Consider the following half-reactions and their standard reduction potential values to answer the following questions. Cu2+ (aq) + e- Cu+ (aq) E° = 0.15 V
Br2 (l) + 2e- 2Br- (aq) E° = 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 E°cell. [2 marks]
iv) Calculate the value ΔG° [2 marks]
v) What is the value of Kc at 25°C? [2 marks]

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

Cu2+ (aq) + e- >Cu+ (aq) E° = 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)

E°cell= E°cathode - E°anode

E°cathode= 1.08 V

E°anode= 0.15V

E°cell = 1.08-0.15 = 0.93 V

But

∆G°= -nFE°cell

n= 2, F=96500C, E°cell= 0.93V

∆G° = -(2× 96500× 0.93)

∆G= -179490 J

But;

∆G = -RTlnK

R=8.314 JK-1

T= 25+273= 298K

Kc= the unknown

∆G° = -179490 J

Substituting values and making lnK the subject of the formula

lnK= ∆G/-RT

lnK= -( -179490/8.314 × 298)

lnK= 72.45

K= e^72.45

K= 2.91×10^31

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2

idk

explanation: