I read in a book that "Sometimes, both the electrodes dip in the same electrolyte solution & in such cases we do not require a salt bridge".
How is this possible?
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$\begingroup$ If both electrodes are in the same vessel of the same solution, ions are free to flow between the electrodes. The purpose of a salt bridge is to allow for a complete circuit by enabling the ions to move freely, so would be extraneous in this case because they can already move freely. $\endgroup$– EllCommented Jun 5, 2014 at 23:30
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2 Answers
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It depends where and how the electrodes are positioned. The UC Davis ChemWiki page Electrochemistry 2: Galvanic cells and electrodes describes the possible arrangements.
For example, if the electrodes are placed in the same vessel, a salt bridge is not required (as shown below). Ions can pass through the porous barrier.
If, however, the electrodes are are placed in separate vessels (such as in the diagram below), then a salt bridge is required to facilitate ion movement.
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You need a salt bridge / barrier to maintain electrical contact (without metal) but to prevent copper from plating onto the $\ce{Zn}$.
A super-saturation of metallic copper would produce a small voltage in a homogenous electrolyte with but $\ce{Zn}$ and $\ce{Cu}$ ions. For any $\ce{Cu}$-positive voltage it is be energetically favorable for $\ce{Cu}$ to plate onto the $\ce{Zn}$. However, there is activation energy in nucleating a new phase ($\ce{Cu}$ metal). Thus we will get super-saturated: Conceptually, the "concentration" of $\ce{Cu}$ metal at the Zn electrode exceeds the concentration of $\ce{Cu}$ metal at the copper electrode before the $\ce{Cu}$ phase appears! However, once $\ce{Cu}$ plating begins, the voltage would drop to zero since the cell is now "shorted". Copper will continue to build up and zinc will erode away but no voltage will appear. If this nucleation voltage never is reached (preventing voltage-buildup requires an electrical connection between the two strips), you can get a small amount of energy from a cell in a homogenous solution, but it probably will be much less than 1.1V, and maintaining a low enough voltage would require current to keep flowing.
