Homework 13.35- Calculating Standard Cell Potential

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Melissa Trieu 1K
Posts: 6
Joined: Fri Sep 26, 2014 2:02 pm

Homework 13.35- Calculating Standard Cell Potential

Postby Melissa Trieu 1K » Fri Feb 06, 2015 9:12 pm

It says to determine the equilibrium constant

Mn(s) + Ti2+(aq) <> Mn2+(aq) + Ti(s)

The answer book says that Ti2+(aq) + 2e- > Ti(s) is the cathode E = -1.63V and
Mn2+(aq) + 2e- > Mn(s) is the anode E = -1.18V

Why is is like this?

Kayla Denton 1A
Posts: 106
Joined: Fri Sep 26, 2014 2:02 pm

Re: Homework 13.35

Postby Kayla Denton 1A » Fri Feb 06, 2015 10:39 pm

From the balanced equation, Mn goes from an oxidation state of 0 to an oxidation state of +2 so it loses electrons (oxidation). Therefore, it must be the anode. Ti goes from an ox state of +2 to 0 so it gains electrons (reduction); it must be the cathode.
When you're calculating Ecell, you ALWAYS look at the E values of the reduction reactions given in the appendix. You don't switch the sign! So that's why they're both written in the reduction form.

You'd do Ecell = E(species that reduces) - E(species that oxidizes), where the E values are those reduction values from the appendix.


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