## 7th Edition #6.O.3

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

Max Hayama 4K
Posts: 63
Joined: Fri Sep 28, 2018 12:16 am

### 7th Edition #6.O.3

Aqueous solutions of (a)Mn+2, (b)Al+3, (c)Ni+2, (d)Au+3 with concentrations of 1M are electrolyzed at pH=7. For each solution, determine whether the metal ion or water will be reduced at the cathode.

The solutions manual says that the species with the most positive E value will be reduced at the cathode. But because this is an electrolytic cell, shouldn't it be the opposite since a charge is being applied to force the less favorable reaction?

For example (a), Mn+2 has an E value is -1.18V, while water's E value is -.42. The solution manual indicates that water will be reduced, but shouldn't Mn+2 be reduced since it will create a negeative E cell value (unfavorable), and will require a charge to force this unfavorable reaction to proceed?

MaanasO 1A
Posts: 72
Joined: Fri Sep 28, 2018 12:26 am

### Re: 7th Edition #6.O.3

Nah because when you add the voltages, you flip the sign for the reaction that is oxidized. So for your example, you will have -0.42 + 1.18 > 0, so that means the overall reaction will be spontaneous.

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