Answer :
Answer:
∆G°= -55005J or -55KJ
Explanation:
The first step is to determine E°cell= E°cathode - E°anode
2Cl-(aq)/Cl2(g) system is the cathode while 2Br-(aq)/Br2(g) is the anode
E°cell= 1.360-1.075
E°cell= 0.285V
From
∆G°= -nFE°cell
n= 2 from the balanced reaction equation, two electrons were transferred.
F= 96500
E°cell=0.285V
∆G°= -(2×96500×0.285)
∆G°= -55005J or -55KJ
The ΔG° for a cell is 55000 J
The equation for the reaction is given as:
[tex]\mathbf{Cl_{2(g)} + 2Br_{(aq)} \to 2Cl_{(aq)} + Br_{2(l)}}[/tex]
In the reaction, the reduction of chlorine takes place at the cathode;
i.e.
[tex]\mathbf{Cl_2+2e^- \to 2Cl^-}[/tex]
The oxidation of Bromine takes place at the anode;
i.e.
[tex]\mathbf{2Br \to Br_2+2e^-}[/tex]
The Cell potential [tex]\mathbf{E^0_{Cell} }[/tex] can be calculated as:
[tex]\mathbf{E^0_{Cell}= E^0_{cathode} - E^0_{anode} }[/tex]
[tex]\mathbf{E^0_{Cell}=(1.360 - 1.075) volts }[/tex]
[tex]\mathbf{E^0_{Cell}=0.285 \ volts }[/tex]
∴
The ΔG° for a cell can be determined by using the expression;
[tex]\mathbf{\Delta G^0 = -nFE^0_{cell}}[/tex]
where;
- F = faraday's constant
[tex]\mathbf{\Delta G^0 = -(2) \times 96485 C/mol \times 0.285 J/C}[/tex]
[tex]\mathbf{\Delta G^0 =54996.45 \ J}[/tex]
[tex]\mathbf{\Delta G^0 \simeq 55000 \ J}[/tex]
Learn more about Faraday's Law here:
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