Need help on Galvanic Cells

[Frais]

I'm new to this so I'll be asking questions now or then
So there's this multiple-choice question that I'm struggling to understand and i hope to find someone to explain it
Question
When a piece of metal M is placed into a blue solution of QCl₂, the metal develops a dark coating and the colour of the solution becomes less intense. It can be concluded from these observations that:
a) Metal Q is more reactive than Metal M
b) Metal M is reduced more easily than metal Q
c) The Q²⁺ ions are more easily reduced than M²⁺ ions
d) Chloride ions, Cl^- are generally inert (unreactive) during chemical reactions
Thanks

[jakesilove]

I'm new to this so I'll be asking questions now or then
So there's this multiple-choice question that I'm struggling to understand and i hope to find someone to explain it
Question
When a piece of metal M is placed into a blue solution of QCl₂, the metal develops a dark coating and the colour of the solution becomes less intense. It can be concluded from these observations that:
a) Metal Q is more reactive than Metal M
b) Metal M is reduced more easily than metal Q
c) The Q²⁺ ions are more easily reduced than M²⁺ ions
d) Chloride ions, Cl^- are generally inert (unreactive) during chemical reactions
Thanks

Hey! Welcome to the forums!

So, let's think about what's happening here. We've got some sort of metal, M, and a solution, QCl2. The solution is actually ionised, so will contain



and



Now, the solution is getting LESS intense, and a dark colouring is formed on the metal. This suggests that whatever was in the solution (ie. Q) is turning into a solid, and plating itself around the metal. Hopefully it's fairly clear why this is so.

That means that, somehow, Q is turning into a solid. The reaction outlining this would look like



This is because we need Q to go from an ion (with a charge) to a solid metal (without a charge). Thus, we add two electrons. We use the pneumonic OIL RIG (Oxidation It Loses, Reduction It Gains) to note that, as Q has gained electrons, it has REDUCED into a metal.

However, if a reduction equation occurs, there must be a corresponding oxidation equation. This would be M losing electrons, and turning into an ion. Now, if that ion was MORE LIKELY to reduce back into metal than Q, it just would. So, no reaction would occur if Q reduced, creating M ions, but then M just reduced back into metal again. From this, we can deduce that Q ions are more easily reduced than M ions.

Note that a more active metal will replace a less active metal in solution; therefore, M is clearly more reactive than Q, making a) incorrect. Also, M does not reduce, making c) false, and d) cannot be figured out from the description.

[QC]

 

Also, M does not reduce, making c) false

Do you mean b)?