5.2.3 (f- i) Electrode Potentials

Syllabus

(f) use of the term standard electrode (redox) potential, Eo including its measurement using a hydrogen electrode

{Eo data will be provided on examination papers. }

(g) the techniques and procedures used for the measurement of cell potentials of:

(i) metals or non-metals in contact with their ions in aqueous solution

(ii) ions of the same element in different oxidation states in contact with a Pt electrode

{For measurement of standard cell potentials, ions of the same element can have concentrations of 1 mol dm–3 or be equimolar.}

(h) calculation of a standard cell potential by combining two standard electrode potentials

(i) prediction of the feasibility of a reaction using standard cell potentials and the limitations of such predictions in terms of kinetics and concentration

What does this mean?

If we place a Copper strip in a solution of Copper ions and connect it to a Zinc solution containing a Zinc strip w can measure a voltage.

Image result for zinc copper voltaic cell

Note that there must be some way for ions to migrate from one side to the other to complete the circuit - moving charged particles are a current even if they are not electrons.

We could measure a voltage - but this would only be meaningful if we define some standard conditions so that all voltages can be usefully compared to each other.

Standard Conditions.

Atmospheric pressure

1moldm-3 solutions

25oC = 298 K

In addition, even if we measured the voltage for the cell above under standard conditions we would have an EMF for this particular cell and would need to measure a similar EMF for every other cell.

This is less useful than measuring a "half-cell" (one side of a full cell) compared to a reference half-cell with an electrode potential (Eo) of 0 volts.

Image result for electrode potential fe2+ Fe3+

This reference half-cell is that of 2H+ + e- --> H2

The problem with the Hydrogen half-cell is that it's not possible to have a solid electrode of Hydrogen.

So the conductor is inert Platinum (won't react and change the voltage)

The Hydrogen surrounds this inert electrode at 1 atm pressure.

In this way, the voltage we can read on the voltmeter equals the EMF of the Iron half-cell only (since the Hydrogen is defined as 0v and so doesn't contribute).

The salt-bridge is usually a piece of filter paper soaked with saturated solution of Potassium Nitrate (or another very soluble substance that won't disrupt the overall reaction).

But it can be an inverted glass tube full of the solution

If we needed to measure an EMF for Fe3+ + e- --> Fe2+ , for example, we have the problem that neither ion is a metal and so won't conduct.

The solution is another platinum electrode.

Image result for electrode potential fe2+ Fe3+

Calculating an EMF

In simple terms: EMF = Most Positive - Most negative

So, for a reaction between the Lithium and Nickel half cells EMF = -0.26 - - 3.05 = 2.81 v

Even though the Lithium half cell has a negative electrode potential it is still more positive than the Nickel one.

If the voltmeter was removed from the cell then the reaction would happen.

But which reaction?

Li+ (aq) + e- --> Li(s) -3.05v

Ni2+ (aq) + 2e- --> Ni(s) -0.26v

All half cells are written as reductions so one of the two equations must be reversed - this will be the most negative one.

1 - Reverse the most negative

Li(s) --> Li+ (aq) + e- -3.05v

Ni2+ (aq) + 2e- --> Ni(s) -0.26v

2 - Balance the electrons

2Li(s) --> 2Li+ (aq) + 2e-

Ni2+ (aq) + 2e- --> Ni(s)

3 - Add the half equations and cancel the electrons

2Li(s) + Ni2+ (aq) + 2e- --> Ni(s) + 2Li+ (aq) + 2e-

Predicting Feasibility

A level examiners like to ask you whether a reaction can happen or not.

Or sometimes which reaction can happen, or how it can be forced to happen.

Given the list below again:

Q. Is it feasible for there to be a reaction between Cadmium and Magnesium Nitrate solution?

i) The two relevant half equations are:

Cd2+(aq) + 2e- --> Cd(s) -0.40v

Mg2+(aq) + 2e- --> Mg(s) -2.37v

ii) For there to be a reaction involving Magnesium Nitrate it would be necessary to start with aqueous Magnesium ions which agrees with the way the half-equation is written.

Mg2+(aq) + 2e- --> Mg(s)

But it would be necessary to reverse the first so that it starts with solid Cadmium.

Cd(s) --> Cd2+(aq) + 2e-

So the question really is do the Eo values make this possible.

Applying the principle of keep it simple stupid - the more negative half equation is the one that gets reversed, the more positive goes forwards.

In this case the Magnesium half-equation should be reversed. The reaction between Cadmium and Magnesium ions is not feasible.

Q. Is it feasible for there to be a reaction between Alumium and Zinc Chloride solution?

i) The two relevant half equations are:

Zn2+(aq) + 2e- --> Zn(s) -0.76v

Al3+(aq) + 3e- --> Al(s) -1.66v

ii) For there to be a reaction involving Zinc Chloride it would be necessary to start with aqueous Zinc ions which again agrees with the way the half-equation is written.

Zn2+(aq) + 2e- --> Zn(s)

But it would be necessary to reverse the first so that it starts with solid Aluminium.

Al(s) --> Al3+(aq) + 3e-

So the question still is do the Eo values make this possible.

In this case it is the Aluminium half-equation that is more negative. So reversing it works. this reaction is Feasible

Another Examiner favourite is to ask for an Reducing Agent that can reduce one substance but not another

For instance what could be used to reduce Iron (II) ions to Iron but not Aluminium ions to Aluminium.

With any luck the list you're given on the exam paper will be in ascending or descending order when arranged by Eo value.

In which case, pick the half-equation between the two mentioned in the question.

Fe2+(aq) + 2e- -->Fe(s) -0.44v & Al3+(aq) + 3e- --> Al(s) -1.66v

The only half equation in between is Zn2+(aq) + 2e- --> Zn(s) -0.76v

To reduce Fe2+ to Fe (make its half-equation go forwards) we need another half-equation more negative (one "happy" to go backwards).

But it mustn't be more negative than the Aluminium half-equation or it would drive that forward too.

This is why the middle half-equation is chosen but we must pick the substance on one side or the other.

Since we want the Iron (II) equation to be driven forwards our choice must go backwards so we must choose the substance on the right - in other words, Zinc metal rather than Zinc ions.