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docker41 [41]
2 years ago
5

Write a balanced half-reaction for the product that forms at each electrode in the aqueous electrolysis of the following salts:(

a) FeI₂;
Chemistry
1 answer:
Semmy [17]2 years ago
6 0

The balanced half-reaction for the product that forms at anode is Fe⁺² + 2e⁻ → Fe(s) and 2H₂O + 2e⁻ → H₂ + 2OH⁻, the product that forms at cathode is 2I⁻  → I₂ + 2e- and 2H₂O  → O₂ + 4H⁺ + 4e⁻  

<h3>What is Balanced Chemical Equation ?</h3>

The equation during which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation is called balanced chemical equation.

Now write the equation for FeI₂

At cathode:

Fe⁺² + 2e⁻ → Fe(s) Eo = - 0.44 V

2H₂O + 2e⁻ → H₂ + 2OH⁻ Eo = - 0.827 V

It is easy to decrease Fe⁺² ions than the water, the product which is formed at cathode is Iron.

At anode:

2I⁻  → I₂ + 2e-    Eo = - 0.54 V

2H₂O  → O₂ + 4H⁺ + 4e⁻    Eo = -1.23 V

O₂ gas formed at anode.

Thus from the above conclusion we can say that The balanced half-reaction for the product at anode is Fe⁺² + 2e⁻ → Fe(s) and 2H₂O + 2e⁻ → H₂ + 2OH⁻, the product that forms at cathode is 2I⁻  → I₂ + 2e- and 2H₂O  → O₂ + 4H⁺ + 4e⁻.

Learn more about the Balanced chemical equation here: brainly.com/question/26694427

#SPJ4

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Show dipole dipole force of water and ethyll alcohol.​
Alexeev081 [22]

Answer:

Water: H2O --> [H+] + [OH-]

Ethyl alcohol: C2H5OH --> [C2H5O-] + [H+]

Explanation:

Dipole-dipole force is observed when an ionic compound ionizes (forms ions), revealing its slightly positive ion and slightly negatively ion.

The chemical formula of Water is H2O.

H2O --> [H+] + [OH-]

The chemical formula of ethyl alcohol is C2H5OH.

C2H5OH --> [C2H5O-] + [H+]

Note the positive and negative parts of both water and ethyl alcohol as expressed in the equations above.

6 0
3 years ago
How many grams are there in 3.34 mol HCl?
just olya [345]

Answer:

121.78 g

Explanation:

You multiply the mols by the grams in one mol.

4 0
3 years ago
A soluble iodide was dissolved in water. Then an excess of silver nitrate, AgNO3, was added to precipitate all of the io- dide i
Sergeeva-Olga [200]

Answer:

m_{I^-}=1.18gI^-

Explanation:

Hello,

In this case, the reaction is given as:

I^-+AgNO_3\rightarrow AgI+NO_3^-

Thus, starting by the yielded grams of silver iodide, we obtain:

n_{I^-}=2.185gAgI*\frac{1molAgI}{234.77gAgI}*\frac{1molI}{1molAgI}=9.31x10^{-3}molI^-

Which correspond to the iodide grams in the silver iodide. In such a way, by means of the law of the conservation of mass, it is known that the grams of each atom MUST remain constant before and after the chemical reaction whereas the moles do not, therefore, the mass of iodine from the silver iodide will equal the mass of iodine present in the soluble iodide, thereby:

m_{I^-}=9.31x10^{-3}molI^-\frac{127gI^-}{1molI^-} =1.18gI^-

And the rest, correspond to the iodide's metallic cation which is unknown. Such value has sense since it is lower than the initial mass of the soluble iodide which is 1.454g, so 0.272 grams correspond to the unknown cation.

Best regards.

7 0
4 years ago
The mixing of which pair of reactants will result in a precipitation reaction csi(aq) + naoh(aq) hcl+ca(oh)2
Fudgin [204]
<span>Pb(NO3)2(aq) + K2SO4(aq) → PbSO4(s) + 2 KNO3(aq) is </span>
7 0
3 years ago
What is the theoretical yield of vanadium, in moles, that can be produced by the reaction of 1.0 mole of V2O5 with 4.0 mole of c
yuradex [85]

Answer:

Theoretical moles of V are 1.6 moles

Explanation:

The theoretical yield of a reaction is defined as the amount of product you would make if all of the limiting reactant was converted into product.

In the reaction:

V2O5(s) + 5Ca(i) → 2V(i) + 5CaO(s)

Based on the reaction, 1 mol of V2O5 needs 5 moles of Ca for a complete reaction. As there are just 4 moles, <em>limiting reactant is Ca.</em> As there are produced 2 moles of V per 5mol of Ca, Theoretical moles of V are:

4 moles of Ca × (2mol V / 5Ca) = <em>1.6 moles of V</em>

<em></em>

I hope it helps!

5 0
3 years ago
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