1
$\begingroup$

A typical Galvanic Cell, say a Daniell cell, consitsts of two separate beaker, one containing Zn rod dipped inside aq $\ce{ZnSO4}$ and the other beaker containing Cu rod dipped inside $\ce{CuSO4}$. Does it matter what electrolyte we dip the metal in?

If we change the electrolyte of the beaker containing Zn rod to say $\ce{AgNO3}$. I would expect a displacement reaction to take place where a redox reaction occur between the Zinc rod and the Ag+ ions. No electrons will flow through the external wire. What if I replace the electrolyte with a solution of a metal that is less reactive then the metal dipped in? eg. $\ce{NaCl}$, $\ce{MgCl2}$, $\ce{KCl}$(aq). Would my cell still work?

Also, I don't understand the need of a porous pot OR salt bridge in a galvanic cell. Why would it matter if the solutions are mixed together? Why not just have one single beaker, containing an electrolyte $\ce{KCl}$(aq), and have the two different metals dipped into it. In fact, if the solutions are mixed together, there will not even be unbalance charge accumulation, and the porous pot/salt bridge would not be necessary?

$\endgroup$
  • $\begingroup$ Hello and welcome to StackExchange! Take a look at the help center to learn how to use this site better. Cheerio! $\endgroup$ – Pritt Balagopal May 25 '17 at 4:38
1
$\begingroup$

The electrolyte is very important for the functioning of the galvanic cell. The emf that the cell generates depends on the type and the concentration of the electrolyte used. In the Daniell cell, the reaction would be:

$$\ce{Zn + CuSO4 -> ZnSO4 + Cu}$$

As per the Nernst's equation, you get the emf of the cell to be:

$$\text{E}=\text{E}^\text{o}-\frac{2.303\text{RT}}{\text{nF}}\log_{10}{\left(\frac{\ce{[Zn^{2+}]}}{\ce{[Cu^{2+}]}}\right)}$$

Clearly, the emf of the cell depends on the concentrations of the cell. Dipping the rods in $\ce{NaCl}$ would not make the cell work.

And for why the salt bridge is needed, think about why you would build a cell in the first place. To generate power in a usable way right? If you let the solutions mix, they would react directly, and the energy would be released as heat instead, and not as electrical power. You want the ions to move towards the electrodes and get discharged there.

$\endgroup$
  • $\begingroup$ I don't want to add another answer so I'm commenting here. A salt bridge also: (1) completes the internal circuit of the galvanic cell. (2) establishes charge neutralization of the 2-electrode solutions by supplying them with appropriate ions. (3) minimizes liquid-liquid junction potential. $\endgroup$ – Berry Holmes May 25 '17 at 7:10
  • $\begingroup$ Also see: Ivan Neretin's comment to my question. $\endgroup$ – Berry Holmes May 25 '17 at 7:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.