ConceptConceptDocsDocuments

CAIE A-Level Chemistry 24.2.8 Concentration and Electrode Potential

Practise predicting how aqueous-ion concentration changes an electrode potential and the resulting cell voltage.

Syllabus
2028–2030
Course
Chemistry 9701
Level
A2

Exam points

  • write the reduction half-equation and use Le Chatelier’s principle when an ion concentration changes
  • for M^n+ + ne− ⇌ M, lowering [M^n+] shifts left and makes the reduction potential less positive
  • combine changes in the affected half-cell with its cathode or anode role to predict the change in Ecell

24.2.8—How the value of an electrode potential, E question 1

[Maximum number: 2]

Hypophosphorous acid is an inorganic acid.
The conjugate base of hypophosphorous acid is H2PO2−\mathrm{H}_{2} \mathrm{PO}_{2}^{-}.

H2PO2−\mathrm{H}_{2} \mathrm{PO}_{2}^{-}is a strong reducing agent. It can be used to reduce metal cations without the need for electrolysis.
equation 1HPO32−+2H2O+2e−⇌H2PO2−+3OH−1 \quad \mathrm{HPO}_{3}^{2-}+2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{H}_{2} \mathrm{PO}_{2}^{-}+3 \mathrm{OH}^{-}E⊖=−1.57 VE^{\ominus}=-1.57 \mathrm{~V}

In an experiment, an alkaline HPO32−/H2PO2−\mathrm{HPO}_{3}{ }^{2-} / \mathrm{H}_{2} \mathrm{PO}_{2}{ }^{-}half-cell is constructed with [H2PO2−]=0.050 moldm−3\left[\mathrm{H}_{2} \mathrm{PO}_{2}^{-}\right]=0.050 \mathrm{~mol} \mathrm{dm}^{-3}.

All other ions are at their standard concentration.

Predict how the value of E of this half-cell differs from its E⊖E^{\ominus} value.

Explain your answer.

All question bank results loaded