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# Galvanic Cell Tutorial

(AKA Electrochemical Cell or Voltaic Cell)

A galvanic cell is a spontaneous electrochemical cell that produces electricity by a spontaneous redox reaction.

To complete the circuit, a salt bridge is require for "ION FLOW".

What you need to know:

Oxidation Occurs at the Anode, reduction occurs at the Cathode.

Table J is an oxidation table.

Oxidation is on top (AN OX)

Reduction is on the bottom (Red Cat)

Anodes are located above Cathodes

Electrons flow through the wire (from lost to gained) Oxidation to reduction or Anode to Cathode.

Red Cat  An Ox

Electrodes must be metal (no ions), if there are only ions present an inert electrode (ex. platinum) should be used.

The salt bridge or porous membrane is used to allow "Ions Flow". This allows electrons to continuously move through the wire while replacing these lost negatives charges.

In other words an electron goes left and a negative ion goes right. Everything is still neutral.

Past Regents Questions

June 2010-47 Given the balanced ionic equation representing the reaction in an operating voltaic cell:
Zn(s) + Cu2+(aq)==> Zn2+(aq) + Cu(s)
The flow of electrons through the external circuit in this cell is from the
(1) Cu anode to the Zn cathode
(2) Cu cathode to the Zn anode
(3) Zn anode to the Cu cathode
(4) Zn cathode to the Cu anode

42 Given the balanced equation representing the reaction occurring in a voltaic cell:
Zn(s) + Pb2+(aq)==> Zn2+(aq) + Pb(s)
In the completed external circuit, the electrons flow from
(1) Pb(s) to Zn(s)
(2) Pb2+(aq) to Zn2+(aq)
(3) Zn(s) to Pb(s)
(4) Zn2+(aq) to Pb2+(aq)

26 Where does oxidation occur in an electrochemical cell?
(1) at the cathode in both an electrolytic cell and a voltaic cell
(2) at the cathode in an electrolytic cell and at the anode in a voltaic cell
(3) at the anode in both an electrolytic cell and a voltaic cell
(4) at the anode in an electrolytic cell and at the cathode in a voltaic cell

27 Which statement is true for any electrochemical cell?

(1) Oxidation occurs at the anode, only.
(2) Reduction occurs at the anode, only.
(3) Oxidation occurs at both the anode and the cathode.
(4) Reduction occurs at both the anode and the cathode.

 How do you use this Reduction table? When determining oxidation and reduction reactions find both possible reduction reactions. Which ever is high is the actual reduction reaction that will take place. The lower reaction is actually oxidation and must be reversed. To find the voltage of the cell add the 2 voltages together, remembering that you reverse the lower reaction so the Eo must also be reversed (+ becomes - or vice versa). Example Determining Oxidation and Reduction Here are the 2 possible reduction reactions (reduction charge is goes down) Cu2+ + 2e- --> Cu Zn2+ + 2e- --> Zn Now look for them on the table. Since  Cu2+ + 2e- --> Cu is higher on the table, the other reaction is reversed. So the reactions are written like this: Cu2+ + 2e- --> Cu    (Red.)   Cu is the cathode Zn--> Zn2+ + 2e-         (Ox.) Zn is the anode Determining Voltages Cu2+ + 2e- --> Cu  Eo= +0.34v Zn--> Zn2+ + 2e-   Eo= -(-0.76v) Note- since we reversed Zn we reverse the sign Now just add them up 0.34v + 0.76v=1.10v ***Never to be multiply cell potential (voltages) even when you balance you reaction*** See below for explanation

Here is a voltaic cell lab simulation.

Try it- you have to pick a metal and that metals ion in the solution or it won't work. Se if you can predict the voltage.

From-http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/flashfiles/electroChem/volticCell.html