Voltage Chemistry problem

[silveryhair]

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For a cell, in which the reaction is:
Mg(s) + Fe2+ --> Mg2+ + Fe(s)


One may increase the voltage obtained by:

A) decreasing the Mg2+
B) decreasing the Fe2+
C) increasing the size of Mg Electrode
D) increasing the side of Fe Electrode

Can you please explain it? Thanks!

 

[txlonghorn]

Is it A? You would just use Le Chatelier's principle.

 

[namiie]

Is the answer A?

This is asking for voltage or cell potential instead of just regular changes in product concentration so you'd want use the Nernst equation for this (change in cell potential with changing concentrations of reactants) instead of regular LeChatelier, I believe.

The Nernst equation is Ecell = Ecell standard - (0.591/n)(log Q) where Q is the reaction quotient. So here, it'd be [Mg2+]/[Fe2+]. If you decrease the [Mg2+], then Q would be smaller, and so log Q would also be smaller. If you have to subtract by a smaller quantity, then you end up with a bigger number (bigger cell potential/voltage) than if you have to subtract by a larger quantity. That's how I'd approach it.

 

[mashinka]

This seems like a good question. What's the answer?
Doesn't increasing the voltage mean that more electrons are traveling from the anode to the cathode.

So I just split it up into half reactions.
Mg----> Mg2+ + 2e. (oxidation-anode)
Fe2+ + 2e ----> Fe (reduction-cathode)

So if we want to increase the production of electrons to increase the voltage, we want to increase the amount of Mg produced. And by looking at the half reaction and just based in the answer choices given, we can deduce by le Chatlier that decreasing Mg2+ shift the equilibrium towards the production of Mg and subsequent production of more electrons.

Can someone confirm this explanation. It seems simplier to me than the one above, but then again it's late and i may be talking out of my ass haha

 

[Cool Beans]

Let's assume that the reaction has reached equilibrium. At equilibrium both Ecell and delta G = 0.
Let's say we now increase the concentration of Fe2+ ions in the solution. This results in a shift to the right according to Le Chatelier's Principle; this means the rxn is now spontaneous to the right with means that Ecell is now positive and delta G is now negative.
We can now make some generalizations:
Anything that shifts the equilibrium to the right increases Ecell (more positive) but decreases delta G (more negative).
Anything that shifts the equilibrium to the left decreases Ecell (more negative) and increases delta G (more positive).

 

[silveryhair]

Yup the Answer is A. Thanks for the help guys!