5.2.3 (f- i) Electrode Potentials

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.

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 E^o 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 E^o 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 E^o 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.