If the standard reduction potentials for four divalent elements $X, Y, Z,$ and $W$ are $-1.46 \ V, -0.36 \ V, 0.15 \ V,$ and $-1.24 \ V$ respectively,then:

  • A
    $X$ will displace $Z^{+2}$ from aqueous solution
  • B
    $Y$ will displace $Z^{+2}$ from aqueous solution
  • C
    $W$ will displace $Z^{+2}$ from aqueous solution
  • D
    All the above statements are correct

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The value of $x$ from the given data is

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 oxidation potentials for the half-reactions are given as $Zn \to Zn^{2+} + 2e^{-}; E^o = +0.76 \ V$ and $Fe \to Fe^{2+} + 2e^{-}; E^o = +0.41 \ V$. The $EMF$ for the cell reaction $Fe^{2+} + Zn \to Zn^{2+} + Fe$ is ............ $V$.

Given the standard electrode potentials,what is the standard electrode potential for the reaction $Fe^{3+}_{(aq)} + e^{-} \rightarrow Fe^{2+}_{(aq)}$ in $V$?
$Fe^{3+}_{(aq)} + 3e^{-} \rightarrow Fe_{(s)}$ ; $E^o = -0.036 \ V$
$Fe^{2+}_{(aq)} + 2e^{-} \rightarrow Fe_{(s)}$ ; $E^o = -0.440 \ V$

For the cell $Zn_{(s)} | Zn^{2+}_{(aq)} || M^{x+}_{(aq)} | M_{(s)}$,different half cells and their standard electrode potentials are given below:
$M^{x+}_{(aq)} / M_{(s)}$$Au^{3+}_{(aq)} / Au_{(s)}$$Ag^{+}_{(aq)} / Ag_{(s)}$$Fe^{3+}_{(aq)} / Fe^{2+}_{(aq)}$$Fe^{2+}_{(aq)} / Fe_{(s)}$
$E^o M^{x+} / M (V)$$1.40$$0.80$$0.77$$-0.44$

If $E^o Zn^{2+}/Zn = -0.76 \ V$,which cathode will give a maximum value of $E^o_{cell}$ per electron transferred?

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