1 / 13

Electrochemical Cells – Voltage (Electric potential)

Electrochemical Cells – Voltage (Electric potential). The half cells Standard electrode potentials Calculating voltages Examples. Half Cells. Previously we have learnt that every redox reaction has an oxidation half and a reduction half

webster
Télécharger la présentation

Electrochemical Cells – Voltage (Electric potential)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Electrochemical Cells – Voltage (Electric potential) The half cells Standard electrode potentials Calculating voltages Examples

  2. Half Cells • Previously we have learnt that every redox reaction has an oxidation half and a reduction half • Cells also have two halves – (oxidation (left) and reduction (right)). We call these half cells – strangely enough • Every half cell has a rated ability hold on to its electrons: • Some grab electrons very well e.g. • F2 → F- • Cl2 → Cl- • Some find it very hard to hold their electrons e.g. • Li → Li+ • Na → Na+ • This ability is called a standard electrode potential and is measured in volts

  3. Standard Electrode Potential • A half cell doesn’t have any ability to give or take electrons by itself – it needs to have something else to give the electrons to (or to take them from). In this case we compare every half cell with its ability to take electrons from a hydrogen half cell: H2 → H+ • The hydrogen half cell is the oxidation (left) half cell • This potential to push (or pull) electrons is measured in volts (as mentioned earlier) • It is called a standard electrode potential because it is measured under standard conditions: • Temperature: 25°C • Concentrations: 1.0 molL-1 • Atmospheric pressure: 1 atmosphere (101.3kPa)

  4. Calculating Voltages of Cells • Because all of the half cells potentials are relative to Hydrogen, we can work out the difference in potential when we combine any two half cells • Because each half cell has a different potential to give or take electrons there will be a flow from the best giver to the best taker of electrons • We use the following formula to calculate the difference in voltage between two half cells: • E°(cell) = E°(RHE) – E° (LHE) • In other words the total Electrode potential for the cell = the Electrode potential of the right hand electrode – the Electrode potential of the left hand electrode • If the E°(cell) value is a positive voltage then the electrons are flowing in the correct direction therefore we say that the reaction is spontaneous (it happens without us having to add energy)

  5. Examples of Standard Electrode Potentials • Note that the potentials of the following reactions are all for the oxidation reactions – this is because they were all compared to the reduction half cell of H+ • MnO4-,Mn2+ = 1.51V • Au3+/Au = 1.50V • Cl2/Cl- = 1.44V • Ag+/Ag = 0.80V • Fe3+,Fe2+ = 0.77V • Cu2+/Cu = 0.34V • H+/H2 = 0.00V • H+/H2 = 0.00V • Pb2+/Pb = -0.13V • Fe2+/Fe = -0.47V • Zn2+/Zn = -0.76V • Al3+/Al = -1.66V • Mg2+/Mg = -2.36V • Na+/Na = -2.71V

  6. Exercises: Calculating The Voltage MnO4-,Mn2+ = 1.51V Au3+ /Au = 1.50V Cl2/Cl- = 1.44V Fe3+,Fe2+ = 0.77V Cu2+/Cu = 0.34V H+/H2 = 0.00V Fe2+/Fe = -0.47V Zn2+/Zn = -0.76V Al3+/Al = -1.66V Mg2+/Mg = -2.36V Na+/Na = -2.71V

  7. Examples of Electrochemical Cells – The Dry Cell • The dry cell consists of three main parts: • The zinc case – which gets oxidised • The black paste which gets reduced • The carbon electrode • The black (oxidant) paste contains two main chemicals which undergo the following reaction: MnO2 + NH4+ + H2O + e- → Mn(OH)3 + NH3 • The zinc half equation is: Zn → Zn2+ + 2e- Oxidant paste Carbon anode Zinc cathode

  8. Example II – Lead Acid Cells • A lead-acid battery supplies electrical current through the following reactions: • Lead oxidising Pb → Pb2+ + 2e- • Lead oxide being reduced PbO2 + 4H+ + 2e→ Pb2+ + 2H2O • In each half equation the Pb2+ builds up on the outside of the sheets as PbSO4 • Each cell generates about 2 volts so 6 cells are placed in series to create a 12V battery • This type of cell is also ‘rechargeable’ - the reactions can be reversed by running current through the cell in the opposite direction to what it normally travels The diagram below shows alternating lead and lead dioxide sheets in a “pool” of H2SO4

  9. Exam Practice - 2008 Can’t see the exam paper below? Go to the NCEA website and search for 90696 • Have a go at questions:

  10. Exam Practice - 2007 Can’t see the exam paper below? Go to the NCEA website and search for 90696 • Have a go at questions:

  11. Exam Practice - 2006 Can’t see the exam paper below? Go to the NCEA website and search for 90696 • Have a go at questions:

  12. Exam Practice - 2005 Can’t see the exam paper below? Go to the NCEA website and search for 90696 • Have a go at questions:

  13. Exam Practice - 2004 Can’t see the exam paper below? Go to the NCEA website and search for 90696 • Have a go at questions:

More Related