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sergey [27]
3 years ago
15

What is the total number of moles (n) of electrons exchanged between the oxidizing agent and the reducing agent in the overall r

edox equation: 5 Ag (aq) +Mn2 (aq) +4H20()5Ag(s)+ MnO4 (aq) +8H' (aq)
a. 1
b. 2
c. 3
d. 5 (
e.?
Chemistry
1 answer:
Sergeu [11.5K]3 years ago
4 0
Answer:
             5 moles of electrons

Explanation:

The balance equation is as follow,

<span>                   5 Ag</span>⁺ + Mn⁺²<span> + 4 H</span>₂O    →<span>    5 Ag + MnO</span>₄⁻<span> + 8 H</span>⁺

Reduction of Ag:

                         Ag⁺ + 1 e⁻    →    Ag
Or,
                         5 Ag⁺  +  5 e⁻    →    5 Ag

Oxidation of Mn:

                          Mn⁺²   →    MnO₄⁻  +  5 e⁻

Result:
          Hence 5 moles of Ag⁺ accepts 5 electrons from 1 mole of Mn⁺².
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Read 2 more answers
Equal molar quantities of Ca2 and EDTA (H4Y) are added to make a 0.010 M solution of CaY2- at pH 10. The formation constant for
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Answer:

the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

Explanation:

Given the data in the question;

Ca^{2+ + y^{4- ⇄  CaY^{2-

Formation constant Kf

Kf = CaY^{2- / ( [Ca^{2+][y^{4-] ) = 5.0 × 10¹⁰

Now,

[y^{4-] = \alpha _4CH_4Y; ∝₄ = 0.35

so the equilibrium is;

Ca^{2+ + H_4Y ⇄  CaY^{2- + 4H⁺

Given that; CH_4Y = Ca^{2+     { 1 mol Ca^{2+  reacts with 1 mol H_4Y  }

so at equilibrium, CH_4Y = Ca^{2+ = x

∴

Ca^{2+ + y^{4- ⇄  CaY^{2-

x        + x         0.010-x

since Kf is high, them x will be small so, 0.010-x is approximately 0.010

so;

Kf = CaY^{2- / ( [Ca^{2+][y^{4-] ) =  CaY^{2- / ( [Ca^{2+][\alpha _4CH_4Y] )  = 5.0 × 10¹⁰

⇒ CaY^{2- / ( [Ca^{2+][\alpha _4CH_4Y] )  = 5.0 × 10¹⁰

⇒ 0.010 / ( [x][ 0.35 × x] )  = 5.0 × 10¹⁰

⇒ 0.010 / 0.35x²  = 5.0 × 10¹⁰

⇒ x² = 0.010 / ( 0.35 × 5.0 × 10¹⁰ )

⇒ x² = 0.010 / 1.75 × 10¹⁰

⇒ x² = 0.010 / 1.75 × 10¹⁰

⇒ x² = 5.7142857 × 10⁻¹³

⇒ x = √(5.7142857 × 10⁻¹³)

⇒ x = 7.559 × 10⁻⁷

Therefore, the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

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