2Cr(s) + 3Cu2+(aq) → 2Cr3+(aq) + 3Cu(s)
Which half reaction occurs at the cathode?
(1 point)
A. 2Cr(s) → 2Cr3+(aq) + 3e–
B. 3Cu2+(aq) + 2e– → 3Cu(s)
C. 2Cr(s) → 2Cr3+(aq) + 6e–
D. 3Cu2+(aq) + 6e– → 3Cu(s)
D. 3Cu2+(aq) + 6e– → 3Cu(s)
The half reaction that occurs at the cathode is the reduction half reaction, where electrons are gained.
In the given reaction: 2Cr(s) + 3Cu2+(aq) → 2Cr3+(aq) + 3Cu(s)
The reduction half reaction is: 3Cu2+(aq) + 2e– → 3Cu(s)
Therefore, the correct answer is B. 3Cu2+(aq) + 2e– → 3Cu(s)
To determine which half reaction occurs at the cathode, we need to understand the cathode's role in the overall redox reaction.
The cathode is where reduction occurs, which means that electrons are gained. In other words, the species being reduced accepts electrons and undergoes a reduction reaction.
Let's examine the given options:
A. 2Cr(s) → 2Cr3+(aq) + 3e–
B. 3Cu2+(aq) + 2e– → 3Cu(s)
C. 2Cr(s) → 2Cr3+(aq) + 6e–
D. 3Cu2+(aq) + 6e– → 3Cu(s)
In the reaction, chromium (Cr) is being oxidized, which means it is losing electrons and undergoing an oxidation reaction. Option A and Option C involve the oxidation of chromium (Cr), so they cannot be the answer.
On the other hand, copper (Cu) is being reduced, which means it is gaining electrons and undergoing a reduction reaction. Option B and Option D involve the reduction of copper (Cu).
Comparing Option B and Option D, we can see that the only difference is the number of electrons involved. The balanced equation tells us that 3Cu2+(aq) reacts with 6e–, so the reduction half reaction that occurs at the cathode is:
D. 3Cu2+(aq) + 6e– → 3Cu(s)
Therefore, the answer is D.