Question

How many mole of electrons are required for the reduction of 1 mole of Cr³⁺ to Cr_(s)?

• A. 1
• B. 6.022 x 10²³
• C. 3
• D. 6

Answer 1

The Correct Option is C

Concept:
The reduction of metal ions to their elemental form is a fundamental concept in electrochemistry. In this case, we are considering the reduction of chromium ions (Cr³⁺) to solid chromium metal (Cr_(s)).

Reduction Half-Reaction:
The balanced half-reaction for this process is:
Cr³⁺_(aq) + 3e⁻ \(\rightarrow\) Cr_(s)

This equation tells us that for every mole of Cr³⁺ ions reduced, 3 moles of electrons are required. Electrons act as reducing agents by supplying the negative charge needed to neutralize the positive charge on the Cr³⁺ ion, converting it to its metallic (zero oxidation state) form.

Explanation:
• Chromium in the +3 oxidation state (Cr³⁺) gains 3 electrons.
• This electron gain reduces Cr³⁺ to Cr_(s), which is the metallic form of chromium.
• The number of electrons added must match the charge of the ion being reduced to balance both mass and charge.

Conclusion:
Since the reduction of one mole of Cr³⁺ requires exactly 3 moles of electrons, the correct and balanced stoichiometric value is 3.

Therefore, the correct answer is: 3.