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Taya2010 [7]
3 years ago
14

13. Write the following balanced reactions using cell notation. Use platinum as an inert electrode, if needed. (a) Mg(s) + Ni2+(

aq) ⟶ Mg2+(aq) + Ni(s) (b) 2Ag+(aq) + Cu(s) ⟶ Cu2+(aq) + 2Ag(s) (c) Mn(s) + Sn(NO3 ) 2 (aq) ⟶ Mn(NO3 ) 2 (aq) + Au(s) (d) 3CuNO3 (aq) + Au(NO3 ) 3 (aq) ⟶ 3Cu(NO3 ) 2 (aq) + Au(s)
Chemistry
1 answer:
sp2606 [1]3 years ago
5 0

Answer:

A) Mg ₍s₎ | Mg²⁺₍aq₎║Ni²⁺₍aq₎ | Ni₍s₎

B)  Cu₍s₎ | Cu²⁺₍aq₎║ Ag⁺  |  Ag⁰₍s₎

C)  Cu₍s₎ | Cu²⁺₍aq₎║ Ag⁺  |  Ag⁰₍s₎

D) Pt₍s₎ | CuNO₃₍ aq₎ , Cu(NO₃)₂₍aq₎ ║ Au(NO₃)₃₍aq₎ | Au₍s₎

Explanation:

A) Mg + Ni²⁺ → Mg²⁺ + Ni

Mg⁰→ Mg²⁺ + 2e⁻    oxidation   anode

Ni²⁺ + 2e⁻ → Ni⁰       reduction   cathode

Mg ₍s₎ | Mg²⁺₍aq₎║Ni²⁺₍aq₎ | Ni₍s₎

anode        salt bridge       cathode

B)   2 Ag⁺₍aq₎ + Cu₍s₎ → Cu²⁺₍aq₎ + 2 Ag₍s₎

2 Ag⁺ + 2e⁻ → 2 Ag⁰   cathode

Cu →  Cu²⁺ + 2e⁻          anode

 Cu₍s₎ | Cu²⁺₍aq₎║ Ag⁺  |  Ag⁰₍s₎

anode       salt bridge       cathode

C) Mn ₍s₎  + Sn(NO₃)₂ ₍aq₎ → Mn (NO₃)₂ ₍aq₎  + Au₍s₎

Mn⁰ → Mn²⁺+ 2e⁻       anode

Sn²⁺ + 2e⁻ → Sn⁰₍s₎      cathode

Mn₍s₎ | Mn(NO₃)₂ ₍aq₎ ║ Sn(NO₃)₂ ₍aq₎ | Sn₍s₎

anode                salt bridge                   cathode

D) 3CuNO₃₍ aq₎ + Au(NO₃)₃₍aq₎  → 3Cu(NO₃)₂₍aq₎  + Au₍s₎

3Cu⁺ → 3Cu²⁺ + 3e⁻  anode

Au³⁺ + 3e⁻ → Au⁰                cathode

Pt₍s₎ | CuNO₃₍ aq₎ , Cu(NO₃)₂₍aq₎ ║ Au(NO₃)₃₍aq₎ | Au₍s₎

       anode                           salt bridge                   cathode

Semi-reactions are noted to determine oxidation and reduction. The oxidation reaction occurs at the anode and the reduction reaction at the cathode. In the case of subsections a, b, c the electrodes may be the solid metal, while in the case d the species are in an accusative medium and require the platinum electrode.

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Answer:

V=43.46mL

Explanation:

Hello!

In this case, since the reaction between sulfuric acid and aluminum hydroxide is:

3H_2SO_4+2Al(OH)_3\rightarrow Al_2(SO_4)_3+6H_2O

Whereas the ratio of sulfuric acid to aluminum hydroxide is 3:2; thus, we first compute the moles of sulfuric acid that complete react with 3.209 g of aluminum hydroxide:

n_{H_2SO_4}=3.209gAl(OH)_3*\frac{1molAl(OH)_3}{78.00gAl(OH)_3} *\frac{3molH_2SO_4}{2molAl(OH)_3} \\\\n_{H_2SO_4}=0.0617molH_2SO_4

Then, given the molarity, it is possible to obtain the milliliters as follows:

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Best regards!

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