Question

The difference of potential of two electrodes in a galvanic cell is known as:
[A] EMF
[B] Potential difference
[C] Electrode difference
[D] Ionic difference

Answer 1

Hint: In a galvanic cell there are two electrodes. One is the anode and other is the cathode. Oxidation takes place at the anode and reduction takes place at the cathode. The difference between the electrodes gives us the maximum potential difference between the two cells.

Complete answer:
Before answering this question, let us discuss what a galvanic cell is.
We also know Galvanic cells by the name of Voltaic cells. It is an electrochemical cell which converts chemical energy into electrical energy. Inside the cell we have active chemicals that take part in redox reactions simultaneously and is the main working principle of the cell.
Generally in a Galvanic cell we find two sections where two different metals are immersed in an electrolytic solution. The electrolytic solution consists of the ions of the same metal that is immersed in it.
In this cell, one metal acts as the anode and the other acts as the cathode. Anode is the negative portal and oxidation takes place there and cathode is positive and reduction takes place at the cathode.
We join the two cells by a salt bridge which consists of generally an amalgam or through a porous membrane.
We can calculate the EMF of the cell by subtracting the anode electrode potential from the cathode electrode potential-
     \[{{E}_{cell}}={{E}_{cathode}}-{{E}_{anode}}\]
As we can see from the above equation, the difference between the two cells is the EMF of the cell.

Therefore, the correct answer of option [A] EMF.

Note: A common example of a Galvanic cell is the Daniel cell which contains a zinc half-cell with zinc metal dipped in a zinc sulphate solution and a copper half-cell with a copper metal dipped in a copper sulphate solution. The two half cells are joined through a salt bridge. The electrons flow through the external circuit. Zinc is the anode and copper is the cathode. Zinc metals are oxidised to $Z{{n}^{2+}}$ releasing 2 electrons and $C{{u}^{2+}}$ ions are reduced to copper metal by taking up the electrons. The half-cell reactions can be written as-
     \[\begin{align}
  & Zn(s)\to Z{{n}^{2+}}(aq.)+2{{e}^{-}} \\
 & C{{u}^{2+}}(aq.)+2{{e}^{-}}\to Cu(s) \\
\end{align}\]