Com relaçao a pilha mg/mg2+//cu2+/cu,indique

Chemistry

1 answer:

meriva3 years ago

6 0

Answer:

A: Cu; B: Mg; C: Mg + Cu²⁺ ⟶ Mg²⁺ + Cu

Explanation:

The left-hand side of the cell diagram is where oxidation occurs (the anode).

Mg ⟶ Mg²⁺ + 2e⁻

So, Mg is the anode.

The right-hand side of the cell diagram is where reduction occurs (the cathode)

Cu²⁺ + 2e⁻ ⟶ Cu

As the copper ions hit the copper cathode, they remove electrons from the metal and become copper atoms. Since the metal has lost electrons, Cu is the positive electrode.

To get the overall reaction, we add the two half-reactions,

Mg ⟶ Mg²⁺ + 2e⁻

Mg + Cu²⁺ ⟶ Mg²⁺ + Cu

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A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi

Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of $Na$ = 5.0 g

Mass of $Cl_2$ = 10.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

First we have to calculate the moles of $Na$ and $Cl_2$ .

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$