1. A voltaic cell made of a Cr electrode in a solution of 1.0 M in CR 3+ and a g

Question

1. A voltaic cell made of a Cr electrode in a solution of 1.0 M in CR3+ and a gold electrode in a solution that is 1.0 M in Au3+.

.Provide the half cell reactions for each of the electrodes.

.Which is the anode and which is the cathode? How did you decide?

.Give an overall balanced equation for the cell.

.Calculate the E0cellat 25C.

We havent gone over this in lecture yet but are suposed to understand how to do it in lab... I am completely lost. If you could explain it to me that would be awesome! Thanks!

Explanation / Answer

Half cell reaction for Anode:

Cr -----> Cr3+ + 3e-

For Cathode:

Au3+ + 3e- ----> Au

Cr electrode is the Anode and gold electrode is the Cathode. We decide this with the help of E0

E0 Au3+/ Au = 1.50 V and E0 Cr3+/Cr = -0.74.

E0 reduction for Au3+ > E0 reduction for Cr3+. Hence Au3+ gets reduced and Cr gets oxidezed.

Overall Reaction:

Cr + Au3+ ------------> Cr3+ + Au

Overall E0 = E0 (cathode) - E0 (anode)

=> Overall E0 = 1.5 - (-0.74) = 2.24 V

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