Standard electrode potentials are measured by comparison with the standard hydrogen electrode - AQA - A-Level Chemistry - Question 6 - 2020 - Paper 3

To conduct the experiment, first prepare the acidified TiO²⁺(aq) solution by dissolving titanium(IV) sulfate (Ti(SO₄)₂) in 0.50 mol dm⁻³ sulfuric acid. Use a 50 cm³ solution.

Next, set up a standard electrochemical cell by placing the prepared TiO²⁺(aq)/Ti(s) and the Cu²⁺(aq)/Cu(s) electrodes in a voltmeter to measure the cell potential.

The standard electrode potential (E°) can be calculated using the measured potential difference and the known potential of the copper electrode using the equation:

$E_{cell} = E_{Cu^{2+}/Cu} - E_{TiO^{2+}/Ti}$

Rearranging gives:

$E_{TiO^{2+}/Ti} = E_{Cu^{2+}/Cu} - E_{cell}$

Substituting the known values, if the measured cell voltage is 0.34 V, we obtain:

$E_{TiO^{2+}/Ti} = 0.34 V - E_{cell}$

The expected result confirms that E° for the TiO²⁺(aq)/Ti(s) electrode is -0.88 V.

The half-equation for the electrode reaction in the TiO²⁺(aq)/Ti(s) electrode in acidic conditions is:

$TiO^{2+} + 2e^- + 4H^+ \rightarrow Ti(s) + 2H_2O$

From Table 2, the standard electrode potential for copper is +0.34 V, while the potential for the nitric acid reaction (as a strong oxidizing agent) is +0.96 V. Copper does not react with dilute acids like sulfuric or hydrochloric acid because those acids do not provide a potential enough for displacement.

However, nitric acid can oxidize copper because its potential is significantly higher, allowing the reaction to proceed. Copper can oxidize nitric acid, making it reactive.

The equation for the reaction between copper and nitric acid can be written as:

$Cu(s) + 4HNO_{3}(aq) \rightarrow Cu(NO_{3})_{2}(aq) + 2NO_2(g) + 2H_2O(l)$