Answer:
See below explanation
Explanation:
Checking the reduction potencials:
ZnāŗĀ²/Zn° , E° = -0.76 V
FeāŗĀ²/Fe° , E° = -0.44 V
For which FeāŗĀ² will be the ion that will reduce, and Zn° will lose electrons
ANODE: Ā Ā Zn° ā ZnāŗĀ² + 2eā»
CATODE: FeāŗĀ² + 2eā» ā Fe°
NET CELL REACTION: FeSOā (ac) + Zn° ā Fe° + ZnSOā (ac)
In the external circuit, electrons will migrate from anode (Zn|ZnāŗĀ²) to catode (Fe|FeāŗĀ²), and in the salt bridge; anions migrate from FeāŗĀ²|Fe compartment to Zn|ZnāŗĀ² compartment