REDOX and ELECTROCHEMISTRY Notes
REDOX AND ELECTROCHEMISTRY
Oxidation-Reduction Reactions
- Definition: Chemical reactions involving a change in the oxidation state of one or more substances and a transfer of electrons.
- Oxidation: Loss of one or more electrons, leading to an increase in positive oxidation state.
- Example: Mgo \rightarrow Mg^{+2} + 2e^{-}
- Mgo is oxidized and acts as the reducing agent (reductant).
- Example: Mgo \rightarrow Mg^{+2} + 2e^{-}
- Reduction: Gain of one or more electrons, leading to a decrease in positive oxidation state.
- Example: Cl^{+5} + 6e^{-} \rightarrow Cl^{-1}
- Cl^{+5} is reduced and acts as the oxidizing agent (oxidant).
- Example: Cl^{+5} + 6e^{-} \rightarrow Cl^{-1}
- Spectator Ion: An ion that does not change its oxidation state during a reaction.
- Example: Mg + 2HCl \rightarrow MgCl2 + H2; Cl^{-1} is the spectator ion.
Oxidation Numbers
- Rules:
- The sum of oxidation numbers in a neutral compound is always zero.
- Example: In MgCl_2, Mg is +2 and each Cl is -1.
- Example: In H2SO4, each H is +1 and each O is -2, so 2(+1) + 4(-2) = -6, thus S must be +6.
- Determine oxidation numbers of elements with multiple possibilities by using known oxidation numbers of other elements in the compound.
- Use the Periodic Table to aid in determining oxidation numbers (charges).
- The sum of oxidation numbers in a neutral compound is always zero.
Oxidation-Reduction Reactions (Continued)
- In a redox reaction, both oxidation and reduction must occur.
- One or more electrons are transferred from the reductant to the oxidant.
- Question: In the reaction, Cl2 + H2O \rightarrow HClO + HCl, the Cl_2 is:
- (c) both oxidized and reduced
- Question: In the half-reaction, Bao \rightarrow Ba^{+2} + 2e^{-}, the barium atom:
- (d) loses electrons
- Question: In the reaction, Ca + NiCl2 \rightarrow CaCl2 + Ni, the oxidation number of chlorine:
- (b) remains the same
Examples of Identifying Redox Components from Equations
- Example 1: MnO2 + 4HCl \rightarrow MnCl2 + 2H2O + Cl2
- Oxidation half-reaction:
- Reduction half-reaction:
- Substance oxidized:
- Substance reduced:
- Reducing agent:
- Oxidizing agent:
- Spectator ion(s):
- Example 2: 2NaClO3 \rightarrow 2NaCl + 3O2
- Oxidation half-reaction:
- Reduction half-reaction:
- Substance oxidized:
- Substance reduced:
- Reducing agent:
- Oxidizing agent:
- Spectator ion(s):
Standard Electrode Potentials
- Electromotive force (emf) is the sum of the two half-cell potentials.
- E{cell} = E{ox} + E_{red}
- Under standard conditions (1M concentrations, 298K, 1 atm):
- E^{\circ}{cell} = E^{\circ}{ox} + E^{\circ}_{red}
- Coefficients in half-reactions are ignored because electrode potentials are intensive properties (do not depend on the amount of reactant, only on molar concentrations).
- E^{\circ} is derived from a comparison to a standard reduction potential designated with a value of 0.00 V (Hydrogen half-cell).
- Standard Hydrogen Electrode: Consists of a platinum wire encased in a glass sleeve through which hydrogen gas is passing at 1 atmosphere. The platinum wire is attached to a platinum foil coated with finely divided platinum that serves as a catalyst. This assembly is immersed in a 1M acid solution.
Table of Standard Reduction Potentials
- The more positive the E^{\circ} value, the greater the tendency for the reduction to occur.
- F_2 is the most easily reduced and is the strongest oxidizing agent.
- F_2 can gain electrons from any substance below it in the table.
- Li^+ is the least easily reduced and is the weakest oxidizing agent.
- F_2 is the most easily reduced and is the strongest oxidizing agent.
- Common oxidants: Cr2O7^{-2}, NO_3^{-}, metals in high oxidation states
- Reading the half-reaction in reverse gives the oxidation half-reaction.
- Reactions with very negative E^{\circ}{red} have very positive E^{\circ}{ox} and are easily oxidized.
- Li^{\circ} is very easily oxidized and is a strong reducing agent.
- Li loses electrons to all atoms or ions above it.
- Li^{\circ} is very easily oxidized and is a strong reducing agent.
- Common reducing agents: H_2, metals in low oxidation states (difficult to keep because of oxygen in the air)
Activity Series of Metals
- See the activity series of metals for ease of oxidation (all written as reduction half-reactions).
- If the substance lower on the table is the substance oxidized, the reaction is spontaneous.
- Example: Tell whether the reaction is spontaneous: 3Ca + 2AlCl3 \rightarrow 3CaCl2 + 2Al
- Oxidation: 3(Ca^{\circ} \rightarrow Ca^{+2} + 2e^-)
- Reduction: 2(Al \rightarrow Al^{+3} + 3e^-)
- If the sum of the E^{\circ} values is positive, the reaction is spontaneous (reverse the sign of the oxidation E^{\circ}).
- Here, Ca is +2.87 and Al is -1.66, resulting in an overall value of +1.21, indicating spontaneity.
Balancing Equations
Rules:
- Write the half-reactions.
- Balance the electrons.
- Finish balancing by inspection.
Example: Balance the following equation: Al + ZnO \rightarrow Al2O3 + Zn
- Oxidation (O):
- Reduction (R):
- So, _ Al + _ ZnO \rightarrow _ Al2O3 + _ Zn
- Oxidation (O):
- Reduction (R):
Example: Balance the following equation: H3AsO3 + NaIO3 \rightarrow NaI + H3AsO_4
Example: Balance the following equation: Cu + HNO3 \rightarrow Cu(NO3)2 + H2O + NO
Solution: Cu + HNO3 \rightarrow Cu(NO3)2 + H2O + NO
- But NO_3^{-1} are spectator ions, so now must balance them.
- _ Cu + _ HNO_3 \rightarrow _ Cu(NO3)2 + _ H_2O + _ NO
Example: Balance the following equation: Cu + HNO3 \rightarrow Cu(NO3)2 + H2O + NO_2
Galvanic (Voltaic) Cells
- The transfer of electrons in a spontaneous reaction can produce energy in the form of heat (sum of E^{\circ} is +).
- Energy released in a spontaneous redox reaction can be used to perform electrical work.
- The electron transfer is forced to take place through an external pathway rather than directly between the reactants.
- Example: Net ionic equation: Zns + Cu^{+2} \rightarrow Zn^{+2} + Cus
- The substance lower on the chart of reduction potentials will transfer electrons to the substance higher on the chart.
- The reaction is spontaneous.
- Oxidation(occurs at anode; - electrode): Zn_s \rightarrow Zn^{+2} + 2e^{-}
- Reduction(occurs at cathode; + electrode): Cu^{+2} + 2e^{-} \rightarrow Cu_s
- Electrons flow from anode (Zns) to cathode (Cus) through an external wire.
- Zn electrode loses mass; [Zn^{+2}] increases.
- Cu electrode gains mass; [Cu^{+2}] decreases.
- Salt Bridge: Contains a solution of ions that will not react with either electrode; ions flow through the salt bridge to neutralize the buildup of positive charge (extra ions) in the anode and negative charge (decrease in + ions) and complete the circuit.
- The circuit is not completed without the salt bridge.
- Negative ions (anions) flow through the salt bridge towards the anode.
- Positive ions (cations) flow through the salt bridge towards the cathode.
- Problem: Given the cell, Al/Al^{+3}//Ag/Ag^{+1}
- Oxidation half-reaction:
- Reduction half-reaction:
- Net ionic reaction:
- Anode Cathode
- Direction electrons flow: from to
- Positive ions flow toward _ electrode
- Negative ions flow toward electrode
- Ions that increase in solution
- Mass of electrode that increases __
- Maximum Voltage is __
- Voltage at Equilibrium is _
- Solution: Problem: Given the cell,
- Oxidation half-reaction:
- Reduction half-reaction:
- Net ionic reaction:
- Anode ____ Cathode ______ (electrodes – metals)
- Direction electrons flow: from to
- Positive ions flow toward _
- Negative ions flow toward _
- Ions that increase in solution
- Mass of electrode that increases __
- Maximum Voltage is __
- Voltage at Equilibrium is _
Electrolysis
- Using electrical energy to cause non-spontaneous redox reactions to occur.
- Example: Molten salts (2NaCll \rightarrow 2Nas + Cl_{2g})
- Description: Two electrodes in a molten salt; a battery or other source of direct current pushes electrons into one electrode (-) and pulls electrons from another (+).
- Na^+ ions pick up electrons at the negative electrode (cathode) and are reduced.
- Cl^- ions migrate to the positive electrode (anode), lose electrons, and are oxidized.
- Anode: 2Cl^- \rightarrow Cl_2 + 2e^{-}
- Cathode: 2Na^+ + 2e^- \rightarrow 2Na
- The signs of the anode and cathode are reversed from that of an electrochemical (voltaic) cell.
- Used commercially in the production of active metals.
Electrolysis of Aqueous Solutions
- Electrolysis of aqueous solutions (NaCl_{aq}, brine)
- Ionic equation: 2Na^+ + 2Cl^- + 2H2O \rightarrow 2Na^+ + 2OH^- + Cl2 + H_2
- Overall reaction: 2Cl^- + 2H2O \rightarrow 2OH^- + Cl2 + H_2
- Since 2H2O + 2e^- \rightarrow H2 + 2OH^- E^{\circ} = -0.83V (cathode), Na^+ + e^- \rightarrow Na E^{\circ} = -2.71V so the water is more easily reduced.
- Since 2Cl^- \rightarrow Cl2 + 2e^- E^{\circ} = -1.36V (anode), 2H2O \rightarrow 4H^+ + O_2 + 4e^- E^{\circ} = -1.23V. Should produce oxygen but actually produces chlorine gas due to overvoltage (greater voltage required than predicted). Overvoltage is sufficiently high for oxygen, so chlorine is oxidized. Overvoltage is high for gases and low for metals.
- Used commercially to prepare hydrogen, chlorine, and sodium hydroxide.
Electrolysis with Active Electrodes
- Electrodes can participate in electrolysis, allowing for “silver plating.”
Electroplating
- Metal electrodes will be oxidized if their oxidation potential is more positive than water
- Ni electrode: Nis \rightarrow Ni^{+2} + 2e^- E^{\circ} = 0.28V while 2H2O \rightarrow 4H^+ + O_2 + 4e^- E^{\circ} = -1.23V, so the nickel electrode will be oxidized and then nickel ions will then be reduced at the cathode.
- Cathode: Ni^{+2} + 2e^- \rightarrow Ni_s
- As the current flows, nickel dissolves from the anode and then deposits on the cathode.
- This method is used to purify crude metals such as Cu, Zn, Co, and Ni.
- This is also used to electroplate a metal onto another (e.g., silver and gold-plated tableware and jewelry).
- The substance to be electroplated is the cathode.
- The metal used in the electroplating is the anode.
Quantitative Aspects of Electrolysis
- The stoichiometry of the half-reactions tells how many electrons are needed.
- Example: Na^+ + 1e^- \rightarrow Na_s, so 1 mole of Na is deposited for every one mole of electrons.
- Example: Al^{+3} + 3e^- \rightarrow Al_s, so 1 mole of aluminum is deposited for every three moles of electrons.
- The amount of a substance oxidized or reduced is therefore directly proportional to the number of electrons passing through the cell.
- Example: How many moles of Mg are needed to reduce 1.5 moles of Sn^{+4}?