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Vikentia [17]
3 years ago
12

What is the final, balanced equation that is formed by combining these two half reactions?

Chemistry
1 answer:
OLga [1]3 years ago
4 0

Answer:

Cu + 2NO3- + 4H+ ---------> Cu2+ + 2NO2 + 2H20

Explanation:

The half reactions are;Cu ----------> Cu2+ + 2e-

NO3 + 2e + 2H* —------> N02 + H20

It is worthy to note that oxidation is the an increase in oxidation number while reduction is a decrease in oxidation number.

In the first half reaction involving Copper, copper has been oxidized as its oxidation number increased from 0 to +2.

IN the reduction half reaction, NO3- has been reduced as its oxidation number decreased from +5 to +4.

In balancing half reactions, the following must be observed;

1. split the reactions into half equations, one for oxidation and the other for reduction.

Cu ------> Cu2+ + 2e-   ( oxidation)

NO3- + 2e- -----> NO2   (reduction).

2. balance on half with respect to atoms and charges.

3. balance the other half with respect to atoms and charges

4. balance by adding H+ ions in acidic solution

NO3- + e- -------> NO2

(-1 )       (-1)             (0)

NO3- + 2H+ +e- -------> NO2

(-1)       (+2)     (-1)            (0)

5. balance hydrogen by adding water molecules

NO3 - + 2H+ + e - -------->  NO2 + H2O

6. combine both equations

Cu ---> Cu2+ + 2e-                                  *1

NO3- + 2H+ + e- -------> NO2 + H2O      *2

= Cu -----> Cu2+ + 2e-

2NO3- +4H+ + 2e- -------> 2 NO2 + 2H20

Combined equation  = Cu + 2NO3- + 4H+ ---------> Cu2+ + 2NO2 + 2H20

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Explanation:

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6 0
2 years ago
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if a steel spoon were to be plated with silver, state what would be suitable as the anode, cathode, electrolyte​
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Answer:

Therefore, the amount of heat produced by the reaction of 42.8 g S = <u>(-5.2965 × 10²) kJ = (-5.2965 × 10⁵) J</u>

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Given reaction: 2S + 3O₂ → 2 SO₃

Given: The enthalpy of reaction: ΔH = - 792 kJ

Given mass of S: w₂ = 42.8 g, Molar mass of S: m = 32 g/mol

In the given reaction, the number of moles of S reacting: n = 2

As, Number of moles: n = \frac{mass\: (w_{1})}{molar\: mass\: (m)}

∴  mass of S in 2 moles of S: w_{1} = n \times m = 2\: mol \times 32\: g/mol = 64\: g

<em>Given reaction</em>: 2S + 3O₂ → 2 SO₃

<em>In this reaction, the limiting reagent is S</em>

⇒ 2 moles S produces (- 792 kJ) heat.

or, 64 g of S produces (- 792 kJ) heat.

∴ 42.8 g of S produces (x) amount of heat

⇒ <u><em>The amount of heat produced by 42.8 g S:</em></u>

x = \frac{(- 792\: kJ) \times 42.8\: g}{64\: g} = (-529.65)\: kJ

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(\because 1 kJ = 10^{3} J)

<u>Therefore, the amount of heat produced by the reaction of 42.8 g S = (-5.2965 × 10²) kJ = (-5.2965 × 10⁵) J</u>

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