## 6N.1

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

Moderators: Chem_Mod, Chem_Admin

keyaluo4C
Posts: 54
Joined: Fri Sep 28, 2018 12:17 am

### 6N.1

The half reactions for 6N.1 (a) are
Mn2+ + 2e- = Mn (s) E = -1.18 V
Ti2+ + 2e- = Ti (s) E = -1.63 V
Why does the answer key say that Ti is the one being reduced? Shouldn't it be Mn because TI has lower reduction potential?

MariahClark 2F
Posts: 60
Joined: Wed Nov 15, 2017 3:04 am

### Re: 6N.1

Look at the order of the elements in the actual equation that is given to you. The half reactions displayed are backwards in a sense, which is how you can get the cell potential value. Mn is oxidized to become Mn2+ and Ti2+ is reduced to become Ti.

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

### Who is online

Users browsing this forum: No registered users and 1 guest