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:
![Q = \dfrac{\text{[Fe}^{2+}]^{3}}{ \text{[Cr}^{3+}]^{2}} = \dfrac{0.25^{3}}{ 0.75^{2}} =\dfrac{0.0156}{0.562} = 0.0278\\\\E = 0.33 - \left (\dfrac{8.314 \times 298}{6 \times 96485}\right ) \ln(0.0278)\\\\=0.33 -0.00428 \times (-3.58) = 0.33 + 0.0153 = \textbf{0.35 V}\\\text{The cell potential is }\large\boxed{\textbf{0.35 V}}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BFe%7D%5E%7B2%2B%7D%5D%5E%7B3%7D%7D%7B%20%5Ctext%7B%5BCr%7D%5E%7B3%2B%7D%5D%5E%7B2%7D%7D%20%3D%20%5Cdfrac%7B0.25%5E%7B3%7D%7D%7B%200.75%5E%7B2%7D%7D%20%3D%5Cdfrac%7B0.0156%7D%7B0.562%7D%20%3D%200.0278%5C%5C%5C%5CE%20%3D%200.33%20-%20%5Cleft%20%28%5Cdfrac%7B8.314%20%5Ctimes%20298%7D%7B6%20%5Ctimes%2096485%7D%5Cright%20%29%20%5Cln%280.0278%29%5C%5C%5C%5C%3D0.33%20-0.00428%20%5Ctimes%20%28-3.58%29%20%3D%200.33%20%2B%200.0153%20%3D%20%5Ctextbf%7B0.35%20V%7D%5C%5C%5Ctext%7BThe%20cell%20potential%20is%20%7D%5Clarge%5Cboxed%7B%5Ctextbf%7B0.35%20V%7D%7D)
The neutrons take up most of the space of an atom
Answer:
the particles of matter are arranged in different ways for the different states
Explanation:
because solid liquid and gas all three matters have different states for example the particles in a solid are closely packEd and form of movement is vibration
The grams of oxygen that are produced is 228.8 grams
<em>calculation</em>
2H₂O₂ → 2H₂O +O₂
Step 1: use the mole ratio to determine the moles of O₂
from equation above H₂O₂:O₂ is 2:1
therefore the moles of O₂ = 14.3 moles ×1/2 = 7.15 moles
Step 2: find mass of O₂
mass = moles × molar mass
= 7.15 moles × 32 g/mol =228.8 g
