"How does concentration affect cell potential" explanation

$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$

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Yinhan_Liu_1D
Posts: 51
Joined: Sat Sep 24, 2016 3:00 am

"How does concentration affect cell potential" explanation

Hi, all,

We mentioned La Chatelier's Principle today.

I understand that adding Mn2+ would result in the RXN going to the right (more Al3+ and Mn will be produced). And I kinda understand that adding more "positive charges" to the right means that there will be a higher difference in E.

But how does this change the Ecell exactly? This equation appears as though they are reacting with each other but since Mn2+ and Al are not in direct contact, how will more Al be oxidized( which means more e- will be moving from the left to the right) ? How does the salt bridge change in order to help more redox RXN in this process?

Can you please explain this by step for me in detail?

I would appreciate your help; )

abram_wassily_1G
Posts: 10
Joined: Wed Sep 21, 2016 2:56 pm

Re: "How does concentration affect cell potential" explanation

In the reaction you are referring to, Mn2+ is being reduced to Mn and Al is being oxidized to Al3+. So instead of thinking that you are adding more “positive charges,” understand that Mn2+ is gaining electrons and Al is donating electrons.

Also, using Le Chatelier’s principle, an increase in concentration of Mn2+ causes equilibrium to shift as the reaction goes toward the products, producing a higher positive voltage.

As the reaction shifts toward the right, E(cathode) increases which, therefore, increases Ecell since Ecell= E(cathode) - E(anode).

Return to “Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)”

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