Oxidation occurs at the anode, so your answer is (2) loss of electrons
Answer:
The answer to your question is 40 L of NH₃
Explanation:
Data
Volume of NH₃ = x
mass of N₂ = 25 g
mass of H₂ = excess
Balanced chemical reaction
N₂ + 3H₂ ⇒ 2NH₃
Process
1.- Find the molar mass of N₂ and NH₃
N₂ = 14 x 2 = 28g
2NH₃ = 2[ 14 + 3] = 34 g
2.- Write a proportion to solve this problem
28 g of N₂ --------------- 34 g of NH₃
25 g of N₂ ------------- x
x = (25 x 34)/28
x = 30.36 g of NH₃
3.- Calculate the volume of NH₃
17 g of NH₃ -------------- 22.4 L
30.36 g of NH₃ -------- x
x = (30.36 x 22.4) / 17
x = 40 L
Answer:
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
<span>To generate all of the energy for the cell, store substances such as food, water, and wastes, and engulf pathogens that have invaded the cell.</span>
