Question

Write the net balanced ionic equation for the Fe metal with HCl in the absence of air.
A.$2Fe + 6{H^ + } \to 2F{e^{3 + }} + 3{H_2}$
B.$Fe + 2{H^ + } \to 2F{e^{2 + }} + {H_2}$
C.$Fe + 2{H^ + } \to 2F{e^{3 + }} + {H_2}$
D.None of these

Answer 1

Hint:We know that for writing the net ionic equation for the unbalanced reaction, first we need to be able to identify strong electrolytes, weak electrolytes, and insoluble compounds. Strong electrolytes dissociate entirely into their ions in water. Weak electrolytes yield very few ions in solution, so they are represented by their molecular formula.

Complete answer:
Given in the question is Fe reacts with HCl in the absence of air.
First, we need to be able to identify strong electrolytes, weak electrolytes, and insoluble compounds. Secondly, we need to separate the net ionic equation into the two half-reactions. This means we identify and separate the reaction into an oxidation half-reaction and a reduction half-reaction. For one of the half-reactions, balance the atoms except for O and H. We want the same number of atoms of each element on each side of the equation. Now we have to repeat this with the other half-reaction. We add ${H_2}O$ to balance the O atoms. Add ${H^ + }$ to balance the H atoms. Then, we balance charge i.e.,
add electrons to one side of each half-reaction to balance charge. Lastly, we add the two half-reactions together.
The net ionic equation is as shown:
$Fe + 2{H^ + } \to 2F{e^{2 + }} + {H_2}$

Therefore, the correct answer is option (B).

Note:
Rust is formed from a complex oxidation–reduction reaction involving dilute acid solutions that contain $C{l^ - }$ ions (they are effectively dilute HCl), iron metal, and oxygen. When an object rusts, iron metal reacts with HCl (aq) to produce iron(II) chloride and hydrogen gas.