Terms:
Electrochemical Cell: a system consisting of electrodes that dip into an electrolyte and in which a chemical reaction either uses or generates an electric current.
Voltaic/Galvanic Cell: an electrochemical cell in which a spontaneous reaction generates an electric current.
Electrolytic Cell: an electrochemical cell in which an electric current drives an otherwise non-spontaneous reaction.
Galvanic/Voltaic Cell:
A voltaic cell consists of two half-cells.
Half-cell: the portion of an electrochemical cell in which a half reaction takes place, electrically connected to the other half-cell.
Here, one is a zinc/zinc ion half-cell where a strip of zinc metal is in a solution of a zinc salt. The other is a copper metal strip in a solution of a copper salt. These are called a zinc electrode (the anode) and copper electrode (the cathode), respectively.
An external circuit must connect the two zinc and copper half-cells two allow electron flow between the metal electrodes, represented in the diagram by the placement of the voltmeter.
The zinc electrode is dissolved as it's oxidized (as the zinc metal atom loses two e–). These e– flow through the zinc electrode to the external circuit. From the external circuit, they reach the copper electrode and the other half-cell. This gives the copper electrode a negative charge, which attracts the positively charged copper cations in solution. Copper is then deposited on the copper electrode as copper cations become reduced to copper metal.
One metal must be able to reduce the cation of the other without allowing the two solutions to mix. The charges in the solution need to remain in contact without allowing the metals to react. The solution to this problem is a salt bridge, the peach-colored tube filled with KCl in the diagram.
Salt Bridge: a tube of an electrolyte in a gel that is connected to the two half-cells of a voltaic cell; the salt bridge allows the flow of ions but prevents the mixing of the different solutions that would allow direct reaction of the cell reactants.
The two half-cell reactions are:
Zn (s) → Zn2+ (aq) + 2e–
(oxidation half-reaction, a species loses electrons)
Cu2+ (aq) + 2e– → Cu (s)
(reduction half-reaction, a species gains electrons)
Anode: the electrode at which oxidation occurs. Electrons are given up by the anode and flow to the cathode. Represented with a negative sign to show that electrons flow from it.
Cathode: the electrode at which reduction occurs. Electrons flow from the anode to the cathode. Represented with a positive sign.
Cell Reaction: the sum of the two half-reactions, the net reaction that occurs in the voltaic cell.
Cell Reaction:
Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s)
Note: electrons are never aqueous, don't give them phase labels.
To sketch and label a voltaic cell, it helps to remember the key information:
1. Remember the basic setup of two half-cells and their electrodes connected by an external current and an internal salt bridge.
2. Recognize any information in the problem that will help you figure out which half-cell reaction is the reduction and which is the oxidation. If one half-cell is known to be reduced, then we know that it is the cathode and the other is the anode.
3. Remember the electrons flow away from the anode to the cathode.
Notation:
Example:
For the voltaic cell in the above diagram, it is written as:
Zn (s) │ Zn2+ (aq) ║ Cu2+ (aq) │ Cu (s)
-The anode (oxidation half-cell) is written on the left, the cathode (reduction half-cell) is written on the right.
-The cell terminals are written on the far right and far left of the notation. The salt solutions are written on the inner right and inner left of the notation.
-The single vertical line represents a phase boundary, for example between the solid and the solution.
-The double vertical line represents the salt bridge.
-When the half reaction involves a gas, an inert material such as platinum may be used as a surface for the reaction. In this case, Pt will be added to the outside end of that electrode half-reaction.
-Coefficients are not included.
Example:
Write the following reaction in cell notation:
2Tl (s) + Sn2+ (aq) → 2Tl+ (aq) + Sn (s)
First write the half-reactions and find out what is reduced and what is oxidized.
2Tl (s) → 2Tl+ (aq) + 2e– (oxidation)
2e– + Sn2+ (aq) → Sn (s) (reduction)
Apply notation:
Tl (s) │ Tl+ (aq) ║ Sn2+ (aq) │ Sn (s)
Example:
Write the following reaction in cell notation:
Zn (s) + 2Fe3+ (aq) → Zn2+ (aq) + 2 Fe2+ (aq)
(with platinum used as a catalyst).
First write the half-reactions and find out what is reduced and what is oxidized.
Zn (s) → Zn2+ (aq) + 2e–
Fe3+ (aq) + e– → Fe2+ (aq)
Apply notation:
Zn (s) │ Zn2+ (aq) ║ Fe3+ (aq) , Fe2+ (aq) │ Pt
Note: There is a comma between Fe3+ and Fe2+ on the right side because they are both in the same phase, and platinum appears on the end.
To write a cell reaction from the cell notation, write each half-reaction from the notation and add them together.
Example:
Tl (s) │ Tl+ (aq) ║ Sn2+ (aq) │ Sn (s)
Write half-reactions:
2Tl (s) → 2Tl+ (aq) + 2e–
2e– + Sn2+ (aq) → Sn (s)
Take the sum of these reactions:
2Tl (s) +
(Simplify)
2Tl (s) + Sn2+ (aq) → 2Tl+ (aq) + Sn (s)
Note: These cell notations show products and reactants in standard state:
aq → 1.00 M
gas → 1 atm
When not in standard state, the information for concentration or pressure goes in parentheses just to the right for each species.
Example:
Zn (s) │ Zn2+ (aq) ║ H+ (aq) (0.200 M) │ H2 (g) │ Pt
Cell Potential:
Potential Difference: the difference in electric potential (electrical pressure) between two points. Measured in volts (V).
The electrical work expended on a charge moving through an electric potential:
Electrical work = charge x potential difference
Using corresponding units:
Joules = coulombs x volts → J = C x V
Faraday Constant: Denoted F, the magnitude of the charge on one mole of electrons; equal to 96,485 C/mol e–. The faraday is a unit of charge equal to 96,485 C.
Cell Potential: the maximum potential difference between the electrodes of a voltaic cell. Denoted Ecell. Ecell must be positive for a spontaneous process. Larger values for Ecell are desirable. The larger the Ecell, the greater the energy we get out of the system.
Maximum work obtainable form a voltaic cell:
Wmax = ∆Go = –nFEcell
where:
n = number of moles of electrons transferred in the overall cell equation.
F = the Faraday constant
Ecell = the cell potential
Note: ∆Go will be less than zero for a spontaneous process.
thnk u
ReplyDeleteby vishal srivastava.
mumbai
mumbai?! what a wonderful thing to see on my birthday! thank you.
ReplyDeleteso does the least oxidised (most reduced species) appear closest to the platinum inert electrode?
ReplyDeletesdfghjkl
DeleteI was taught that it is used as a surface of reaction, so when you have two components of a half cell that are either both gas or both aqueous, the platinum provides a surface.
ReplyDeleteFrom Ebbing's book, page 780: "When the half-reaction involves a gas, an inert material such as platinum serves as a terminal and as an electrode surface on which the half-reaction occurs. The platinum catalyzes the half-reaction but otherwise is not involved in it."
So I believe is has more to do with phase than with oxidation/reduction.
at last something useful
ReplyDeleteuzair abro from Karachi, pakistan
THANK YOU! I have a test on this today and this was the most helpful thing I have ever come across. You are appreciated.
ReplyDeleteIts really helping me
ReplyDelete