Answer:
The correct answer is : 6 electrons
Explanation:
From the reduction-oxidation reaction, we identify the half-reactions:
Oxidation reaction: Cr (oxidation number= 0) loses 3 electrons and forms the cation Cr³⁺.
Cr → Cr³⁺ + 3 e-
Reduction reaction: the cation Ni²⁺ gains 2 electrons to form Ni (oxidation number= 0).
Ni²⁺ + 2 e- → Ni
We write the two half-reactions together and we multiply the reduction reaction by 3 and the oxidation reaction by 2 to obtain the same quantity of electrons in each half-reaction: 6.
(Ni²⁺ + 2 e- → Ni) x 3
(Cr → Cr³⁺ + 3 e-) x 2
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3 Ni²⁺ + 2 Cr → 3 Ni + 2 Cr³⁺ (balanced reaction)
Therefore, in the balanced reaction a total of 6 electrons are transferred.
