Galvanic or Voltaic Cells

20.4 Galvanic or Voltaic Cells

Chemical Concepts Demonstrated: Voltaic/galvanic cells, relative half-cell potentials

Demonstration:

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) ---> Zn²⁺ (aq) + 2 e^-	E^o= 0.76 V
cathode:	2 H⁺ (aq) + 2 e^- ---> H₂ (g)	E^o= 0.00 V
	Zn (s) + 2 H⁺ (aq) ---> Zn²⁺ (aq) + H₂ (g)	E^o_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) ---> Cu²⁺ (aq) + 2 e^-	E^o= - 0.34 V
cathode:	2 H⁺ (aq) + 2 e^- ---> H₂ (g)	E^o= - 0.00 V
	Cu (s) + 2 H⁺ (aq) ---> Cu ²⁺ (aq) + H₂ (g)	E^o_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) ---> Cu²⁺ (aq) + 2 e^-	E^o= - 0.34 V
cathode:	Zn²⁺ (aq) + 2 e^----> Zn (s)	E^o= - 0.76 V
	Cu (s) + Zn²⁺ (aq) ---> Cu²⁺ (aq) + Zn (s)	E^o_cell = -1.10 V

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