Electrochemistry Sam Pomichter

18.2 Introduction Oxidation- the loss of electrons Reduction- the gain of electrons We can identify oxidation-reduction reactions through changes in oxidation states: oxidation corresponds to an increase in oxidation state and reduction corresponds to a decrease in oxidation state. The oxidation state is the difference between the number of electrons associated with an atom in a compound compared as with the number of electrons in an atom of the element. In ions, the oxidation sate is the is the ionic charge. In covalent compounds the oxidation state corresponds to the formal charge. Elements are assumed to have an oxidation state of 0

Introduction Continued… As you can see the Carbon was reduced because it decreased in oxidation state from a +4 to a +2 The Hydrogen was oxidized because it increased in oxidation state from 0 to +1

Balancing Redox Reactions Both mass and charge must be balanced Thankfully, we can balance redox reactions occurring in aqueous solution with an easier way called the half-reaction method of balancing In the procedure we break down the overall equation into two half-reactions: one for oxidation and then one for reduction Then we balance the half reactions individually and add the together These steps may differ slightly for reactions occurring in acidic or basic solutions

Problem #1 Half-Reaction Method of Balancing Aqueous Redox Equations in Acidic solution Balance the redox equation Al(s) + Cu 2+ (aq)  Al 3+ (aq) + Cu(s)

Problem #1 Solution Al(s) + Cu 2+ (aq)  Al 3+ (aq) + Cu(s) Oxidation: Al(s)  Al 3+ (aq) Reduction: Cu 2+  Cu(s) All elements are balanced with respect to mass, so we proceed to the next step. Al(s)  Al 3+ (aq) + 3e - Balance each half-reaction’s charges 2e - + Cu 2+ (aq)  Cu(s) Continue…

Problem #1 Solution Continued Next we make the number of electrons in both half-reactions equal 2[Al(s)  Al 3+ (aq) + 3e - ] 2Al(s)  2Al 3+ (aq) + 6e - 3[2e - + Cu 2+ (aq)  Cu(s)] 6e - + 3Cu 2+ (aq)  3Cu(s) And finally we add the two half-reactions together, cancelling out the electrons an other species necessary 2Al(s)  2Al 3+ (aq) + 6e - 6e - + 3Cu 2+ (aq)  3Cu(s) Answer: 2Al(s) + 3Cu 2+ (aq)  2Al 3+ (aq) + 3Cu(s)

When Balancing the Mass When balancing each half-reaction with respect to mass you should do so in the following order: 1.Balance all other elements besides H and O 2.Balance the Oxygen by adding H 2 O 3.Balance H by adding H + Now Here’s a problem that requires balancing. Balance the redox equation: Fe 2+ (aq) + MnO (aq)  Fe 3+ (aq) + Mn 2+ (aq)

Problem #2 Mass Ex. Fe 2+ (aq) + MnO (aq)  Fe 3+ (aq) + Mn 2+ (aq) Oxidation: Fe 2+ (aq)  Fe 3+ (aq) Reduction: MnO (aq)  Mn 2+ Now we balance reaction with respect to mass. All elements besides H and O are balanced so we go straight to them Fe 2+ (aq)  Fe 3+ (aq) MnO (aq)  Mn 2+ (aq) + 4H 2 O(l) 8H + (aq) + MnO (aq)  Mn 2+ (aq) + 4H 2 O(l) Now y’all solve the rest…

Problem #2 Solution Fe 2+ (aq)  Fe 3+ (aq) + 1e — Balancing the charges 5e H + (aq) + MnO 4 -- (aq)  Mn 2+ (aq) + 4H 2 O(l) 5[Fe 2+ (aq)  Fe 3+ (aq) + 1e — ] Make sure the amounts of electrons are equal 5Fe 2+ (aq)  5Fe 3+ (aq) + 5e — 5e H + (aq) + MnO 4 -- (aq)  Mn 2+ (aq) + 4H 2 O(l) 5Fe 2+ (aq)  5Fe 3+ (aq) + 5e — Add the half-reactions together 5e H + (aq) + MnO 4 -- (aq)  Mn 2+ (aq) + 4H 2 O(l) Answer: 5Fe 2+ (aq) + 8H + (aq) + MnO 4 -- (aq)  5Fe 3+ (aq) + Mn 2+ (aq) + 4H 2 O(l)

When Balancing the Mass in Basic Solution When balancing the mass in a problem involving a basic solution there is one additional step. You must neutralize H + by adding enough OH –. Remember to add the same number of OH – ions to each side of the equation in order to keep the mass balanced. Don’t fret… There is an example on the next slide!!!

Electrochemical Cells The process of creating energy via redox reactions is normally carried out in a device called an electrochemical cell. Voltaic (or Galvanic) Cell- an electrochemical cell that produces electrical current from a spontaneous chemical reaction. Electrolytic Cell- an electrochemical cell that consumes electrical current to drive a nonspontaneous chemical reaction. Electrode- conductive surfaces through which enter or leave the half- cells, a half-cell is one of two electrodes which function within a voltaic cell Half-cells pair up to create an oxidizing-reducing couple

What Happens in a Voltaic Cell Ex: Zn(s) + Cu 2+  Zn 2+ + Cu(s) So in this voltaic cell a solid strip of zinc is placed in a Zn(NO 3 ) 2 solution to form a half- cell. Then a solid strip of copper is placed in a Cu(NO 3 ) 2 solution to from the second half-cell. The strips in each solution act as electrodes and each strip reaches equilibrium in with its ions in each solution. Zn(s)  Zn 2+ (aq) + 2e — Cu(s)  Cu 2+ (aq) + 2e -- The two half-cells are connected by running a wire from the zinc, to an electrical device like a light bulb, then meeting with the copper. As electrons flow away from the zinc electrode, the Zn/Zn 2+ equilibrium shifts to the right (products), thus causing oxidation. As electrons flow toward the copper electrode, the Cu/Cu 2+ equilibrium shifts to the left (reactants) causing reduction. This creates an electrical current that triggers the electrical device.

Anodes, Cathodes, and Bridges oh my Although the video does not feature a voltaic cell in action (its actually a mercury cell), I believe it defines some key features of batteries. Anode- electrode where oxidation occurs. In a voltaic cell the anode is the more negatively charged electrode. Cathode- electrode where reduction occurs. In a voltaic cell the cathode is the positively charged electrode. ELECTRONS ALWAYS FLOW FROM THE ANODE TO THE CATHODE through the wires connecting the electrodes As electrons flow out of the anode, positive ions form in the oxidation half-cell, resulting in the buildup of positive charge in solution. As electrons flow into the cathode, positive ions are reduced at the reduction half-cell, resulting in the buildup of negative charge in solution. Salt Bridges are inverted U-shaped tubes which contain strong electrolytes such as KNO 3 and link the two half cells together. Salt bridges allow counterions to flow from each half-cell without mixing the solutions, they essentially reduce the buildup of both positive and negative charges. SALT BRIDGES ENABLE THE NEUTRALIZATION OF CHARGE WITHOUT MIXING THE HALF-CELL SOLUTIONS THEY ALLOW ENERGY TO FLOW!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Measuring Electrical Currents Artsy Analogy… “The rate of electrons flowing through a wire is analogous to the rate of water moving through a stream” Electrical Current- the flow of the electric charge Electrical currents are measured in Amperes (A), also called Amps. One ampere is represents the flow of one coulomb per second. 1A = 1C/s 1A corresponds to the flow of x electrons per second Electrical current is also driven by a difference in potential energy called potential difference. Potential difference is typically measured in Volt (V), which is equal to 1 joule per coulomb 1V = 1 J/C So basically, a potentially difference of one volt indicates that a charge of one coulomb experiences an energy difference of one joule between two electrodes.

Measuring Electrical Currents Continued Large potentially energy difference corresponds to a large difference in charge between the two electrodes and therefore a strong tendency for electron flow. Potential difference is also also called electromotive force (emf) In voltaic cells, the potential difference between the two electrodes is the cell potential (E cell ) or cell emf. Under standard conditions the cell potential is called standard cell potential or standard emf. The trend states that combining stuff with a high tendency to oxidize and a high tendency to reduce creates a larger charge difference between two electrodes, therefore increasing the cell potential. “The cell potential is a measure of the overall tendency of the redox reaction to occur spontaneously– the lower the cell potential, the lower the tendency to occur. A negative cell potential indicates that the forward reaction is not spontaneous.”

Electrochemical Cell Notation You can represent electrochemical cells with a compact notation called a cell diagram or line notation. So for the reaction… Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s)

Tips for Line Notation Write the oxidation half-reaction on the left and the reduction on the right. A double vertical line, indicating the salt bridge, separates the two half-reactions. Substances in difference phases are separated by a single vertical line, which represents the boundary between the phases. For some redox reactions, reactants and products of one or both of the half-reactions may be in the same phase. In these cases, we separate the reactants and products from each other with a comma in the line diagram.

Line Notation Problem 5Fe(s) + 2MnO 4 — (aq) + 16H +  5Fe 2+ (aq) +2Mn 2+ (aq) + 8H 2 O(l) Oxidation: Fe(s)  Fe 2+ (aq) + 2e – Reduction: MnO 4 — (aq) + 5e — + 8H +  Mn 2+ (aq) + 4H 2 O(l)

Line Notation Solution Fe(s)|Fe 2+ (aq)||MnO 4 --, H + (aq), Mn 2+ (aq)|Pt(s)

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