Question

Use the galvanic cell reaction to answer the question.

2Cr(s) + 3Cu2+(aq) → 2Cr3+(aq) + 3Cu(s)

Which half reaction occurs at the cathode? (1 point) plzz helpp

3Cu2+(aq) + 6e– → 3Cu(s)

2Cr(s) → 2Cr3+(aq) + 3e–

3Cu2+(aq) + 2e– → 3Cu(s)

2Cr(s) → 2Cr3+(aq) + 6e–

Answer 1

Answer: 3Cu2+(aq) + 6e– → 3Cu(s)

Explanation:

I took the test:)

Answer 2

The galvanic cell reaction is

2Cr(s) + 3Cu⁺²(aq) → 2Cr⁺³(aq) + 3Cu(s)  then the half-reaction occurs at the cathode is - 3Cu⁺²(aq) + 2e⁻ → 3Cu(s)

A galvanic cell reaction is a redox reaction that takes place in two half-reactions oxidation and reduction. At anode oxidation reaction and at cathode reduction reaction takes place.

to identify the oxidized and reduced atoms we must assign oxidation states to all atoms.

reactant side:

For copper, it is in its ionic form: Cu⁺²  

For chromium, it is in its atomic form: Cr⁰

Product side:

For copper, it is in its atomic form = Cu⁰

For chromium, it is in its ionic form = Cr⁺³

• Each copper atom is reduced because its oxidation state decreases from +2 in the simple cation to zero in the metal.
• It means copper undergoes reduction.

Thus, the Reduction half-reaction at cathode would be -

3Cu⁺²(aq) + 2e⁻ → 3Cu(s)

Learn more:

https://brainly.com/question/24257688