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 concentration of ClO₂⁻ at equilibrium if the initial concentration of HClO₂ is 0.0654.
<h3>What is concentration?</h3>
The concentration of any substance is the quantity of that substance in per square of the space or container.
The reaction is
HClO₂ + H₂O <=> H₃O⁺ + ClO₂⁻
The pH is 0.454 M
Ka = [H₃O⁺][ClO₂⁻ ] / [HClO₂]
2. 25 × 10⁻² m = [x][x] / 0.454-x]
2 + 0.011 - 0.004994 = 0
solve the quadratic equation
x = 0.0654 = [H3O+] = [ClO2-]
pH = -log (H3O+)
pH = -log(0.0654)
pH = 1.2
equilibrium concentrations of
[HClO2] = 0.454 -x = 0.454 -0.0654 = 0.3886 M
[ClO2- ] = x = 0.0654
Thus, the equilibrium concentrations is 0.0654.
To learn more about concentration, refer to the link:
brainly.com/question/16645766
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Answer:
Explanation:
To write the chemical symbol for an atom or cation with 22 electrons, we must understand some few things about an atom.
An atom is made up of positively charged paticles called protons which are located in the nucleus. The neutrons do not have any charges and are they reside together in the nucleus with the protons.
Electrons are negatively charged particles which orbits the nucleus.
Titanium in its neutral state is the 22nd element on the periodic table and it has 22 electrons.
Vanadium is the the 23rd element on the periodic table. If it loses an electron, it becomes positively charged and it is a cation.
Chromium is the 24th element on the periodic table. A loss of two electrons makes the net electrons remaining on it 22.
The symbols are:
₂₂Ti
₂₃V¹⁺
₂₄Cr²⁺
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