cu(s) half cell is positive in comparison to the 2H+(aq) / H2(g) half cell.
Cu2+(aq) + 2e– ⇌ Cu(s)
This tells us that the position of equilibrium for the Cu2+(aq) / Cu(s) half reaction lies to the right and so the Cu2+(aq) / Cu(s) half cell must be the reduction half cell. Electrons are flowing from the 2H+(aq) / H2(g) half cell (the oxidation half cell in this case) and if they were not blocked by the high resistance voltmeter, they would be gained by Cu2+ ions.
We can use a data sheet to look up the standard electrode potential, E⦵, of numerous half reactions as compared to the standard hydrogen electrode.
Na+ (aq) + e– ⇌ Na (s) – 2.71V
Fe2+ (aq) + 2e– ⇌ Fe (s) – 0.44V
2H+ (aq) + 2e– ⇌ H2 (g) 0.00V
Fe3+ (aq) + e– ⇌ Fe2+ (aq) +0.77V
Cl2 (g) + 2e– ⇌ 2Cl– (aq) +1.36V
The more negative (less positive) E⦵, the greater the tendency of that half reaction to release electrons. Na(s) is a very strong reducing agent – the equilibrium position lies strongly to the left in this half equation, which means that Na(s) readily loses electrons (oxidised) forming Na+(aq).
The more positive (less negative) E⦵, the greater the tendency of that half reaction to gain electrons. Cl2(g) is a very strong oxidising agent – the equilibrium position lies strongly to the right in this half equation which means that Cl2(g) will readily gain electrons (reduced) forming 2Cl–(aq) .
And finally, what is a salt bridge?
A salt bridge is a way of electrically connecting / completing a circuit between two half cells. The simplest is a strip of filter paper soaked in saturated potassium nitrate. The movement of K+ and NO3– ions competes the electrical connection.
