Consider the reaction $M_{(aq)}^{n+} + n e^{-} \to M_{(s)}$. The standard reduction potential values of the elements $M_1$,$M_2$,and $M_3$ are $-0.34 \ V$,$-3.05 \ V$,and $-1.66 \ V$ respectively. The order of their reducing power will be

  • A
    $M_1 > M_2 > M_3$
  • B
    $M_3 > M_2 > M_1$
  • C
    $M_1 > M_3 > M_2$
  • D
    $M_2 > M_3 > M_1$

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Similar Questions

The electrode potentials for $Cu^{2+}_{(aq)} + e^- \rightarrow Cu^{+}_{(aq)}$ and $Cu^{+}_{(aq)} + e^- \rightarrow Cu_{(s)}$ are $+0.15 \ V$ and $+0.50 \ V$ respectively. The value of $E^o_{Cu^{2+}/Cu}$ will be $........ \ V$.

The standard electrode potentials are: $K^{+}/K = -2.93 \ V$,$Ag^{+}/Ag = 0.80 \ V$,$Hg^{2+}/Hg = 0.79 \ V$,$Mg^{2+}/Mg = -2.37 \ V$,$Cr^{3+}/Cr = -0.74 \ V$. Arrange these metals in the increasing order of their reducing power.

The electrode potentials are given as follows:
$Fe_{(aq)}^{3+} + e^- \to Fe_{(aq)}^{2+}$; $E^o = 0.771 \, V$
$I_{2(s)} + 2e^- \to 2I_{(aq)}^-$; $E^o = 0.536 \, V$
For the cell reaction $2Fe_{(aq)}^{3+} + 2I_{(aq)}^- \to 2Fe_{(aq)}^{2+} + I_{2(s)}$,the value of $E^o_{cell}$ is:

The standard electrode potentials $(E^{0}_{Red})$ of four elements $A, B, C,$ and $D$ are $-3.05 \ V, -1.66 \ V, -0.40 \ V,$ and $0.80 \ V$ respectively. Which of the following is the most reactive?

If the reduction potential is more,then

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