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3 years ago

What are the half-reactions for a galvanic cell with Zn and Mg electrodes?

Chemistry

1 answer:

Alona [7]3 years ago

7 0

the half-reactions

cathode : Zn²⁺ (aq) + 2e⁻ ---> Zn (s)

anode : Mg (s) → Mg²⁺ (aq) + 2e−

a balanced cell reaction

Zn²⁺(aq) + Mg(s)→ Zn(s) + Mg²⁺ (aq)

<h3 /><h3>Further explanation</h3>

Given

Zn and Mg electrodes

Required

The half-reactions for a galvanic cell

Solution

To determine the reaction of a voltaic cell, we must determine the metal that serves as the anode and the metal that serves as the cathode.

To determine this, we can either know from the standard potential value of the cell or use the voltaic series

1. voltaic series

The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent

So the metal on the left will easily undergo oxidation and function as anode

Since Mg is located to the left of Zn, then Mg functions as anode and Zn as a cathode

2. Standard potentials cell of Mg and Zn metals :

Mg2+ + 2e– → Mg E° = -2,35 V

Zn2+ + 2e– → Zn E° = -0,78 V

The anode has a smaller E°, then Mg is the anode and Zn is the cathode.

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Ammonia (NH3) reacts with oxygen to form nitric oxide (NO) and water vapor: 4NH3 + 502 4NO + 6H2O b) When 20.0 g NH3 and 50.0 g

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Answer: a) . Ammonia is the limiting reagent

b. Oxygen is left over and 0.1375 g of oxygen is left over.

c. The theoretical yield of $NO$ is 35.29 g.

d. The theoretical yield of $H_2O$ is 31.74 g.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For $NH_3$

Given mass of ammonia = 20.0 g

Molar mass of ammonia = 17.031 g/mol

Putting values in equation 1, we get:

$\text{Moles of ammonia}=\frac{20.0g}{17.031g/mol}=1.17mol$

For $O_2$

Given mass of oxygen gas = 50.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{50.0g}{32g/mol}=1.6mol$

The chemical equation for the reaction is

$4NH_3+5O_2\rightarrow 4NO+6H_2O$

By Stoichiometry of the reaction:

4 moles of ammonia reacts with = 5 moles of oxygen

So 1.17 moles of ammonia will react with = $\frac{5}{4}\times 1.17=1.4625mol$ of oxygen

As, given amount of oxygen is more than the required amount. So, it is considered as an excess reagent and (1.6-1.4625)= 0.1375 g of oxygen is left unreacted.

Thus ammonia is considered as a limiting reagent because it limits the formation of product.

1. By Stoichiometry of the reaction:

4 moles of ammonia produces = 4 moles of $NO$

1.17 moles of ammonia will produce = $\frac{4}{4}\times 1.17=1.17moles$ of $NO$