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Molodets [167]
2 years ago
10

Are any of these equations redox reactions? if so what are their reducing and oxidizing agents?

Chemistry
1 answer:
MaRussiya [10]2 years ago
8 0

Yes, Fe(s)+CuSO4→FeSO4(aq)+Cu(s) is a redox reaction. The reducing agent is Fe(s), while the oxidizing agent is CuSO4.

<h3>Oxidation-reduction reaction (redox)</h3>

Oxidation-reduction reaction is the type of reaction in which electrons are transferred from one element to another. A compound or an element either gains are looses an electron in redox reactions.

Fe(s)+CuSO4→FeSO4(aq)+Cu(s)

From the equation above, Fe(s) is the reducing agent because it gives away 2 electrons and undergoes oxidation reaction.

From the equation above, CuSO4 is an oxidizing agent because it gives oxygen to Fe(s) and undergoes a reduction reaction.

Therefore, the reducing agent is Fe(s), while the oxidizing agent is CuSO4.

Learn more about redox reactions here:

brainly.com/question/26263317

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Find the standard enthalpy of formation of ethylene, C2H4(g), given the following data: C2H4(g) + 3 O2(g) --&gt; 2CO2(g) + 2 H2O
Brums [2.3K]

Answer : The standard enthalpy of formation of ethylene is, 52.4 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of C_2H_4 will be,

2C(s)+2H_2(g)\rightarrow C_2H_4(g)    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

(1) C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)     \Delta H_1=-1411kJ

(2) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.5kJ

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.8kJ

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equation, we get :

(1) 2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)     \Delta H_1=+1411kJ

(2) 2C(s)+2O_2(g)\rightarrow 2CO_2(g)    \Delta H_2=2\times (-393.5kJ)=-787kJ

(3) 2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=2\times (-285.8kJ)=-571.6kJ

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(+1411kJ)+(-787kJ)+(-571.6kJ)

\Delta H=52.4kJ

Therefore, the standard enthalpy of formation of ethylene is, 52.4 kJ

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