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Galvanic or Voltaic Cells

 

Chemical Concepts Demonstrated

  • Voltaic/galvanic cells, relative half-cell potentials

Demonstration

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  • One of the dishes is filled with ZnSO4 and the other with HCl.
  • A strip of Zn metal is attached at one end to the posts of the electrochemistry template and is placed at the other end into the dish filled with Zn2+.
  • The hydrogen electrode is attached, placed into the HCl solution, and H2 gas is bubbled in. Insert the salt bridge.
  • The Zn2+/Zn half-cell is replaced with a Cu2+/Cu half-cell.
  • The H+/H2 half-cell is replaced with a Zn2+/Zn. (picture 2)   
     

Observations

The potential in the absence of the salt bridge is 0.00 V.  After the salt bridge is inserted, the potential of the first set up is around + 0.76 V and the cell is a galvanic or voltaic cell.  The Zn2+/Zn half-cell is the anode the H+/H2 is the cathode.

In the second set up, both the magnitude and the sign of the potential change.   The potential is now roughly - 0.34 V.

Picture 2 shows the third set up.  The potential is now - 1.10 V. If the leads are changed, the cell potential becomes + 1.10 V and the cell becomes a galvanic or voltaic cell.

Explanations (including important chemical equations)

With the leads connected so as to produce a cell potential of + 0.76 V, the half reactions are:

anode: Zn (s) ---> Zn 2+ (aq) + 2 e- E= 0.76 V
cathode: 2 H + (aq) + 2 e - ---> H2 (g) E= 0.00 V
  Zn (s) + 2 H +  (aq) ---> Zn 2+ (aq) + H2 (g) Eo cell = 0.76 V

If the standard-state potential for the H+/H2 half-cell is assumed to be 0.00 V, and the potential for the anode half-reaction is equal in magnitude but opposite in sign to the standard-state potential for the Zn2+/Zn couple, then the standard-state reduction potential for the Zn2+/Zn half-cell must be - 0.76 V.

If the Zn2+/Zn half-cell is replaced with a Cu2+/Cu half-cell without reversing the leads to the voltmeter, the overall cell potential is - 0.34 V and the standard-state reduction potential for the Cu2+/Cu couple is therefore + 0.34 V.

anode: Cu (s) ---> Cu 2+ (aq) + 2 e- E=  - 0.34 V
cathode: 2 H + (aq) + 2 e - ---> H2 (g) E=  - 0.00 V
  Cu (s) + 2 H +  (aq) ---> Cu 2+ (aq) + H2 (g) Eo cell  = -0.34 V

If the H+/H2 half-cell is replaced with a Zn2+/Zn half-cell, the overall cell potential should be - 1.10V. 

anode: Cu (s) ---> Cu 2+ (aq) + 2 e- E=  - 0.34 V
cathode: Zn 2+ (aq) + 2 e---> Zn (s) E=  - 0.76 V
  Cu (s) + Zn 2+ (aq) ---> Cu 2+ (aq) + Zn (s) Eo cell  = -1.10 V

To set up a voltaic cell using these half reactions, one would have to reverse the leads to the voltmeter.

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