Answer:
B) Cu2+ + 2e– → Cu.
Step-by-step explanation:
Hello there!
In this case, in the regard of redox reactions, we must keep in mind that half-reactions show us how the flow of electrons when an element undergoes a change in its oxidation state, say increasing or decreasing. In such a way, if the element is oxidized, increase the oxidation state, we add electrons on the right side and the contrary if the element is reduced, decrease the oxidation state.
In such a way, we can see that the reaction B) Cu2+ + 2e– → Cu, shows how the Cu2+ cations are reduced by adding electrons to produce Cu and thus, conserve the charge.
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