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A galvanic cell is powered by the following redox reaction: Br2 H2 2OH Answer the following questions about this cell. If you need any electrochemical data, be sure you get it from the ALEKS Data tab. Write a balanced equation for the half-reaction that takes place at the cathode. Write a balanced equation for the half-reaction that takes place at the anode. Calculate the cell voltage under standard conditions. Round your answer to decimal places.

User CBredlow
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Final answer:

In a galvanic cell, half-reactions occur at separate electrodes: reduction at the cathode and oxidation at the anode. The cathode is the positive electrode, while the anode is the negative electrode. Standard cell potential can be calculated but requires the standard reduction potentials for the specific half-reactions.

Step-by-step explanation:

Galvanic Cell Half-Reactions, Electrodes, and Potentials

In a galvanic cell, a spontaneous redox reaction occurs that drives the flow of electrons from the anode to the cathode through an external circuit. The original question does not provide a complete redox reaction, but assuming it involves bromine (Br2), hydrogen (H2), and hydroxide ions (OH-), the following can be considered:

For the provided redox reaction, the half-reaction that occurs at the cathode (reduction) might be:

2H+ + 2e- → H2

And the half-reaction at the anode (oxidation) could be:

2Br- → Br2 + 2e-

The cathode is the positive electrode, as it undergoes reduction, and the anode is the negative electrode, as it undergoes oxidation. Without actual potential values given, the cell voltage under standard conditions cannot be calculated here. However, the standard cell potential can be found by using standard reduction potential tables and subtracting the anode potential from the cathode potential.

User Jegadesh B S
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Answer:

1.91 V

Step-by-step explanation:

The correct cell reaction. equation is;

Br2(l) + H2(g) + 2OH-(aq) -> 2Br-(aq) + 2H2O(l)

Anode half reaction;

H2(g) + 2OH^-(aq) --------> 2H2O(l) +2e- -0.82 V

Cathode half reaction;

Br2(l) + 2e- -----> 2Br^-(aq). +1.09 V

Hence

E° cell= E°cathode - E°anode

E° cell= 1.09 - (-0.82)

E° cell= 1.91 V

User TimeString
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