Question

Depict the galvanic cell in which the following reaction takes place:
$Zn(s){\text{ + }}2A{g^ + }(aq.){\text{ }} \to {\text{ }}Z{n^{2 + }}(aq.){\text{ + }}2Ag(s)$
Further show:
A.Which of the electrode is negatively charged?
B. The carriers of the current in the cell.
C. Individual reaction at each electrode.

Answer 1

Galvanic cell is one of the electrochemical cells which is used to supply the current with help of transfer of electrons through a redox reaction taking place at each electrode respectively. Thus we will identify the electrodes at which oxidation and reduction takes place respectively. The flow of carriers in the external circuit will go from positive electrode to negative electrode.

Complete Step By Step Answer:
$Zn(s){\text{ + }}2A{g^ + }(aq.){\text{ }} \to {\text{ }}Z{n^{2 + }}(aq.){\text{ + }}2Ag(s)$
The above cell reaction can be represented as:
$Zn(s)|{\text{ }}Z{n^{2 + }}(aq.){\text{ || }}A{g^ + }(aq.){\text{ | }}Ag(s)$
With the help of the above reaction we can identify that there are two electrodes namely zinc electrode and silver electrode. Since the Zinc atom undergoes oxidation and the silver atom undergoes reduction we can say that the zinc electrode is anode (oxidation takes place at anode) and the silver electrode will behave like cathode (reduction takes place at cathode). Now we will write these cell reaction separately as:
$\Rightarrow$ At anode (zinc electrode):
$Zn(s) \to {\text{ }}Z{n^{2 + }}(aq.){\text{ + 2}}{{\text{e}}^ - }$ ___________ $(1)$
The oxidation of zinc takes place and it will lose two electrons to become $Z{n^{2 + }}(aq.)$ .
$\Rightarrow$ At cathode (silver electrode):
${\text{ }}A{g^ + }(aq.){\text{ + 1}}{{\text{e}}^ - }{\text{ }} \to {\text{ }}Ag(s)$ ____________ $(2)$
Here reduction of metal takes place by accepting one electron. We can write the cell reaction by multiplying equation $(2)$ by two and then adding with equation $(1)$ .
$Zn(s){\text{ + }}2A{g^ + }(aq.){\text{ + 2}}{{\text{e}}^ - }{\text{ }} \to {\text{ }}Z{n^{2 + }}(aq.){\text{ + }}2Ag(s){\text{ + 2}}{{\text{e}}^{{ - ^{}}}}$
$Zn(s){\text{ + }}2A{g^ + }(aq.){\text{ }} \to {\text{ }}Z{n^{2 + }}(aq.){\text{ + }}2Ag(s)$
A. Since at anode oxidation of zinc metal takes place then it has an excess of free electrons. Due to more electronic charge anode will behave like a negative electrode. Therefore we can say that zinc electrodes are negatively charged electrodes.
B. The carriers of current in the cell are free electrons which move from anode to cathode and thus produce the current in the external circuit from cathode to anode.
C. The individual cell reactions are represented as:
At anode: $Zn(s) \to {\text{ }}Z{n^{2 + }}(aq.){\text{ + 2}}{{\text{e}}^ - }$
At cathode: ${\text{ }}A{g^ + }(aq.){\text{ + 1}}{{\text{e}}^ - }{\text{ }} \to {\text{ }}Ag(s)$

Note:
Since the flow of electric current is always in the opposite direction of flow of electrons. This is because electric current will flow from cathode to anode indeed electrons flow from anode to cathode. Here anode is a negative electrode because it has excessive electrons as compared to cathode. Galvanic cell reaction is always a redox reaction.