Construct a Al-Br2 electrochemical cell, write down the anode reaction and cathode reaction and determine the cell voltage.
Ande reaction :
Oxidation half reaction :
Al (s) ------------------> Al+3 + 3 e- Eo = - 1.662 V
cathode reaction :
reduction half reaction ;
Br2 + 2e- ---------------> 2 Br - Eo = 1.0873 V
Overall reaction :
3 Br2 + 2 Al --------------> 2 Al+3 + 6 Br -
Eo cell = Eo cathode - Eo anode
= 1.0873 - (- 1.662)
= 2.75 V
cell voltage.= 2.75 V
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