Answer:
0.35 V
Explanation:
(a) Standard reduction potentials
<u> E°/V</u>
Fe²⁺ + 2e- ⇌ Fe; -0.41
Cr³⁺ + 3e⁻ ⇌ Cr; -0.74
(b) Standard cell potential
<u> E°/V</u>
2Cr³⁺ + 6e⁻ ⇌ 2Cr; +0.74
<u>3Fe ⇌ 3Fe²⁺ + 6e-; </u> <u>-0.41
</u>
2Cr³⁺ + 3Fe ⇌ 2Cr + 3Fe²⁺; +0.33
3. Cell potential
2Cr³⁺(0.75 mol·L⁻¹) + 6e⁻ ⇌ 2Cr
<u>3Fe ⇌ 3Fe²⁺(0.25 mol·L⁻¹) + 6e-
</u>
2Cr³⁺(0.75 mol·L⁻¹) + 3Fe ⇌ 2Cr + 3Fe²⁺(0.25 mol·L⁻¹)
The concentrations are not 1 mol·L⁻¹, so we must use the Nernst equation
(a) Data
E° = 0.33 V
R = 8.314 J·K⁻¹mol⁻¹
T = 298 K
z = 6
F = 96 485 C/mol
(b) Calculations:
<span>The atoms or molecules attain enough kinetic energy to overcome any intermolecular attractions they have. Since there are no longer any attractive forces between the particles, they are free to drift away into space. The same sort of thing happens in ordinary evaporation, but only at the surface. </span>
Answer:
The carbocation intermediate reacts with a nucleophile to form the addition product.
Explanation:
The reaction of benzene with an electrophile is an electrophillic substitution reaction. Here the electrophile replaces hydrogen. There is no formation of carbocation as intermediate in the reaction. Infact there is transition state where the electorphile attacks on benzene ring and at the same time the hydrogen gets removed from the benzene. So a transition carbocation is formed.
The general mechanism is shown in the figure.
i) Attack of the electrophile on the benzene (which is the nucleophile)
ii) The carbocation intermediate loses a proton from the carbon bonded to the electrophile.
iii) the carbocation formation is the rate determining step.
iv) There is no formation of addition product.
Thus the wrong statement is
The carbocation intermediate reacts with a nucleophile to form the addition product.