Oxidant half reaction:
2H+ + 2e- --> H2
Reductant half reaction:
Fe --> Fe2+ + 2e-
Tribune ions:
2Cl- --> 2Cl-
I will assume iron II. Fe + 2HCl -> FeCl2 + H2
Fe(s) + 2HCl (aq) --> fecl2(aq) + H2 (g)
Fe + 2HCl = FeCl2 + H2 , so hydrogen gas will be made.
FeOH3CL Fe(OH)2 + 2 HCl => FeCl2 + 2 H2O
Assuming the iron forms iron (II) (and iron ion with two positive charges), it will form iron chloride and hydrogen gas. Fe (s) + 2HCl (aq) ---> FeCl2 (aq) + H2 (g) Iron = Fe Hydrochloric acid = HCl Chlorine = Cl Hydrogen = H (aq) = aqueous - in water (s) = solid (g) = gas
Fe + 2HCl --> FeCl2 + H2
fe + 2HCL --- FeCl2 + H2 Iron :)
I will assume iron II. Fe + 2HCl -> FeCl2 + H2
Very fast corrosion e.g Fe + 2HCl -----> FeCl2 + H2
Fe(s) + 2HCl (aq) --> fecl2(aq) + H2 (g)
Fe + 2HCl = FeCl2 + H2 , so hydrogen gas will be made.
FeOH3CL Fe(OH)2 + 2 HCl => FeCl2 + 2 H2O
Assuming the iron forms iron (II) (and iron ion with two positive charges), it will form iron chloride and hydrogen gas. Fe (s) + 2HCl (aq) ---> FeCl2 (aq) + H2 (g) Iron = Fe Hydrochloric acid = HCl Chlorine = Cl Hydrogen = H (aq) = aqueous - in water (s) = solid (g) = gas
The oxidation half-reaction is: Fe => Fe+3 + 3e-, and the reduction half-reaction is: F2 + 2e- => 2 F-1. For a complete equation, the oxidation half-reaction as written must be multiplied by 2 and added to the reduction half-reaction as written multiplied by 3 to result in an overall reaction of 2 Fe + 3 F2 = 2 FeF3.
Idealy. Fe + 2HCl --> FeCl2 + H2 You get iron II chloride and hydrogen gas.
The full equation is: Fe(s) + 2HCl(aq) -> FeCl2(s) + H2(g) The Net Ionic Equation is: Fe2+(s) + Cl1-(aq) -> FeCl2(s) (made necessary corrections, however this is not the net Ionic equation....) Hope that helps!
Since Fe is a transition metal, the equation varies. If Fe has the oxidation state of +2: Fe + Cl2 -> FeCl2 If Fe has the oxidation state of +3: 2 Fe + 3 Cl2 -> 2 FeCl3