Redox Reactions in Transition Metals

Transition metals can exist in various oxidation states, and redox reactions can change these oxidation states. The species that loses electrons is oxidized and acts as a reducing agent, whereas the species that gains electrons is reduced and acts as an oxidizing agent.

We can use the electrochemical series to write out the redox reaction. Let's use the example of reacting Mg with \(\text{Cr}_2\text{O}_7^{2-}/\text{H}^+\) Here's how:

  1. List out the half-cell equations:

    \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\), E=+1.33
    \(\text{Mg}^{2+} + 2e^- \rightarrow \text{Mg}\), E=-2.38

    Inverting the second half-equation,
    \(\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^-\), E=+2.38
  2. Identify the reducing and oxidising agent: Mg has a more positive E value (+2.38 vs +1.33). Thus, it is the reducing agent and \(\text{Cr}_2\text{O}_7^{2-}\) is the oxidising agent. Mg releases electrons and \(\text{Cr}_2\text{O}_7^{2-}\) accepts them.
  3. Balancing out the electrons: We know that one Mg atom will release 2 electrons and one \(\text{Cr}_2\text{O}_7^{2-}\) will accept 6 electrons. Therefore, three Mg atoms are needed to release 6 electrons. 

    \(\text{3Mg} \rightarrow \text{3Mg}^{2+} + 6e^-\)
  4. Cancelling out: We put both half-equations side by side and cancel out electrons, H+ and H2O. 

    \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\)
    \(\text{3Mg} \rightarrow \text{3Mg}^{2+} + 6e^-\)

    \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 3\text{Mg} \rightarrow 3\text{Mg}^{2+} + 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\)
  5. Calculate the voltage of the cell: We add up the E values of both half-equations. 

    +2.38 + (+1.33) = +3.71V
Download PDF

© Copyright 2025 Lecturemate - All Rights Reserved