Question

Question #be12b

Answer 1

The answer is b) one-third of its atomic mass

Explanation:

You're actually dealing with a redox reaction in which iron is being oxidized and chlorine is being reduced

`2stackrel(color(blue)(0))(Fe_((s))) + 3stackrel(color(blue)(0))(Cl_(2(g))) -> 2stackrel(color(blue)(+3))(Fe) stackrel(color(blue)(-1))(Cl_(3(s)))`

The equivalent mass of a substance that participates in a redox reaction depends on the number of electrons lost or gained by that substance in said redox reaction.

`"eq. mass" = "molar mass"/("no. of electrons lost or gained")`

More specifically, the equivalent mass of iron will be equal to its molar mass divided by the number of electrons it loses (it's being oxidized( in the reaction.

This means that you have to write the oxidation half-reaction and see how many atoms are being lost by iron

`stackrel(color(blue)(0))(Fe_((s))) -> stackrel(color(blue)(+3))(Fe)Cl_(3(s)) + color(red)("x")e^(-)`

Notice that iron's oxidation state goes from 0 on the reactants' side, to +3 on the products' side. This means that each atom of iron will lose 3 electrons, so that the half-reaction becomes

`stackrel(color(blue)(0))(Fe_((s))) -> stackrel(color(blue)(+3))(Fe)Cl_(3(s)) + color(red)(3)e^(-)`

This means that iron's equivalent mass will be equal to

`"eq. mass" = "molar mass"/(color(red)(3)e^(-)) = 1/3 * "molar mass"`

Therefore, iron's equivalent mass will be equal to one-third of its molar mass.

SIDE NOTE You could have use the balanced chemical equation for the half-reaction, with the important distinction that you would have had

`2stackrel(color(blue)(0))(Fe_((s))) -> 2stackrel(color(blue)(+3))(Fe) Cl_(3(s)) + 6e^(-)`

However, you need to get the number of electrons lost per atom, and since 6 electrons are being lost by 2 atoms of iron, the result would have been the same, 3 electrons lost per atom.