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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) EĀ° = 0.15 V
Br2 (l) + 2e- 2Br- (aq) EĀ° = 1.08 V

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

Respuesta :

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Answer:

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

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

Explanation:

We have to first state the fact that the reaction having the most positive reduction potential occurs at the cathode in any spontaneous electrochemical cell. The half reaction with the less positive electrode potential usually occurs at the anode.

The overall reaction equation is;

2Cu2+ (aq) + Br2 (l) ----->2Cu+ (aq) + 2Br- (aq)

EĀ°cell= EĀ°cathode - EĀ°anode

EĀ°cathode= 1.08 V

EĀ°anode= 0.15V

EĀ°cell = 1.08-0.15 = 0.93 V

But

āˆ†GĀ°= -nFEĀ°cell

n= 2, F=96500C, EĀ°cell= 0.93V

āˆ†GĀ° = -(2Ɨ 96500Ɨ 0.93)

āˆ†G= -179490 J

But;

āˆ†G = -RTlnK

R=8.314 JK-1

T= 25+273= 298K

Kc= the unknown

āˆ†GĀ° = -179490 J

Substituting values and making lnK the subject of the formula

lnK= āˆ†G/-RT

lnK= -( -179490/8.314 Ɨ 298)

lnK= 72.45

K= e^72.45

K= 2.91Ɨ10^31

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