Question

The removal of salt bridge from Daniell cell brings in the following changes:

• A. $E_{Zn/Z{n}^{2+}}^o$ changes
• B. $E_{Cu/C{u}^{2+}}^o$ changes
• C. $E_{cell}^o$ drops to zero
• D. none of the above

Answer 1

The correct option is C $E_{cell}^o$ drops to zero

Salt bridge connects the solutions of two half-cells and completes the cell circuit. It prevents transference or diffusion of the solutions from one half-cell to the other.

It keeps the solutions in two half-cells electrically neutral. In an anodic half cell, positive ions pass into the solution and there shall be an accumulation of extra positive charge in the solution around the anode which will prevent the flow of electrons from anode. This does not happen because negative ions are provided by salt bridge.

Similarly, in cathodic half-cell, negative ions will accumulate around cathode due to deposition of positive ions by reduction. To neutralize these negative ions, a sufficient number of positive ions is provided by a salt bridge. Thus, the salt bridge maintains electrical neutrality.

It prevents liquid-liquid junction-potential, i.e., the potential difference which arises between two solutions when in contact with each other.

So in absence of salt-bridge, liquid-liquid junction-potential forms and $E^{0}cell$ drops to zero.