The Student Room Group

Electrode potentials a level chem

Water is added to the beaker containing the magnesium chloride solution. What is the effect on the EMF of the cell?

The mark scheme says the EMF increases but shouldn’t it decrease?

If you add water you lower the conc of Mg2+ ions so the equilibrium should shift to make more Mg2+, simultaneously making more electrons and making the cell more negative. So shouldn’t EMF decrease?

(The Mg half cell is connected to standard hydrogen electrode in the question)
Reply 1
Original post by Aeshakhan
Water is added to the beaker containing the magnesium chloride solution. What is the effect on the EMF of the cell?
The mark scheme says the EMF increases but shouldn’t it decrease?
If you add water you lower the conc of Mg2+ ions so the equilibrium should shift to make more Mg2+, simultaneously making more electrons and making the cell more negative. So shouldn’t EMF decrease?
(The Mg half cell is connected to standard hydrogen electrode in the question)

Not entirely sure, but Mg is a stronger reducing agent (more negative E cell value) than hydrogen

When you add water, you dilute the concentration of Mg2+ so the reaction shifts to the left which produces more electrons,thus giving it a more negative e cell value

You calculate EMF by doing most positive takeaway the most negative. However, since the Mg E cell value is already a negative value, you're essentially adding it ( Just going to use random values here)

Original: 0.00 - (-2.34) = + 2.34
After adding water, E cell value becomes more negative (lets say it goes from -2.34 to -3.05)
After adding water: 0.00 - (-3.05) = +3.05

So whilst the E cell value for magnesium alone decreases, because you're subtracting a negative number, the EMF value ends up increasing.
Reply 2
Please could you link the question in. I've tried searching for it online and the one I've found asks about the effect on the magnitude of the cell...
Reply 3
Original post by alem135
Not entirely sure, but Mg is a stronger reducing agent (more negative E cell value) than hydrogen
When you add water, you dilute the concentration of Mg2+ so the reaction shifts to the left which produces more electrons,thus giving it a more negative e cell value
You calculate EMF by doing most positive takeaway the most negative. However, since the Mg E cell value is already a negative value, you're essentially adding it ( Just going to use random values here)
Original: 0.00 - (-2.34) = + 2.34
After adding water, E cell value becomes more negative (lets say it goes from -2.34 to -3.05)
After adding water: 0.00 - (-3.05) = +3.05
So whilst the E cell value for magnesium alone decreases, because you're subtracting a negative number, the EMF value ends up increasing.

Strictly speaking for calculating (cell) we do (RHS) - (LHS) when using conventional cell representation. The SHE is always on the left so (cell) would be (RHS) - 0.00 = (RHS)
Reply 4
Original post by alem135
Not entirely sure, but Mg is a stronger reducing agent (more negative E cell value) than hydrogen
When you add water, you dilute the concentration of Mg2+ so the reaction shifts to the left which produces more electrons,thus giving it a more negative e cell value
You calculate EMF by doing most positive takeaway the most negative. However, since the Mg E cell value is already a negative value, you're essentially adding it ( Just going to use random values here)
Original: 0.00 - (-2.34) = + 2.34
After adding water, E cell value becomes more negative (lets say it goes from -2.34 to -3.05)
After adding water: 0.00 - (-3.05) = +3.05
So whilst the E cell value for magnesium alone decreases, because you're subtracting a negative number, the EMF value ends up increasing.

Isn’t EMF supposed to be E right - E left ? Does this not apply when we have a standard Hydrogen electrode
Reply 5
Original post by Methene
Strictly speaking for calculating (cell) we do (RHS) - (LHS) when using conventional cell representation. The SHE is always on the left so (cell) would be (RHS) - 0.00 = (RHS)
My point so emf should be more negative then shouldn’t it? As we’re making more e?
Reply 6
Original post by Methene
Please could you link the question in. I've tried searching for it online and the one I've found asks about the effect on the magnitude of the cell...
8667A263-B00A-4755-930B-A09F09997651.jpeg
Reply 7
Original post by Aeshakhan
8667A263-B00A-4755-930B-A09F09997651.jpeg

Note it says magnitude of the EMF. While it does become more negative, the magnitude increases
Reply 8
Original post by Methene
Note it says magnitude of the EMF. While it does become more negative, the magnitude increases

OMDS that’s so cheeky!! Ofc because magnitude is the mod of a value!! Thank you so much I was soo confused
Reply 9
Original post by Aeshakhan
OMDS that’s so cheeky!! Ofc because magnitude is the mod of a value!! Thank you so much I was soo confused

Haha no worries :smile:
Reply 10
Original post by Methene
Strictly speaking for calculating (cell) we do (RHS) - (LHS) when using conventional cell representation. The SHE is always on the left so (cell) would be (RHS) - 0.00 = (RHS)

Oh lmao, i do ocr so we’ve just learnt most positive takeaway most negative

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