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Triss [41]
3 years ago
12

The purpose of the salt bridge in an electrochemical cell is to ________. Question 10 options: provide oxygen to facilitate oxid

ation at the anode maintain electrical neutrality in the half-cells via migration of ions provide a means for electrons to travel from the anode to the cathode provide a means for electrons to travel from the cathode to the anode provide a source of ions to react at the anode and cathode
Chemistry
1 answer:
svetlana [45]3 years ago
8 0

Answer:The purpose of the salt bridge in an electrochemical cell is to <u>maintain electrical neutrality in the half-cells via migration of ions </u>

Explanation:

A salt bridge is a u-shaped tube or filter paper bridge type  laboratory device used to connect the oxidation and reduction half-cells of a galvanic/voltaic cell. it maintains electrical neutrality in the half cells thereby  preventing the cell from having imbalance of charges in the flow of electrons moving from one half cell to another.

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A solution of NaF is added dropwise to a solution that is 0.0144 M in Ba 2 . When the concentration of F - exceeds __________ M,
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Answer:

When [F⁻] exceeds 0.0109M concentration, BaF₂ will precipitate

Explanation:

Ksp of BaF₂ is:

BaF₂(s) ⇄ Ba²⁺(aq) + 2F⁻(aq)

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The solution will produce BaF₂(s) -precipitate- just when [Ba²⁺] [F⁻]² > 1.7x10⁻⁶.

As the concentration of [Ba²⁺] is 0.0144M, the product [Ba²⁺] [F⁻]² will be equal to  ksp just when:

1.7x10⁻⁶ = [Ba²⁺] [F⁻]²

1.7x10⁻⁶ = [0.0144M] [F⁻]²

1.18x10⁻⁴ = [F⁻]²

0.0109M = [F⁻]

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Answer:

a. 1.7 × 10⁻⁴ mol·L⁻¹; b. 5.5 × 10⁻⁹ mol·L⁻¹

c. 2.3 × 10⁻⁴ mol·L⁻¹;    5.5 × 10⁻⁸ mol·L⁻¹

Explanation:

a. Silver iodate

Let s = the molar solubility.  

                     AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq); Ksp = 3.0 × 10⁻⁸

E/mol·L⁻¹:                               s               s

K_{sp} =\text{[Ag$^{+}$][IO$_{3}$$^{-}$]} = s\times s =  s^{2} = 3.0\times 10^{-8}\\s = \sqrt{3.0\times 10^{-8}} \text{ mol/L} = 1.7 \times 10^{-4} \text{ mol/L}

b. Barium sulfate

                     BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq); Ksp = 1.1 × 10⁻¹⁰

I/mol·L⁻¹:                                0.02             0

C/mol·L⁻¹:                                 +s              +s

E/mol·L⁻¹:                            0.02 + s          s

K_{sp} =\text{[Ba$^{2+}$][SO$_{4}$$^{2-}$]} = (0.02 + s) \times s \approx  0.02s = 1.1\times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{0.02} \text{ mol/L} = 5.5 \times 10^{-9} \text{ mol/L}

c. Using ionic strength and activities

(i) Calculate the ionic strength of 0.02 mol·L⁻¹ Ba(NO₃)₂

The formula for ionic strength is  

\mu = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\\mu = \dfrac{1}{2} (\text{[Ba$^{2+}$]}\cdot (2+)^{2} + \text{[NO$_{3}$$^{-}$]}\times(-1)^{2}) = \dfrac{1}{2} (\text{0.02}\times 4 + \text{0.04}\times1)= \dfrac{1}{2} (0.08 + 0.04)\\\\= \dfrac{1}{2} \times0.12 = 0.06

(ii) Silver iodate

a. Calculate the activity coefficients of the ions

\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(1)^{2}\sqrt{0.06} = -0.51\times 0.24 = -0.12\\\gamma = 10^{-0.12} = 0.75

b. Calculate the solubility

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq)

K_{sp} =\text{[Ag$^{+}$]$\gamma_{Ag^{+}}$[IO$_{3}$$^{-}$]$\gamma_{IO_{3}^{-}}$} = s\times0.75\times s \times 0.75 =0.56s^{2}= 3.0 \times 10^{-8}\\s^{2} = \dfrac{3.0 \times 10^{-8}}{0.56} = 5.3 \times 10^{-8}\\\\s =2.3 \times 10^{-4}\text{ mol/L}

(iii) Barium sulfate

a. Calculate the activity coefficients of the ions

\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(2)^{2}\sqrt{0.06} = -0.51\times16\times 0.24 = -0.50\\\gamma = 10^{-0.50} = 0.32

b. Calculate the solubility

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq

K_{sp} =\text{[Ba$^{2+}$]$\gamma_{ Ba^{2+}}$[SO$_{4}$$^{2-}$]$\gamma_{ SO_{4}^{2-}}$} = (0.02 + s) \times 0.32\times s\times 0.32 \approx  0.02\times0.10s\\2.0\times 10^{-3}s = 1.1 \times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{2.0 \times 10^{-3}} \text{ mol/L} = 5.5 \times 10^{-8} \text{ mol/L}

7 0
3 years ago
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