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Consider the following half reactions and their standard reduction potential values to answer the following questions.

Cu2+ (aq) + e- Cu+ (aq)ā€‹EĀ° = 0.15 V
Br2 (l) + 2e- 2Br- (aq) ā€‹EĀ° = 1.08 V

i) State which reaction occurs at the anode and which at the cathode.ā€‹ [2 marks]
ii) Write the overall cell reaction.ā€‹ [2 marks]
iii) Calculate the value of EĀ°cell.ā€‹ [2 marks]
iv) Calculate the value Ī”GĀ°ā€‹ [2 marks]
v) What is the value of Kc at 25Ā°C?

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Answer:

Cu2+ (aq) + e- -------->Cu+ (aq)ā€‹EĀ° = 0.15 V anode

Br2 (l) + 2e- -------------> 2Br- (aq) ā€‹EĀ° = 1.08 V cathode

Explanation:

The half equation having a more positive reduction potential indicates the reduction half equation while the half equation having the less positive reduction potential indicates the oxidation half equation.

The overall redox reaction equation is;

2Cu^+(aq) + Br2(g) ----> 2Cu^2+(aq) + 2Br^-(aq)

EĀ°cell= EĀ°cathode -EĀ°anode

EĀ°cathode= 1.08 V

EĀ°anode= 0.15 V

EĀ°cell= 1.08V-0.15V

EĀ°cell= 0.93V

From;

āˆ†G=-nFEĀ°cell

n= 2

F=96500C

EĀ°cell= 0.93V

āˆ†G= -(2Ɨ96500Ɨ0.93)

āˆ†G= - 179.49 J

From āˆ†G= -RTlnK

āˆ†G= - 179.49 J

R=8.314Jmol-K-1

T= 25Ā° +273= 298K

K= ???

lnK= āˆ†G/-RT

lnK=-( - 179.49/(8.314Ɨ298))

lnK= 0.0724

K= e^0.0724

Kc= 1.075

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