Question
Standard electrode potentials are used for various processes:
  • It is used to measure relative strengths of various oxidants and reductants.
  • It is used to calculate standard cell potential.
  • It is used to predict possible reactions.
A set of half-reactions (in acidic medium) along with their standard reduction potential, $E^\circ ($in volt$)$ values are given below:
$I_2 + 2e^- \rightarrow 2I^- ; E^\circ = 0.54 V$
$Cl_2 + 2e^- \rightarrow 2Cl^- ; E^\circ = 1.36 V$
$Mn^{3+} +e^- \rightarrow Mn^{2+}; E^\circ = 1.50 V$
$Fe^{3+} + e^- \rightarrow Fe^{2+}; E^\circ = 0.77 V$
$O_2 + 4H^+ + 4e^- \rightarrow 2H_2O ; E^\circ = 1.23 V$
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Which of the following statements is correct?
  1. $Cl^-$ is oxidised by $O_2.$
  2. $Fe^{2+}$ is oxidised by iodine.
  3. $I^-$ is oxidised by chlorine.
  4. $Mn^{2+}$ is oxidised by chlorine.
  1. $Mn^{3+}$ is not stable in acidic medium, while $Fe^{3+}$ is stable because:
  1. $O_2$ oxidises $Mn^{2+}$ to $Mn^{3+}$
  2. $O_2$ oxidises both $Mn^{2+}$ to $Mn^{3+}$ and $Fe^{2+}$ to $Fe^{3+}$
  3. $Fe^{3+}$ oxidises $H_2O$ to $O_2$
  4. $Mn^{3+}$ oxidises $H_2O$ to $O_2$
  1. The strongest reducing agent in the aqueous solution is:
  1. $I^-$
  2. $Cl^-$
  3. $Mn^{2+}$
  4. $Fe^{2+}$
  1. The emf for the following reaction is:
$\text{I}_2+\text{KCl}\rightleftharpoons2\text{KI}+\text{Cl}_2$
  1. $-0.82\ V$
  2. $+0.82\ V$
  3. $-0.73\ V$
  4. $+0.73\ V$
  1. Which of the following statements is correct for the following reaction?
$Fe^{3+} + Mn^{2+} \rightarrow Fe^{2+}+ Mn^{3+}$
  1. The emf of the cell is positive.
  2. $Fe^{3+}$ oxidises $Mn^{2+}.$
  3. The reaction does not occur.
  4. All are correct.

 

Answer

  1. (c) $I^-$ is oxidised by chlorine.
Explanation:

The half cell having the higher reduction potential will undergo reduction process.
  1. (d) $Mn^{3+}$ oxidises $H_2O$ to $O_2$
Explanation:

Electrode potential of $Mn^{3+}$ is higher than $O_2.​​​​​​​$
  1. (a) $I^-$
Explanation:
Due to least electrode potential value.
  1. (a) $-0.82\ V$
Explanation:
Half reactions:
$I_2 2e^- \rightarrow 2I^-​​​​​​​$
$2CI^- \rightarrow CI_2 + 2e^-$​​​​​​​
$\text{Reduction}\text{ E}^\circ= 0.54\text{V}\\\text{Oxidation}\text{ E}^\circ=-1.36\text{V}\\\overline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\\\ \text{e.m.f}=-0.82\text{V}$
  1. (c) The reaction does not occur.
Explanation:
$Fe^{3+} + Mn^{2+} \rightarrow Mn^{3+} + Fe^{2+}$​​​​​​​
Two half reactions:
$Fe^{3+}+ e^- \rightarrow Fe^{2+}$
$Mn^{2+}\rightarrow Mn^{3+} + e^-​​​​​​​$​​​​​​​
$\text{Reduction}\text{ E}^\circ= 0.77\text{V}\\\text{Oxidation}\text{ E}^\circ=-1.50\text{V}\\\overline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\\\ \text{e.m.f}=-0.73\text{V}$
Since, emf is negative, the reaction does not occur i.e., $Fe^{3+}$ does not oxidise $Mn^{2+}.$

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