I don’t know if I’m right??

Chemistry

2 answers:

alexgriva [62]3 years ago

6 0

Yep definitely correct

allochka39001 [22]3 years ago

3 0

Answer:

what is this for? A-level..?!

Explanation:

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What is the name of the following compound? Se5S6

inessss [21]

Answer:

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2 years ago

What would be the voltage (Ecell) of a voltaic cell comprised of Cr (s)/Cr3+(aq) and Fe (s)/Fe2+(aq) if the concentrations of th

zaharov [31]

Answer:

0.35 V

Explanation:

(a) Standard reduction potentials

E°/V

Fe²⁺ + 2e- ⇌ Fe; -0.41

Cr³⁺ + 3e⁻ ⇌ Cr; -0.74

(b) Standard cell potential

E°/V

2Cr³⁺ + 6e⁻ ⇌ 2Cr; +0.74

3Fe ⇌ 3Fe²⁺ + 6e-; -0.41

2Cr³⁺ + 3Fe ⇌ 2Cr + 3Fe²⁺; +0.33

3. Cell potential

2Cr³⁺(0.75 mol·L⁻¹) + 6e⁻ ⇌ 2Cr

3Fe ⇌ 3Fe²⁺(0.25 mol·L⁻¹) + 6e-

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

$E = E^{\circ} - \dfrac{RT}{zF}\ln Q$

(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:

$Q = \dfrac{\text{[Fe}^{2+}]^{3}}{ \text{[Cr}^{3+}]^{2}} = \dfrac{0.25^{3}}{ 0.75^{2}} =\dfrac{0.0156}{0.562} = 0.0278\\\\E = 0.33 - \left (\dfrac{8.314 \times 298}{6 \times 96485}\right ) \ln(0.0278)\\\\=0.33 -0.00428 \times (-3.58) = 0.33 + 0.0153 = \textbf{0.35 V}\\\text{The cell potential is }\large\boxed{\textbf{0.35 V}}$