A Level Chemistry - Electrode Potentials PDF

A LEVEL CHEMISTRY 3.1.11 ELECTRODE POTENTIALS REDOX REMINDER! Throughout this topic, you will be using the REDOX skills that you learned in Year 12. Make sure you remind yourself of: OIL RIG...

A LEVEL CHEMISTRY 3.1.11 ELECTRODE POTENTIALS REDOX REMINDER! Throughout this topic, you will be using the REDOX skills that you learned in Year 12. Make sure you remind yourself of: OIL RIG REDOX in Acidic Conditions Oxidising & Reducing agents Combining Half-Equations Half-Equations THE ELECTROCHEMICAL SERIES You have already crossed paths with REDOX “potentials” in Year 12. You know that: Group 1 metals are more easily oxidised as you go down the group. i.e. they lose their outer electron more easily. AKA they are stronger reducing agents. You know that Group 7 elements are more easily reduced as you go up the group. i.e. they gain an electron more easily. AKA they are stronger oxidising agents. You also know that the Group 7 Halide ions are more easily oxidised as you go down the group. AKA they are stronger reducing agents. The electrochemical series QUANTIFIES this information by comparing the relative oxidising and reducing power of different elements and compounds on a scale. These are known as their electrode potentials, and they are all measured against the electrode potential of hydrogen. So, some elements have a lower electrode potential than hydrogen. These have negative values. Some have a higher electrode potential than hydrogen. These have positive values. The electrochemical series allows us to compare the REDOX powers of different elements and compounds and allows us to predict reactions. These are all measured at standard conditions 𝚹. These are: 298K 100kPa 1.00mol.dm-3 solution of ions AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.11 ELECTRODE POTENTIALS The half-equations are always presented in the same way. That is with the oxidising agents on the left and the reducing agents on the right. Each equation represents a “half cell” (more on these later). Notice that all reactions are also reversible. The direction each equation takes (oxidation or reduction) depends on what other half cell / equation it is paired with. Strongest Reducing Agent ELECTRODE POTENTIAL E𝚹 (V) HALF-EQUATION Li+(aq) + e- ⇌ Li(s) -3.03 Ca2+(aq) + 2e- ⇌ Ca(s) -2.87 Mg2+(aq) + 2e- ⇌ Mg(s) -2.37 Al3+(aq) + 3e- ⇌ Al(s) -1.66 2H+(aq) + 2e- ⇌ H2(s) 0.00 Cu2+(aq) + 2e- ⇌ Cu(s) +0.34 Fe3+(aq) + e- ⇌ Fe2+(aq) +0.77 Ag+(aq) + e- ⇌ Ag(s) +0.80 Cr2O72– + 14H+ + 6e– ⇌ 2Cr3+ + 7H2O +1.33 Strongest Oxidising Agent Another way to think about this series, How To Navigate the is that any element that H+ can oxidise has a negative E𝚹 value. The Electrochemical Series more negative the value, the more readily they are oxidised. Any element that H+ cannot oxidise has a positive value. The more positive the value, the less readily it will oxidise. AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.11 ELECTRODE POTENTIALS THE STANDARD HYDROGEN ELECTRODE The standard hydrogen electrode (SHE) is the half cell against which all other half cells are measured. In other words, all electrode potential values for other half- cells are relative to this. Make sure you remember the various parts to this and the standard conditions. 298K V Wire connected to a second half cell via a high resistance voltmeter. H2(g) at 100kPa Platinum electrode Platinum is ideal for this as: - it conducts electricity H+(aq) - it is unreactive 1.00mol.dm-3 - it is porous e.g. 1.00mol.dm-3 HCl(aq) An equilibrium exists between the gaseous hydrogen, H2(g), and the H+ ions in solution. If a SHE half cell was connected to another SHE half cell, the voltmeter would read 0.00V. 2H+(aq) + 2e- ⇌ 2H2(g) An equilibrium exists between the gaseous hydrogen, H2(g), and the H+ ions in solution. If a SHE half cell was connected to another SHE half cell, the voltmeter would read 0.00V as there is no difference in REDOX power. AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.11 ELECTRODE POTENTIALS COMBINING HALF CELLS WITH THE SHE In order to determine the electrode potential of another half cell, it must be combined with the SHE. The voltmeter then tells us the electrode potential of that half cell. The two cells are connect in two ways: 1. Via the high resistance voltmeter. This is how the electrons are trying to move from one half cell to another. As it is high resistance, the voltmeter measures the potential difference between the two half cells. 2. A salt bridge. This is usually made from a saturated solution of KNO3(aq). This completes the circuit and also provides ions to maintain the electrochemical balance between the two half cells. V e.g. Cu(s) electrode SHE Salt Bridge 1.00mol.dm-3 Cu2+(aq) When the two half cells are connected reduction occurs in one half cell and oxidation occurs in the other. The electrons that are lost in the half cell where oxidation is taking place, move through the wire, to the voltmeter. The half cell where reduction is taking place is trying to gain electrons from the wire. Since the voltmeter is high resistance, it prevents the movement of electrons from one half cell to another. The voltmeter simply measures the “potential difference” in volts. AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.11 ELECTRODE POTENTIALS When combining the SHE with another half cell, it is always shown on the left. Half cells that have a positive electrode potential. e.g. Cu2+(aq) + 2e- ⇌ Cu(s) E𝚹 = +0.34V e- e- V Cu(s) electrode S.H.E. 1.00mol.dm-3 Cu2+(aq) OXIDATION REDUCTION Electrons are lost Electrons are gained 2H2(g) → 2H+(aq) + 2e- Cu2+(aq) + 2e- → Cu(s) In these cases the hydrogen in the SHE is oxidised and the copper half cell is reduced, so the electrons move from left to right causing the voltmeter to give a positive value. This value is the E𝚹 for the copper half cell. AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.11 ELECTRODE POTENTIALS Half cells that have a negative electrode potential. e.g. Mg2+(aq) + 2e- ⇌ Mg(s) E𝚹 = -2.37V e- e- V Mg(s) electrode S.H.E. 1.00mol.dm-3 Mg2+(aq) REDUCTION OXIDATION Electrons are gained Electrons are lost 2H+(aq) + 2e- → H2(g) Mg(s) → Mg2+(aq) + 2e- In these cases the hydrogen in the SHE is reduced and the magnesium half cell is oxidised, so the electrons move from right to left causing the voltmeter to give a negative value. This value is the E𝚹 for the magnesium half cell. How Electrochemical Cells Work AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.11 ELECTRODE POTENTIALS COMBINING HALF CELLS Any two half cells can be combined to find the potential difference between them. The reading on the voltmeter here is known as the E𝚹 cell or Electromotive Force (EMF) of the cell. Unlike when using the SHE, the the conventional way of showing these half cells is to place the cell that undergoes oxidation on the LEFT. That way we always get a positive value on the voltmeter. e.g. Al3+(aq) + 3e- ⇌ Al(s) E𝚹 = -1.66V Cu2+(aq) + 2e- ⇌ Cu(s) E𝚹 = +0.34V V Al(s) electrode Cu(s) electrode 1.00mol.dm-3 Ag+(aq) 1.00mol.dm-3 Cu2+(aq) OXIDATION REDUCTION Electrons are lost Electrons are gained Al(s) → Al3+(aq) + 3e- Cu2+(aq) + 2e- → Cu(s) Left Right Oxidation occurs Reduction occurs Anode Cathode Negative electrode Postive electrode AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.11 ELECTRODE POTENTIALS CALCULATING E𝚹 cell To calculate E𝚹 cell / EMF for a cell, there is one equation, but two different way to think about it. Choose which one works for you: E𝚹 cell / EMF = E𝚹 of reduced half cell - E𝚹 of oxidised half cell OR E𝚹 cell / EMF = Most positive E𝚹 - Most negative E𝚹 e.g. for the aluminium / copper cell on the previous page: E𝚹 cell / EMF = +0.34 - (-1.66) = +2.00V Your answer should always be a positive value! HINTS | TIPS | HACKS Some half cells do not have a solid in the equation to use as an electrode. Fe3+(aq) + e- ⇌ Fe2+(aq) In these cases, a platinum electrode is used and the solution contains a 1.00mol.dm-3 concentration of both ions. Remember L.O.A.N. from the previous page. You can always rely on it to clarify what’s happening in each of the half cells. If the high resistance voltmeter is replaced with another device (e.g. a bulb), the cell is essentially battery. The electrons moving through the circuit power the device. The mass of the electrodes can change! It will How To Use the increase in the half cell that is reduced and Anticlockwise Rule decrease in the half cell that is oxidised! You can predict which half cell is oxidised and which is reduced using the “anticlockwise rule”. AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.11 ELECTRODE POTENTIALS CONVENTIONAL REPRESENTATION There is a short-hand method of showing an electrochemical cell and the parts of the half cells that it is made from. V Cu(s) electrode Al(s) electrode -3 1.00mol.dm-3 Cu2+(aq) 1.00mol.dm Ag+(aq) Cu2+(aq) + 2e- → Cu(s) Al(s) → Al3+(aq) + 3e- Al is oxidised Cu2+ is reduced Al(s) | Al3+(aq) || Cu2+(aq) | Cu(s) This line means that the two This double line represents components are different the salt bridge states of matter It is slightly more complex when one half cell contains two ions in solution. e.g. Al(s) | Al3+(aq) || Fe3+(aq), Fe2+(aq) | Pt(s) Place a comma between Platinum the two ions. electrode They are both (aq) AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.11 ELECTRODE POTENTIALS HOW CHANGING CONC. AFFECTS E𝚹 CELL All E𝚹 cell values are measured under standard conditions. This includes a concentration of 1.00mol.dm-3 for the ions in a solution of a half cell. You need to be able to predict how a change in concentration affects the value of E𝚹 cell. When these two half cells are combined: Mg2+(aq) + 2e- ⇌ Mg(s) Cu2+(aq) + 2e- ⇌ Cu(s) The magnesium is oxidised and the copper is reduced: Mg(s) → Mg2+(aq) + 2e- Cu2+(aq) + 2e- → Cu(s) Combining the two equations gives us the overall equation: Mg(s) + Cu2+(aq) ⇌ Mg2+(aq) + Cu The expected reaction runs rom left to right through the equation. However the ⇌ sign is present as the reaction could still run in the opposite direction. le Chatelier’s From Year 12, we know that a change in concentration can affect the position of an equilibrium! If a change in concentration here causes a shift to the RIGHT, going WITH the expected reaction, this will cause E𝚹 cell to increase. If a change in concentration here causes a shift to the LEFT, going AGAINST the expected reaction, this will cause E𝚹 cell to decrease. Bottom line: Most Negative Most Positive Effect on E𝚹 cell SHIFT Half Cell Conc. Half Cell Conc. Increase or Decrease Right Increase Decrease or Increase Left Decrease AQA www.chemistrycoach.co.uk © scidekick ltd 2024