A voltaic cell made of a Cr electrode in a solution of 1.0 M in Cr

3+
and a gold electrode in a
solution that is 1.0 M in Au
3+
.
a) Provide the half cell reactions for each of the electrodes.
b) Which is the anode and which is the cathode? How did you decide?
c) Give an overall balanced equation for the cell.
d) Calculate the E
o
cell at 25
o
C.

My old text probably doesn't have the same Eo values as yours. What is the Eo value for

Cr^3+ + 3e ==> Cr(s) in your text and
Au^3+ + 3e ==> Au(s) in your text?
Also what are you having trouble with on this problem?

a) The half cell reaction at the Cr electrode is: Cr(s) -> Cr3+(aq) + 3e-

The half cell reaction at the gold electrode is: Au3+(aq) + 3e- -> Au(s)

b) The anode is the electrode where oxidation occurs, and the cathode is the electrode where reduction occurs. In this case, since Cr(s) is being oxidized to Cr3+(aq) at the Cr electrode, it is the anode. On the other hand, Au3+(aq) is being reduced to Au(s) at the gold electrode, making it the cathode.

c) The overall balanced equation for the cell can be obtained by adding the half cell reactions:
Cr(s) -> Cr3+(aq) + 3e- (anode)
Au3+(aq) + 3e- -> Au(s) (cathode)
---------------------
Cr(s) + Au3+(aq) -> Cr3+(aq) + Au(s)

d) To calculate Eo cell at 25oC, we can use the Nernst equation:
Eo cell = Eo cathode - Eo anode
where Eo cathode and Eo anode are the standard reduction potentials for the cathode and anode half-cell reactions, respectively.

The standard reduction potential for the Cr3+|Cr half cell is -0.74V, and for the Au3+|Au half cell is 1.498V.

Eo cell = 1.498V - (-0.74V)
Eo cell = 2.238V

Therefore, the Eo cell at 25oC is 2.238V.

To answer these questions about the voltaic cell, we need to go step by step. The half cell reactions, anode, cathode, and E°cell can be determined by using the standard electrode potentials for each half reaction and applying the following principles:

a) To write the half cell reactions, we need to know the standard reduction potentials for each half reaction. The half reactions are as follows:

At the Cr electrode:
Cr3+(aq) + 3e- → Cr(s)

At the Au electrode:
Au3+(aq) + 3e- → Au(s)

b) To determine which electrode is the anode and which is the cathode, we look at the standard reduction potentials. The electrode with the higher reduction potential will be the cathode, and the electrode with the lower reduction potential will be the anode.

By looking at the standard reduction potentials, we find that the reduction potential for Au3+ is higher than that of Cr3+. Therefore, the Au electrode is the cathode, and the Cr electrode is the anode.

c) The overall balanced equation for the cell can be obtained by adding the two half cell reactions together and canceling out any common species on both sides of the equation.

Adding the two half reactions, we get:

Cr3+(aq) + Au(s) → Cr(s) + Au3+(aq)

d) To calculate the standard cell potential (E°cell) at 25°C, we use the Nernst equation:

E°cell = E°cathode - E°anode

Using the standard reduction potentials for Au3+ and Cr3+ at 25°C, and plugging the values into the Nernst equation, we can calculate E°cell.

Note: The standard reduction potentials can be found in tables or reference materials.

E°Au3+ = 1.50 V
E°Cr3+ = -0.74 V

E°cell = 1.50 V - (-0.74 V) = 2.24 V