Question

Balance the following equation by ion electron method.
$FeC{l_3} + {H_2}S \to FeC{l_2} + HCl + S$

Answer 1

For this, we first need to have some knowledge about the ion electron method. The ion-electron method (also known as the half-reaction method) divides the redox equation into two halves, one for oxidation and the other for reduction. The balanced redox equation is created by balancing each of these half-reactions independently and then combining them. Ion electron method is important in balancing chemical equations.

Complete answer:
This method includes the following steps:
Fill in the blanks with the skeleton equation and the oxidation number of each element above their symbol.
Determine which species have been oxidised and which have been reduced.
Divide the skeleton equation into two half-reactions, one for oxidation and one for reduction.
Separately balance the two half-reactions.
The total number of electrons acquired in the reduction half-reaction is equal to the total number of electrons lost in the oxidation half-reaction when the two half-reactions are multiplied by suitable integers.
Verification.
$FeC{l_3} + {H_2}S \to FeC{l_2} + HCl + S$
$Ionicequation:F{e_3} + {H_2}S \to F{e_3} + {H^ + } + S$
Step I: Splitting the redox reactions into two half reactions:
${H_2}S \to 2{H^ + } + S$ (Oxidation half reaction)
$F{e^{3 + }} + e \to F{e^{2 + }}$ (Reduction half reaction)
Step II: Adding of electrons to the side deficient in electrons:
${H_2}S \to 2{H^ + } + S + 2e$
$F{e^{3 + }} + e \to F{e^{2 + }}$
Step III: Balancing the electrons in both half reactions:
${H_2}S \to 2{H^ + } + S + 2e$
$2F{e^{3 + }} + 2e \to 2F{e^{2 + }}$
Step 4: Adding both the half reactions:
${H_2}S + 2F{e^{3 + }} \to 2{H^ + } + S + 2F{e^{3 + }}$
$2F{e^{3 + }} + 2e \to 2F{e^{2 + }}$
And now finally converting into its molecular form, we have,
${{\text{H}}_{\text{2}}}{\text{S + 2F}}{{\text{e}}^{3 + }} + 6{\text{C}}{{\text{l}}^{\text{ - }}} \to {\text{2}}{{\text{H}}^{\text{ + }}}{\text{ + 2C}}{{\text{l}}^ - } + {\text{S + 2F}}{{\text{e}}^{{\text{2 + }}}} + {\text{4C}}{{\text{l}}^{\text{ - }}}$
Or ${H_2}S + 2FeC{l_3} \to 2HCl + S + 2FeC{l_2}$

Note:
We must be noted that in chemical reactions, a balanced chemical equation precisely specifies the quantities of reactants and products. In a typical chemical reaction, the Law of Conservation of Mass states that no mass is created or destroyed. This indicates that both halves of a chemical equation must have the same number of atoms of each element. The sum of the charges on one side of the equation must equal the sum of the charges on the other side of the equation.