Will lead (#Pb#) react with #HgSO_4#?

Answer 1

We need to combine two distinct reduction reactions for mercury, so let's look at their standard reduction potentials.

#2"Hg"^(2+)(aq) + 2e^(-) -> "Hg"_2^(2+)(aq)# #" "# #E_"red"^@ = "+0.92 V"#
#"Hg"_2^(2+)(aq) + 2e^(-) -> 2"Hg"(l)# #" "# #E_"red"^@ = "+0.85 V"#
#"Pb"^(2+)(aq) + 2e^(-) -> "Pb"(s)# #" "# #E_"red"^@ = "-0.13 V"#

The first two are nullified to obtain:

#2"Hg"^(2+)(aq) + 2e^(-) -> cancel("Hg"_2^(2+)(aq))# #" "" "# #E_"red"^@ = "+0.92 V"# #cancel("Hg"_2^(2+)(aq)) + 2e^(-) -> 2"Hg"(l)# #" "" "# #E_"red"^@ = "+0.85 V"# #"--------------------------------------------"# #color(green)(2"Hg"^(2+)(aq) + 4e^(-) -> 2"Hg"(l))# #" "" "# #color(green)(E_"red"^@ = "+1.77 V")#
Now, when we add it to the iron reaction, in order for the reaction to work overall, #E_"cell"^@# has to be positive because of the following two equations:
#\mathbf(DeltaG^@ = -nFE_"cell"^@)# (1)
#\mathbf(DeltaG = DeltaG^@ + RTlnQ)# (2)

where

and the electrons must be eliminated.

Now, since in standard conditions (#25^@ "C"#, #"1 bar"#), we assume that all concentrations are #"1 M"#, #Q = K# and #DeltaG^@ = DeltaG#.
Therefore, when #E_"cell"^@ > 0#, #DeltaG^@ < 0# and #DeltaG < 0#, so the reaction would be spontaneous in standard conditions.

In this instance, it's not too tough because mercury has a much higher reduction potential, which makes it easier to reduce and a useful oxidizing agent.

#1/2(2"Hg"^(2+)(aq) + cancel(4e^(-)) -> 2"Hg"(l))# #" "# #1/2E_"red"^@ = 1/2xx"+1.77 V"# #"Pb"(s) -> "Pb"^(2+)(aq) + cancel(2e^(-))# #" "# #E_"oxid"^@ = "+0.13 V"# #"--------------------------------------------"# #color(green)("Hg"^(2+)(aq) + "Pb"(s) -> "Pb"^(2+)(aq) + "Hg"(l))# #" "# #color(green)(E_"cell"^@ = "+1.02 V")#

Consequently, to obtain the complete response, add back the spectator anion:

#color(blue)("HgSO"_4(aq) + "Pb"(s) -> "PbSO"_4(aq) + "Hg"(l))#

Remember that mercury (II) sulfate appears to break down in water, so you would likely need to use a different solvent for this.

Although I anticipate that the reduction potentials will vary, their relativities ought to hold steady (that is, mercury(II) should continue to have a greater reduction potential than iron(II).

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Answer 2

No, lead (Pb) will not react with HgSO₄.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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