In a reaction, hydrogen gas and aqueous lead ions were formed when metal lead wa

Question

In a reaction, hydrogen gas and aqueous lead ions were formed when metal lead was added to aqueous hydrochloric acid. What is the overall balanced redox equation for this reaction? (NOTE: Ensure that all reactant and products are in their correct states i.e. Solid, Liquid, Gas or Aqueous states. Also, only consider the H+ ions from the hydrochloric acid. Do not worry about the Cl-!)

Pb(s) + 2H+(aq) -> Pb2+(aq) + H2(g)

?Pb(aq) + H+(aq) -> 2Pb2+(aq) + 4H2(s)

?Pb(l) + 2H+(s) -> Pb2+(aq) + H2(s)

Pb(s) + H+(aq) -> Pb2+(aq) + 2H2(g)

In the following balanced redox reaction equation, which species is the oxidising agent and which is the reducing agent for the forward reaction? (NOTE: When selecting your answer, ensure that BOTH the oxidising and reducing agents have been correctly assigned)

F2(aq) + 2Br-(aq) -> Br2(aq) + 2F-(aq)

Oxidising Agent = Br-(aq),?Reducing Agent = F2(aq)

Oxidising Agent = F2(aq),?Reducing Agent = F-(aq)

Oxidising Agent = Br2(aq),?Reducing Agent = Br-(aq)

Oxidising Agent = F2(aq),?Reducing Agent = Br-(aq)

Consider the following general reaction equation for the reactivity of the Group I metals (M) with water:

2M(s) + 2H2O(l) -> H2(g) + 2MOH(aq)

When the experiment is performed using some Group I metals, you observe that Rubidium (Rb(s)) reacts far more vigorously with water than Sodium (Na(s)). What type of redox process are the two alkali metals undergoing and why is Rb(s) more reactive with water than Na(s)?

The two Group I metals are undergoing reduction. The reason why Rb(s)?is more reactive than Na(s)?is because the potential for Group I metals to undergo reduction increases as you go down the periodic table.

Rb(s) is undergoing oxidation when it reacts with water. Na(s) is undergoing reduction when it reacts with water. This is why Rb(s) is more reactive than Na(s).

The two Group I metals are undergoing oxidation. The reason why Rb(s) is more reactive than Na(s) is because the potential for Group I metals to undergo oxidation increases as you go down the periodic table.

This is not a redox reaction and therefore, neither of the two Group I metals are undergoing oxidation or reduction. Rb(s) is just more reactive than Na(s) because it is a shinier metal.?

Pb(s) + 2H+(aq) -> Pb2+(aq) + H2(g)

?Pb(aq) + H+(aq) -> 2Pb2+(aq) + 4H2(s)

?Pb(l) + 2H+(s) -> Pb2+(aq) + H2(s)

Pb(s) + H+(aq) -> Pb2+(aq) + 2H2(g)

Explanation / Answer

Pb(s) + 2H+(aq) ---> Pb2+(aq) + H2(g)

4Pb(s) + 8H+(aq) ---> 4Pb2+(aq) + 4H2(g)

Pb(s) + 2H+(s) -> Pb2+(aq) + H2(g)

2Pb(s) + 4H+(aq) -> 2Pb2+(aq) + 2H2(g)

F2(aq) + 2Br-(aq) -> Br2(aq) + 2F-(aq)

Oxidising Agent : which involves in reduction easily.

Reducing Agent = which involves in oxidation easily.

answer :

Oxidising Agent = F2(aq),Reducing Agent = F-(aq)

2M(s) + 2H2O(l) ----> H2(g) + 2MOH(aq)

answer : The two Group I metals are undergoing oxidation. The reason why Rb(s) is more reactive than Na(s) is because the potential for Group I metals to undergo oxidation increases as you go down the periodic table.

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