Given the reduction potentials of Na+, Mg2+, Al3+ and Ag+ as E°Na+ /Na =-2.71 V ; E°Mg2+ / Mg=-2.37 V E°Al3+/Al=-1.66V ; E°Ag+/ Ag=0.08 V The least stable oxide is

Question

Given the reduction potentials of Na+, Mg2+, Al3+ and Ag+ as E°Na+ /Na =-2.71 V ; E°Mg2+ / Mg=-2.37 V E°Al3+/Al=-1.66V ; E°Ag+/ Ag=0.08 V The least stable oxide is (A)  Ag2O  (B)  Al2O3  (C)  MgO  (D)  Na2O

Tardigrade Admin · Accepted Answer

As the value of reduction potential of the metal ion increases, the tendency of metal oxide to get reduce into metal increases
Since, reduction potential of only

A g

is positive among the given, thus,

A g_{2} O

readily gets reduced into

A g

metal. In other words, it can be said that

A g_{2} O

is the least stable oxide among the given

Q. Given the reduction potentials of
$N a^{+}, M g^{2 +}, A l^{3 +}$ and $A g^{+}$ as
$E_{N a^{+} / N a}^{\circ} = - 2.71 V; E_{M g^{2 +} / M g}^{\circ} = - 2.37 V$
$E_{A l^{3 +} / A l}^{\circ} = - 1.66 V; E_{A g^{+} / A g}^{\circ} = 0.08 V$
The least stable oxide is

$A g_{2} O$

$A l_{2} O_{3}$

$M g O$

$N a_{2} O$

Q. Given the reduction potentials of Na+,Mg2+,Al3+ and Ag+ as ENa+/Na∘​=−2.71V;EMg2+/Mg∘​=−2.37V EAl3+/Al∘​=−1.66V;EAg+/Ag∘​=0.08V The least stable oxide is

Ag2​O

Al2​O3​

MgO

Na2​O

Q. Given the reduction potentials of
$N a^{+}, M g^{2 +}, A l^{3 +}$ and $A g^{+}$ as
$E_{N a^{+} / N a}^{\circ} = - 2.71 V; E_{M g^{2 +} / M g}^{\circ} = - 2.37 V$
$E_{A l^{3 +} / A l}^{\circ} = - 1.66 V; E_{A g^{+} / A g}^{\circ} = 0.08 V$
The least stable oxide is

$A g_{2} O$

$A l_{2} O_{3}$

$M g O$

$N a_{2} O$