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What is the reaction of lead with dilute sulphuric acid?
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Omar N.
What is the reaction of lead with dilute sulphuric acid?
In terms of reactivity, lead is very unreactive. Its oxidation potential under standard conditions is…
Pb →Pb^(2+) + 2e^(-) : E° = -0.13V
NB This is very close to the standard reference hydrogen potential…
H2 → 2H^(+) + 2e^(-) : E° = ±0.00V
The greater the difference, the greater the reaction…
So, the reaction works, but only with a weak driving force (Free Energy).
Gibbs Free Energy ΔG° = -zE°F
z = number of electrons involved.
E = Electrode potential (standard value, E° is quoted above).
F = Faraday Constant.
To move in the forward direction the value of ΔG° should be negative.
Lead is a solid, and in a solution of sulfuric acid, the acid surrounds the metal and reacts over the whole surface.
Lead sulfate is produced which is insoluble and forms a tight coating over the surface of the lead.
As the lead sulfate coating thickens, it isolates the metal from the acid so that any reaction slows and eventually stops.
The overall reaction (which is probably all you wanted) is…
Pb + H2SO4 → PbSO4 + H2
The reaction is so weak that if you add a piece of (cleaned) lead to bench sulfuric acid, you would have to watch closely for a some time before noticing anything.
Supplementary: Hydrogen does not readily form on the surface of lead unless there is “extra” energy available. This energy is referred to as an “overvoltage”. In effect this overvoltage has the effect of reducing the measured electrode potential making the reaction even harder than it is on paper. As a result, it can appear that lead is unreactive in acid solutions and in sulphuric acid in particular with the insoluble sulfate coating increasing the polarisation (=resistance) at the electrode surface.
Extension: An interesting experiment for school research might be to connect a lead electrode directly to a platinum electrode (low, virtually zero overpotential) and place both in the same beaker. Bubbles of hydrogen should slowly appear on the platinum. Using nitric acid should remove problems of a precipitate on the lead electrode. Repeat and compare with sulfuric, hydrochloric and ethanoic acids for comparison. Collecting data should allow some conclusions to be drawn. Investigate / predict what would happen in more concentrated and less concentrated solutions of each acid. There are several points that would need to be considered in order to make the test fair and accurate but that’s for you to do.
In terms of reactivity, lead is very unreactive. Its oxidation potential under standard conditions is…
Pb →Pb^(2+) + 2e^(-) : E° = -0.13V
NB This is very close to the standard reference hydrogen potential…
H2 → 2H^(+) + 2e^(-) : E° = ±0.00V
The greater the difference, the greater the reaction…
So, the reaction works, but only with a weak driving force (Free Energy).
Gibbs Free Energy ΔG° = -zE°F
z = number of electrons involved.
E = Electrode potential (standard value, E° is quoted above).
F = Faraday Constant.
To move in the forward direction the value of ΔG° should be negative.
Lead is a solid, and in a solution of sulfuric acid, the acid surrounds the metal and reacts over the whole surface.
Lead sulfate is produced which is insoluble and forms a tight coating over the surface of the lead.
As the lead sulfate coating thickens, it isolates the metal from the acid so that any reaction slows and eventually stops.
The overall reaction (which is probably all you wanted) is…
Pb + H2SO4 → PbSO4 + H2
The reaction is so weak that if you add a piece of (cleaned) lead to bench sulfuric acid, you would have to watch closely for a some time before noticing anything.
Supplementary: Hydrogen does not readily form on the surface of lead unless there is “extra” energy available. This energy is referred to as an “overvoltage”. In effect this overvoltage has the effect of reducing the measured electrode potential making the reaction even harder than it is on paper. As a result, it can appear that lead is unreactive in acid solutions and in sulphuric acid in particular with the insoluble sulfate coating increasing the polarisation (=resistance) at the electrode surface.
Extension: An interesting experiment for school research might be to connect a lead electrode directly to a platinum electrode (low, virtually zero overpotential) and place both in the same beaker. Bubbles of hydrogen should slowly appear on the platinum. Using nitric acid should remove problems of a precipitate on the lead electrode. Repeat and compare with sulfuric, hydrochloric and ethanoic acids for comparison. Collecting data should allow some conclusions to be drawn. Investigate / predict what would happen in more concentrated and less concentrated solutions of each acid. There are several points that would need to be considered in order to make the test fair and accurate but that’s for you to do.
Lead reacts slowly with dilute sulfuric acid to make lead (II) sulfate and hydrogen gas. The driving force has to be the loss of hydrogen (leChatelier) as both lead and lead sulfate are solids and not in solution.
In a lead/acid battery, the lead sulfate coats the surface and prevents most further reaction, so for the reaction to make lead sulfate on purpose, the lead should be in small particles, and the mixture well-stirred, and warmed consistent with safe operation when hydrogen gas is evolved.
At the end, the grey lead (grey: b/c it’s usually coated with oxide that quickly reacts) will be replaced by white lead sulfate salt.
Lead reacts slowly with dilute sulfuric acid to make lead (II) sulfate and hydrogen gas. The driving force has to be the loss of hydrogen (leChatelier) as both lead and lead sulfate are solids and not in solution.
In a lead/acid battery, the lead sulfate coats the surface and prevents most further reaction, so for the reaction to make lead sulfate on purpose, the lead should be in small particles, and the mixture well-stirred, and warmed consistent with safe operation when hydrogen gas is evolved.
At the end, the grey lead (grey: b/c it’s usually coated with oxide that quickly reacts) will be replaced by white lead sulfate salt.
In terms of reactivity, lead is very unreactive. Its oxidation potential under standard conditions is…
Pb →Pb^(2+) + 2e^(-) : E° = -0.13V
NB This is very close to the standard reference hydrogen potential…
H2 → 2H^(+) + 2e^(-) : E° = ±0.00V
The greater the difference, the greater the reaction…
So, the reaction works, but only with a weak driving force (Free Energy).
Gibbs Free Energy ΔG° = -zE°F
z = number of electrons involved.
E = Electrode potential (standard value, E° is quoted above).
F = Faraday Constant.
To move in the forward direction the value of ΔG° should be negative.
Lead is a solid, and in a solution of sulfuric acid, the acid surrounds the metal and reacts over the whole surface.
Lead sulfate is produced which is insoluble and forms a tight coating over the surface of the lead.
As the lead sulfate coating thickens, it isolates the metal from the acid so that any reaction slows and eventually stops.
The overall reaction (which is probably all you wanted) is…
Pb + H2SO4 → PbSO4 + H2
The reaction is so weak that if you add a piece of (cleaned) lead to bench sulfuric acid, you would have to watch closely for a some time before noticing anything.
Supplementary: Hydrogen does not readily form on the surface of lead unless there is “extra” energy available. This energy is referred to as an “overvoltage”. In effect this overvoltage has the effect of reducing the measured electrode potential making the reaction even harder than it is on paper. As a result, it can appear that lead is unreactive in acid solutions and in sulphuric acid in particular with the insoluble sulfate coating increasing the polarisation (=resistance) at the electrode surface.
Extension: An interesting experiment for school research might be to connect a lead electrode directly to a platinum electrode (low, virtually zero overpotential) and place both in the same beaker. Bubbles of hydrogen should slowly appear on the platinum. Using nitric acid should remove problems of a precipitate on the lead electrode. Repeat and compare with sulfuric, hydrochloric and ethanoic acids for comparison. Collecting data should allow some conclusions to be drawn. Investigate / predict what would happen in more concentrated and less concentrated solutions of each acid. There are several points that would need to be considered in order to make the test fair and accurate but that’s for you to do.
In terms of reactivity, lead is very unreactive. Its oxidation potential under standard conditions is…
Pb →Pb^(2+) + 2e^(-) : E° = -0.13V
NB This is very close to the standard reference hydrogen potential…
H2 → 2H^(+) + 2e^(-) : E° = ±0.00V
The greater the difference, the greater the reaction…
So, the reaction works, but only with a weak driving force (Free Energy).
Gibbs Free Energy ΔG° = -zE°F
z = number of electrons involved.
E = Electrode potential (standard value, E° is quoted above).
F = Faraday Constant.
To move in the forward direction the value of ΔG° should be negative.
Lead is a solid, and in a solution of sulfuric acid, the acid surrounds the metal and reacts over the whole surface.
Lead sulfate is produced which is insoluble and forms a tight coating over the surface of the lead.
As the lead sulfate coating thickens, it isolates the metal from the acid so that any reaction slows and eventually stops.
The overall reaction (which is probably all you wanted) is…
Pb + H2SO4 → PbSO4 + H2
The reaction is so weak that if you add a piece of (cleaned) lead to bench sulfuric acid, you would have to watch closely for a some time before noticing anything.
Supplementary: Hydrogen does not readily form on the surface of lead unless there is “extra” energy available. This energy is referred to as an “overvoltage”. In effect this overvoltage has the effect of reducing the measured electrode potential making the reaction even harder than it is on paper. As a result, it can appear that lead is unreactive in acid solutions and in sulphuric acid in particular with the insoluble sulfate coating increasing the polarisation (=resistance) at the electrode surface.
Extension: An interesting experiment for school research might be to connect a lead electrode directly to a platinum electrode (low, virtually zero overpotential) and place both in the same beaker. Bubbles of hydrogen should slowly appear on the platinum. Using nitric acid should remove problems of a precipitate on the lead electrode. Repeat and compare with sulfuric, hydrochloric and ethanoic acids for comparison. Collecting data should allow some conclusions to be drawn. Investigate / predict what would happen in more concentrated and less concentrated solutions of each acid. There are several points that would need to be considered in order to make the test fair and accurate but that’s for you to do.
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Lead reacts slowly with dilute sulfuric acid to make lead (II) sulfate and hydrogen gas. The driving force has to be the loss of hydrogen (leChatelier) as both lead and lead sulfate are solids and not in solution.
In a lead/acid battery, the lead sulfate coats the surface and prevents most further reaction, so for the reaction to make lead sulfate on purpose, the lead should be in small particles, and the mixture well-stirred, and warmed consistent with safe operation when hydrogen gas is evolved.
At the end, the grey lead (grey: b/c it’s usually coated with oxide that quickly reacts) will be replaced by white lead sulfate salt.
Lead reacts slowly with dilute sulfuric acid to make lead (II) sulfate and hydrogen gas. The driving force has to be the loss of hydrogen (leChatelier) as both lead and lead sulfate are solids and not in solution.
In a lead/acid battery, the lead sulfate coats the surface and prevents most further reaction, so for the reaction to make lead sulfate on purpose, the lead should be in small particles, and the mixture well-stirred, and warmed consistent with safe operation when hydrogen gas is evolved.
At the end, the grey lead (grey: b/c it’s usually coated with oxide that quickly reacts) will be replaced by white lead sulfate salt.
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