For the electrochemical cell 2 Al(s) + 3 Mn²⁺(aq) ⟶ 2 Al³⁺(aq) + 3 Mn(s) (E° = 0.48 V, [Al³⁺] = 1.0 M), which of the following concentrations of Mn²⁺ would cause an increase in E for the cell?

Question

For the electrochemical cell 2 Al(s) + 3 Mn²⁺(aq) ⟶ 2 Al³⁺(aq) + 3 Mn(s) (E° = 0.48 V, [Al³⁺] = 1.0 M), which of the following concentrations of Mn²⁺ would cause an increase in E for the cell?

Answer 1

Hey Tiffany,

I assume that you're familiar with the Nernst equation? A helpful version is as follows:

E = E^o - (0.592/n)*log(Q)

Here, n is just the number of electrons involved, which will stay the same for different concentrations.

Q is the reaction quotient, which always ignores solids, leaving you with Q = [Al³⁺]² / [Mn²⁺]³

Now, recall from math class that the log of 1 is just 0. What that means for you is that when the numerator and the denominator of that Q expression are equal, you will have exactly E^o, or 0.48 V. Since the concentration of aluminum ions is just 1.0M, this will happen when the concentration of magnesium ions is also 1.0M. However, when the numerator is larger, log(Q) will be positive, meaning that your overall voltage is LESS than E^o. The converse is also true: if your denominator (ie: [Mn²⁺]³) is bigger than the numerator, log(Q) will be negative, resulting in an overall voltage that is GREATER than E^o.

So, I don't know what your choices were, but this should be enough information to figure out the correct answer(s).