Electrode Potentials and Electrochemical Cells

• What is a half cell?

  A half cell is a metal dipped into a solution containing its simple ions, an equilibrium is established between the metal ions and atoms.
• An electrochemical cell uses electron transfer reactions to produce electrical energy from a redox reaction. It is two half cells joined together by a salt bridge.
• Oxidation = Negative electrode
• Reduction = Positive electrode
• Electrons flow externally from the negative electrode to the positive electrode. 

  For example, the zinc forms the negative electrode and the copper forms the positive electrode. The electrons flow externally in the circuit from the zinc side to the copper side, causing the zinc atoms to lose and the copper ions in the solution gain electrons to form copper, this circuit allows electrons to flow. This causes an increase in mass in the copper electrode and a decrease in the mass in the zinc electrode. A high-resistance is connected to measure the potential difference without allowing any electrons to flow in the circuit. 
• What is a salt bridge?

  A salt bridge is either a piece of filter paper soaked in a solution of potassium chloride or potassium nitrate, or potassium chloride dissolved in agar gel and set in a U-tube. It has mobile electrons that complete the circuit and allow for the electrons to transfer externally.
• However, care should be taken with potassium chloride solution for a salt bridge as chloride ions can react with some metal ions such as Cu2+ to form complexes. So, potassium nitrate solution is better to use for the salt bridge for solutions in which the metal ion might form a complex with the chloride ions. 
• Cells are used to measure electrode potentials by reference to the standard hydrogen electrode under specific conditions:

  • All solutions should have an ion concentration of 1.00 mol dm-3.
  • All gases should be under 100kPa.
  • The whole cell should be at 298 K.
• The standard hydrogen electrode is an electrode consisting of hydrogen gas in contact with hydrogen ions ($H^+$) on a platinum surface.
• A high-resistance voltmeter is connected to the standard hydrogen electrode (half cell) and the other electrode, a salt bridge is used to connect the two solutions.

  An oxidation reaction must occur, $H_2\rightarrow2H^+$$+$$2e^-$ , meaning a reduction reaction occurs at the other half cell. 
• The standard electrode potential is the electrode potential of a standard electrode with ion concentration of 1.00 mol dm-3 at 298 K connected to a standard hydrogen electrode using a high-resistance voltmeter and a salt bridge. 
• Electrodes not under standard conditions:

  In this cell, the concentration of $CuSO_4$(aq) on the left-hand side is greater than the concentration of the same solution on the right-hand side. An ammeter can be used to show the flow of current in the cell and in particular to show the direction of electric flow in the cell.

  • As the concentration of $Cu^{2+}$ ions is greater on the left-hand side, this is where $Cu^{2+}+$$2e^-\rightarrow Cu$ occurs, making it the positive electrode.
  • The oxidation, $Cu\rightarrow Cu^{2+}+$$2e^-$ occurs at the right-hand electrode to provide electrons for the reduction of the left-hand side. 

  The current will eventually fall to zero once the concentrations of the solutions are equal.
• Redox potentials are always given as a reduction reaction and the feasibility of the reaction is given by the standard electrode potential. The more positive the electrode proteins, the more feasible the reduction reaction. 

  So, if the EMF is positive, then the redox reaction is feasible and if it is negative, it is not.
• Electrochemical cells can be used as a commercial source of electrical energy. Cells can be non-rechargeable (irreversible), rechargeable or fuel cells. 
• Common examples of non-rechargeable cells include alkaline batteries and dry cells. These are called primary cells which can only be used once and then discarded because the materials in the cell cannot be regenerated by recharging.
• Non-rechargeable cell example: Zinc-Manganese cell

  • The porous separator allows ions to pass through it. 
  • The carbon rod is made of graphite as it conducts electricity and allows the movement of electrons. 
  • The zinc is used as a container for the cell but it may begin to leak after long-term use. The zinc becomes compromised. 

  Oxidation (Negative electrode) = $Zn(s)\rightarrow Zn^{2+}(aq)+$$2e^-$ = -0.76V

  Reduction (Positive electrode) = $MnO_2(s)+$$H_2O(l)+$$e^-\rightarrow MnO\left(OH\right)(s)+$$OH^-(aq)$ = +0.74V

  So, the overall EMF of this cells is: +0.74+0.76=+1.50V

  The overall reaction when the cell discharges is: $Zn(s)+$$2MnO_2(s)+$$2H_2O(l)\rightarrow Zn^{2+}(aq)+$$2MnO\left(OH\right)(s)+$$2OH^-(aq)$

  
• Rechargeable cells, often called secondary cells, can be used many times. Secondary cells are able to be recharged, regenerating the original reagents. Common examples of rechargeable cells include lead-acid batteries used in cars, lithium ion and nickel metal hydride cells used in many electronic devices such as mobile phones. Rechargeable cells are more environmentally friendly as they can be reused and prevent waste. Supplies of the metal and other reagents are not depleted as quickly and less energy is used to extract metals.
• Rechargeable cells examples: Lead-acid cells

  The lead-acid cell is the cell used in cars and other vehicles. It is recharged as the vehicle moves. Sulfuric acid used to provide the acid in these types of batteries.

  Oxidation (Negative electrode) = $Pb(s)+$$HSO_4^-(aq)\rightarrow PbSO_4(aq)+$$H^+$$(aq)+$$2e^-$ = -0.46V

  Pb is oxidised from 0 in Pb to +2 in $PbSO_4$.

  Reduction (Positive electrode) = $PbO_2(s)+$$3H^+$$(aq)+$$2HSO_4^-(aq)+$$Pb(s)\rightarrow2PbSO_4(s)+$$2H_2O(l)$ = +1.69V

  Pb is reduced from 0 in Pb to +2 in $PbSO_4$.

  So, the overall EMF of this cells is: +1.69+0.46=+2.15V

  The overall reaction when the cell discharges is: $PbO_2(s)+$$2H^+$$(aq)+$$2HSO_4^-(aq)+$$Pb(s)\rightarrow2PbSO_4(s)+$$2H_2O(l)$

  The overall reaction when the cell is recharged: $2PbSO_4(s)+$$2H_2O(l)\rightarrow PbO_2(s)+$$2H^+$$(aq)+$$2HSO_4^-(aq)+$$Pb(s)$
• Rechargeable cells example: Lithium ion cells

  Lithium ion cells are rechargeable and often used to provide electrical energy for cameras,laptops, tablets and mobile phones.

  Oxidation (Negative electrode) =  $Li^+$$+$$CoO_2+$$e^-\rightarrow Li\left[CoO_2\right]^-$

  Li is oxidised from 0 in Li to +1 in $Li^+$.

  Reduction (Positive electrode) =  $Li\rightarrow Li^+$$+$$e^{^-}$

  $CoO_2$  is reduced from +4 in $CoO_2$ to +3 in $Li^+$$CoO_2$.

  The overall reaction: $Li+$$CoO_2\rightarrow Li^+$$\left[CoO_2\right]^-$

  The overall cell representation: Li\left|Li^+\right|Li^+$,CoO_2\left|LiCoO_2\right|Pt$
• Rechargeable cells example: Nickel-cadmium cells

  Often used to provide electrical energy for tools, camcorders and portable electronics.

  Oxidation (Negative electrode) = $Cd\left(OH\right)_2+$$2e^-\rightarrow Cd(s)+$$2OH^-(aq)$ = -0.88V

  Cd is oxidised from 0 in Cd to +2 in $Cd\left(OH\right)_2$.

  Reduction (Positive electrode) = $NiO\left(OH\right)(s)+$$H_2O(l)+$$e^-\rightarrow Ni\left(OH\right)_2(s)+$$OH^-(aq)$ = +0.52V

  Ni is reduced from +3 in $NiO\left(OH\right)_2$ to +2 in $Ni\left(OH\right)_2$.

  So, the overall EMF of this cells is: +0.52+0.88=+1.40V

  The overall reaction when the cell discharges is: $Cd(s)+$$2NiO\left(OH\right)(s)+$$2H_2O(l)\rightarrow Cd\left(OH\right)_2+$$2Ni\left(OH\right)_2(s)$

  The overall reaction when the cell is recharged: $Cd\left(OH\right)_2+$$2Ni\left(OH\right)_2(s)\rightarrow Cd(s)+$$2NiO\left(OH\right)(s)+$$2H_2O(l)$
• Fuel cells are used to generate an electric current and do not need to be electrically recharged.

  • A fuel cell is an electrical cell which converts the chemical energy of a redox reaction into electrical energy. This cell will continue to function as long as the fuel and oxygen are supplied to it.
  • The oxidation and reduction reactions are both catalysed. 
  • Common examples of fuel cells are hydrogen or ethanol fuel cells.
  • Oxidation occurs at the negative electrode (anode) and the electrons released travel through a wire and into the external circuit. The central electrolyte allows ions and molecules to move through it but not electrons. The ions react with another substance and this is the reduction reaction which takes electrons from the external circuit. 
• Hydrogen cells can operate in acidic or alkaline conditions. Commercial alkaline hydrogen fuel cells use platinum electrodes and concentrated aqueous sodium hydroxide. Porous platinum gives a larger surface area. Fuel is supplied continuously, so voltage output does not change. Fuels may be from non-carbon neutral sources.
• Acidic conditions: 

  Oxidation (Negative electrode) = $H_2(g)\rightarrow2H^+$$(aq)+$$2e^-$ = +1.23V

  Reduction (Positive electrode) = $O_2(g)+$$4H^+$$(aq)+$$4e^-\rightarrow2H_2O(l)$ = 0.00V

  So, the overall EMF of this cells is: +1.23+0.00=+1.23V

  The overall reaction = $2H_2(g)+$$O_2(g)\rightarrow2H_2O(l)$

  The cell notation = Pt|$H_2(g)$|$H^+$$(aq)$||$O_2(g)$|$H^+$$(aq),H_2O(l)$|Pt
• Alkaline conditions: 

  Oxidation (Negative electrode) = $H_2(g)\rightarrow2OH^-(aq)\rightarrow2H_2O(l)+$$2e^-$ = +0.83V

  Reduction (Positive electrode) = $O_2(g)+$$2H_2O(l)+$$4e^-\rightarrow4OH^-(aq)$ = +0.40V

  So, the overall EMF of this cells is: +0.83+0.40=+1.23V

  The overall reaction = $2H_2(g)+$$O_2(g)\rightarrow2H_2O(l)$ 

  The cell notation = Pt|$H_2(g)$|$OH^-(aq),H_2O(l)$||$O_2(g)$|$H_2O(l),OH^-(aq)$|Pt
• Ethanol fuel cells are used as an alternative to the hydrogen fuel cell. Ethanol fuel cells use ethanol from fermentation of crops, which are carbon neutral as carbon dioxide is captured during photosynthesis to generate the carbohydrates used to make the ethanol. 

  Oxidation (Negative electrode) = $C_2H_5OH(l)+$$3H_2O(l)\rightarrow2CO_2(g)+$$3H_2O(l)$

  Reduction (Positive electrode) = $12H^+$$(aq)+$$3O_2(g)+$$12e^-\rightarrow6H_2O(l)$

  The overall reaction, ethanol is oxidised in the fuel cells to carbon dioxide and water = $C_2H_5OH(l)+$$3O_2(g)\rightarrow2CO_2(g)+$$3H_2O(l)$