Question

A variable, opposite external potential (${{E}_{ext}}$ ) is applied to the cell $Zn/Z{{n}^{2+}}(1M)//C{{u}^{2+}}\left( 1M \right)/Cu$ , of potential 1.1 V. When ${{E}_{ext}}$ < 1.1V and ${{E}_{ext}}$ > 1.1 V respectively electron flow from:
A. Cathode to anode in both cases
B. Anode to cathode and cathode to anode
C. Anode to cathode in both cases
D. Cathode to anode and anode to cathode

Answer 1

Hint: In galvanic cell zinc electrode is dipped in zinc sulphate solution and copper electrode is dipped in copper sulphate solution. Here the zinc electrode acts as anode and copper electrode acts as cathode. The zinc electrode potential is -0.34 V and copper electrode potential is 0.76 V.

Complete step by step answer:
- In the question it is given that a variable, opposite external potential ( ${{E}_{ext}}$ ) is applied to the galvanic cell whose potential is 1.1 V.
- We know that in a galvanic cell the electrons flow from anode to cathode.
- The potential of the galvanic cell is 1.1 V.
- If we are not going to apply any external potential or if we are going to apply the external potential less than 1.1 V then the electrons will flow from anode to cathode.
- If we apply external potential greater than 1.1 V then the cell reaction will be reversed. Means the electrons will flow from cathode to anode.
- Therefore if we apply potential less than 1.1 V to the galvanic cell then the electrons will flow from anode to cathode, if we apply greater than 1.1 V to the galvanic cell then the electrons flow from cathode to anode.

So, the correct option is B.

Note: To stop the electron flow from anode to cathode or cathode to anode in a galvanic cell we should apply an external potential of exactly 1.1 V to the galvanic cell. Zinc acts as anode and copper acts as cathode in a galvanic cell.