1.11 - electrode potentials & electrochemical cells

    Cards (16)

    • what do electrochemical cells do?
      convert chemical energy into electrical energy
      eg: like batteries

      by using 2 metals with different reactivities - redox reactions can be set up - these involve the transfer of electrons
    • electrochemical half-cells vs electrochemical cell
      electrochemical half-cells - are when a metal bar is dipped into a solution of it's own ions, an equilibrium is set up
    • What is an electrochemical half-cell?
      It is a part of the apparatus where a metal bar is dipped into a solution of its own ions.
    • What is the role of the voltmeter in an electrochemical cell?
      The voltmeter is used to calculate the electromotive force (EMF) of the cell.
    • How does the salt bridge function in an electrochemical cell?
      The salt bridge allows ions inside the ionic substance to flow to complete the circuit.
    • What is a common ionic substance used in the salt bridge of an electrochemical cell?
      KNO3(potassium nitrate) is often used in the salt bridge.
    • Why is KNO3 often chosen for the ionic substance in the salt bridge in electrochemical cellrather than anything else?
      KNO3 does not react with any ions in the electrolytes, many other substances aren't applicable as they react with the ions in the electrolyte which creates inaccuracy in the readings from the voltmeter
    • What are the main differences between electrochemical half-cells and electrochemical cells?
      • Electrochemical half-cells are parts of the apparatus with a metal bar in its own ion solution.
      • Electrochemical cells are the complete apparatus including two half-cells, a voltmeter, and a salt bridge.
    • for a cell reaction to go to completion, what must happen?
      the voltmeter must be removed
    • how would an equilibrium equation for an electrochemical half cell be written if the solution had Fe2+ cations?
      the electrolyte solution inside an electrochemical half-cell will always have cations of the same solid metal bar that has been dipped into it
      if its Fe2+ ions therefore the solid metal bar must be Fe and as it's 2+ to create an equilibrium there must be 2e- inside the solution as well

      Fe (s) ⇌ Fe2+ (aq) + 2e-
    • the symbol for electromotive force (EMF), which is created when two electrodes (electrochemical half-cells) are connected together
    • how is the EMF (measured in Eθ values by a voltmeter) of all electrodes calculated?
      it's measured by connecting them to a standard hydrogen electrode which is given the Eθ value of 0v by definition
    • how is EMF created?
      it is created when 2 electrodes are connected - it can't be an electrode on its own, it must be connected
      the EMF is given the symbol (E theta)
    • what are the components of a standard hydrogen electrode?
      • H2 (g) at 100kpa
      • 1.0 mol dm-3
      • 298 K
      • platinum electrode
      • and lastly, the concentration of ions in solution of the attached half-cell should also be 1 mol dm-3
    • the conventional representation of half-cells includes IUPAC conventions which helps it all to be written out in a standard formatt
      • vertical solid line - a phase boundary (between solid & solution)
      • A double vertical line - a salt bridge
      • more positive half cell (the reduction reaction) is written on the right hand side
      • species with the highest oxidation state (therefore the more negative half cell) in each half-cell is written next to the salt bridge
      • EMF produced should be indicated (although often not required by mark schemes)
    • how do you represent a conventional electrochemical cell involving a copper half-cell compared to a standard hydrogen electrode?
      • use a platinum electrode (pt) with the hydrogen electrode
      • set up the hydrogen side as h₂(g) | h⁺(aq)
      • represent the copper half-cell as cu²⁺(aq) | cu(s)
      • include a salt bridge, indicated by || between the two half-cells
      • the cell notation is pt(s) | h₂(g) | h⁺(aq) || cu²⁺(aq) | cu(s)