Question

The standard reduction potential of some electrodes are given as:

$E_{Mg^{2+}|Mg}^o = -2.37 \ V$ $E_{Zn^{2+}|Zn}^o = -0.76 \ V$ $E_{Fe^{2+}|Fe}^o = -0.44 \ V$

Which of the following is correct?

• A. Zn oxidises Fe
• B. Zn will reduce $Fe^{2+}$
• C. Zn will reduce $Mg^{2+}$
• D. Mg oxidizes Fe

Answer 1

Answer: (B)

Zn will reduce $Fe^{2+}$

Answer 2

• Understanding Standard Reduction Potentials:

  • A more negative (or less positive) standard reduction potential ($E^o$) indicates a stronger reducing agent (the species itself is more easily oxidized).
  • A metal with a more negative standard reduction potential can reduce the ions of another metal with a less negative (or more positive) standard reduction potential.
  • Conversely, an ion with a more positive standard reduction potential is a stronger oxidizing agent (the species itself is more easily reduced).

• Given Standard Reduction Potentials:

  • $E_{Mg^{2+}|Mg}^o = -2.37 \ V$
  • $E_{Zn^{2+}|Zn}^o = -0.76 \ V$
  • $E_{Fe^{2+}|Fe}^o = -0.44 \ V$

• Ordering of Reducing Strength (Metals):

  The more negative the $E^o$, the stronger the reducing agent. Therefore, the order of reducing strength is: Mg > Zn > Fe. This means:

  • Magnesium (Mg) can reduce both $Zn^{2+}$ and $Fe^{2+}$ ions.
  • Zinc (Zn) can reduce $Fe^{2+}$ ions but cannot reduce $Mg^{2+}$ ions.
  • Iron (Fe) cannot reduce $Zn^{2+}$ or $Mg^{2+}$ ions.

• Analyzing the Options:

  A. Zn oxidises Fe

  • "Zn oxidises Fe" means Zn (metal) acts as an oxidizing agent, and Fe (metal) acts as a reducing agent.
  • Metals generally act as reducing agents (they get oxidized). Zn metal is a reducing agent.
  • If $Zn^{2+}$ oxidizes Fe, the reaction would be $Fe(s) + Zn^{2+}(aq) \rightarrow Fe^{2+}(aq) + Zn(s)$.
  • For this reaction, $E_{cell}^o = E_{Zn^{2+}|Zn}^o - E_{Fe^{2+}|Fe}^o = -0.76 \ V - (-0.44 \ V) = -0.32 \ V$.
  • Since $E_{cell}^o < 0$, this reaction is non-spontaneous. So, $Zn^{2+}$ cannot oxidize Fe.
  • Therefore, option A is incorrect.

  B. Zn will reduce $Fe^{2+}$

  • "Zn will reduce $Fe^{2+}$" means Zn (metal) acts as a reducing agent, and $Fe^{2+}$ acts as an oxidizing agent.
  • The reaction is: $Zn(s) + Fe^{2+}(aq) \rightarrow Zn^{2+}(aq) + Fe(s)$
  • For this reaction, $E_{cell}^o = E_{Fe^{2+}|Fe}^o - E_{Zn^{2+}|Zn}^o = -0.44 \ V - (-0.76 \ V) = +0.32 \ V$. (Alternatively, $E_{cell}^o = E_{oxidation}^o (Zn \rightarrow Zn^{2+}) + E_{reduction}^o (Fe^{2+} \rightarrow Fe) = (+0.76 \ V) + (-0.44 \ V) = +0.32 \ V$).
  • Since $E_{cell}^o > 0$, this reaction is spontaneous.
  • Therefore, Zn will reduce $Fe^{2+}$. Option B is correct.

  C. Zn will reduce $Mg^{2+}$

  • "Zn will reduce $Mg^{2+}$" means Zn (metal) acts as a reducing agent, and $Mg^{2+}$ acts as an oxidizing agent.
  • The reaction is: $Zn(s) + Mg^{2+}(aq) \rightarrow Zn^{2+}(aq) + Mg(s)$
  • For this reaction, $E_{cell}^o = E_{Mg^{2+}|Mg}^o - E_{Zn^{2+}|Zn}^o = -2.37 \ V - (-0.76 \ V) = -1.61 \ V$.
  • Since $E_{cell}^o < 0$, this reaction is non-spontaneous.
  • Therefore, Zn will not reduce $Mg^{2+}$. Option C is incorrect.

  D. Mg oxidizes Fe

  • "Mg oxidizes Fe" means Mg (metal) acts as an oxidizing agent, and Fe (metal) acts as a reducing agent.
  • Mg metal is a very strong reducing agent, not an oxidizing agent.
  • If $Mg^{2+}$ oxidizes Fe, the reaction would be $Fe(s) + Mg^{2+}(aq) \rightarrow Fe^{2+}(aq) + Mg(s)$.
  • For this reaction, $E_{cell}^o = E_{Mg^{2+}|Mg}^o - E_{Fe^{2+}|Fe}^o = -2.37 \ V - (-0.44 \ V) = -1.93 \ V$.
  • Since $E_{cell}^o < 0$, this reaction is non-spontaneous. So, $Mg^{2+}$ cannot oxidize Fe.
  • Therefore, option D is incorrect.