Electrode potentials $(E^o)$ are given below:
$Cu^{+}/Cu = +0.52 \ V$
$Fe^{3+}/Fe^{2+} = +0.77 \ V$
$\frac{1}{2} I_{2(s)}/I^{-} = +0.54 \ V$
$Ag^{+}/Ag = +0.88 \ V$
Based on the above potentials,the strongest oxidizing agent will be:

  • A
    $Cu^{+}$
  • B
    $Fe^{3+}$
  • C
    $Ag^{+}$
  • D
    $I_2$

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Similar Questions

Standard reduction potentials of the half-reactions are given below:
$F_{2(g)} + 2e^- \rightarrow 2F^-_{(aq)}$; $E^o = +2.85 \ V$
$Cl_{2(g)} + 2e^- \rightarrow 2Cl^-_{(aq)}$; $E^o = +1.36 \ V$
$Br_{2(l)} + 2e^- \rightarrow 2Br^-_{(aq)}$; $E^o = +1.06 \ V$
$I_{2(s)} + 2e^- \rightarrow 2I^-_{(aq)}$; $E^o = +0.53 \ V$
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For the reaction $H_2 (1 \, bar) + 2AgCl_{(s)} \rightleftharpoons 2Ag_{(s)} + 2H^{+} (0.1 \, M) + 2Cl^{-} (0.1 \, M)$,$\Delta G^o = -48,250 \, J$ at $25 \, ^oC$. The standard emf of cell in which the given reaction takes place is ................. $V$

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The standard Gibbs energy change for the Daniel cell reaction is $Zn_{(s)} + Cu^{2+}_{(aq)} \longrightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$ where $E_{\text{cell}}^{\circ} = 1.1 \ V$.

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