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oksano4ka [1.4K]
3 years ago
14

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

e ions in solution were [Cr3+] = 0.75 M and [Fe2+] = 0.25 M at 298K?
Chemistry
1 answer:
zaharov [31]3 years ago
7 0

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

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}}

 

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