Question

What is `Pb(IV) + NiCl_2 ->`?

Answer 1

`"Pb"^(4+)(aq) + 4"Cl"^(-)(aq) -> "PbCl"_2(s) + "Cl"_2(g)`

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This is a "simplified" double-replacement reaction (i.e. we ignore some anion coupled with the `"Pb"^(4+)`), so we initially get---noting that `IV = 4`:

`"Pb"^(4+)(aq) + "NiCl"_2(aq) -> "PbCl"_4(s) + "Ni"^(2+)(aq)`

The `"NiCl"_2(aq)` is rather deliquescent (i.e. hygroscopic), and sops up surrounding moisture to form an aqueous solution.

`"PbCl"_4(s)` is going to be written as a solid for now, but it decomposes in aqueous solution into `"PbCl"_2(s)` and `"Cl"_2(g)`. `"Ni"^(2+)` is evidently going to form, but to balance the charge, we must rebalance the mass.

`"Pb"^(4+)(aq) + 2"NiCl"_2(aq) -> "PbCl"_4(s) + 2"Ni"^(2+)(aq)`

And now, we incorporate the decomposition.

`"Pb"^(4+)(aq) + 2"NiCl"_2(aq) -> cancel("PbCl"_4(s)) + 2"Ni"^(2+)(aq)`

`cancel("PbCl"_4(s)) stackrel("H"_2"O"" ")(->) "PbCl"_2(s) + "Cl"_2(g)`

`"-----------------------------------------------------------------"`

`"Pb"^(4+)(aq) + 2"NiCl"_2(aq) -> "PbCl"_2(s) + 2"Ni"^(2+)(aq) + "Cl"_2(g)`

We expect the nickel(II) chloride to dissociate, and thus we are not done:

`"Pb"^(4+)(aq) + cancel(2"Ni"^(2+)(aq)) + 4"Cl"^(-)(aq) -> "PbCl"_2(s) + cancel(2"Ni"^(2+)(aq)) + "Cl"_2(g)`

This gives us:

`color(blue)("Pb"^(4+)(aq) + 4"Cl"^(-)(aq) -> "PbCl"_2(s) + "Cl"_2(g))`