$MX$ is a sparingly soluble salt that follows the given solubility equilibrium at $298 \ K$: $MX_{(s)} \rightleftharpoons M^{+}_{(aq)} + X^{-}_{(aq)}$; $K_{sp} = 10^{-10}$. If the standard reduction potential for $M^{+}_{(aq)} + e^- \rightarrow M_{(s)}$ is $(E^{\ominus}_{M^{+}/M}) = 0.79 \ V$, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\ominus}_{X^{-}/MX_{(s)}/M}$ is . . . . . . $mV$. (nearest integer) [Given: $\frac{2.303 RT}{F} = 0.059 \ V$]

  • A
    $200$
  • B
    $790$
  • C
    $590$
  • D
    $1380$

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Similar Questions

The $EMF$ of a cell whose half-cell reactions are given below is .......... $V$.
$Mg^{2+} + 2e^- \to Mg_{(s)}; E^o = -2.37 \ V$
$Cu^{2+} + 2e^- \to Cu_{(s)}; E^o = +0.34 \ V$ (in $V$)

Which of the following transformations can be carried out by using $HI$ as a reducing agent,under acidic conditions?
[Given: $I_{2(s)} \rightarrow 2I^- ; E^{\circ} = 0.54 \ V$]
$(i)$ $Cu^+ \rightarrow Cu_{(s)} ; E^{\circ} = 0.52 \ V$
$(ii)$ $Cr^{3+} \rightarrow Cr^{2+} ; E^{\circ} = -0.41 \ V$
$(iii)$ $Fe^{3+} \rightarrow Fe^{2+} ; E^{\circ} = 0.77 \ V$
$(iv)$ $Fe^{2+} \rightarrow Fe_{(s)} ; E^{\circ} = -0.44 \ V$

Given the standard electrode potentials: $E^o_{Cr^{3+}/Cr} = -0.74 \ V$,$E^o_{MnO_4^-/Mn^{2+}} = 1.51 \ V$,$E^o_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \ V$,and $E^o_{Cl_2/Cl^-} = 1.36 \ V$. Based on this information,which is the strongest oxidizing agent?

What is the decreasing order of deposition of metal on electrode if standard reduction potentials are given as: $Ag^{+} | Ag = 0.80 \ V$,$Cu^{2+} | Cu = 0.337 \ V$,$Sn^{2+} | Sn = -0.136 \ V$,$Cd^{2+} | Cd = -0.403 \ V$?

The $E^{\circ}$ values of half-cells are given below. Which combination of two half-cells will result in a cell with the maximum potential?
$(i) \, A^{3-} \rightarrow A^{2-} + e^{-}; E^{\circ} = 1.5 \, V$
$(ii) \, B^{+} + e^{-} \rightarrow B; E^{\circ} = 0.5 \, V$
$(iii) \, C^{2+} + e^{-} \rightarrow C^{+}; E^{\circ} = 0.5 \, V$
$(iv) \, D \rightarrow D^{2+} + 2e^{-}; E^{\circ} = -1.15 \, V$

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