Given below are the standard electrode potentials of a few half-cells. $K^{+} / K = - 2.93 V, {Ag}^{+} / Ag = 0.80 V, {Mg}^{2 +} / Mg = - 2.37 V, {Cr}^{3 +} / Cr = - 0.74 V$ . What will be the correct order of these metals in increasing reducing power?

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Solution

Redox potential
Redox potential is a measure of tendency of chemical species to get reduced or oxidised by gaining or losing electrons to an electrode.
Reducing power
Reducing power is the potential of a substance to reduce another substance.
More the oxidation potential the element will have more tendency to get oxidised.
Reduction potential of half cells are:
$K^{+} / K = - 2.93 V {Ag}^{+} / Ag = 0.80 V {Mg}^{2 +} / Mg = - 2.37 V {Cr}^{3 +} / Cr = - 0.74 V$
The more easily it removes electrons or higher the reduction potential higher is its reducing power, the more it will reduce other substance.
Order of these metals in increasing reducing powers is $Ag < Cr < Mg < K .$
Therefore, potassium has lowest reduction potential, therefore, it can donate electrons easily and reduces other substance.

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Similar questions

Given the standard electrode potentials,

K⁺/K = −2.93V, Ag⁺/Ag = 0.80V,

Hg²⁺/Hg = 0.79V

Mg²⁺/Mg = −2.37 V, Cr³⁺/Cr = − 0.74V

Arrange these metals in their increasing order of reducing power.

Given the standard electrode potentials,
$K^{+} / K = - 2.93 V, A g^{+} / A g = 0.80 V, H g^{2 +} / H g$
$= 0.79 V$
$M g^{2 +} / M g = - 2.37 V . C r^{3 +} / C r = - 0.74 V$
Arrange these metals in their increasing order of reducing power.

Q. For the reduction of

N O_{3}^{-}

ion in an aqueous solution,

E^{0} i s + 0.96 V .

Values of

E^{0}

for some metal ions are given below

V^{2 +} (a q) + 2 e^{-} \to V E^{0} = - 1.19 V

F e^{3 +} (a q) + 3 e^{-} \to F e E^{0} = - 0.04 V

A u^{3 +} (a q) + 3 e^{-} \to A u E^{0} = + 1.40 V

H g^{2 +} (a q) + 2 e^{-} \to H g E^{0} = + 0.86 V

The pair(s) metals that is(are) oxidized by

N O_{3}^{-}

in aqueous solution is(are):

Q. Given are the standard electrode potentials of few half-cells. The correct order of these metals in increasing reducing power will be:

K^{+} / K = - 2.93 V, A g^{+} / A g = 0.80 V

M g^{2 +} / M g = - 2.37 V, C r^{3 +} / C r = - 0.74 V

Given the standard electrode potentials

K⁺| k = -2.93 V; Ag⁺|Ag = +0.8V; Mg²⁺| Mg=-2.37 V

Arrange these metals in order of increasing reducing power.

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Given below are the standard electrode potentials of a few half-cells. K+/K=-2.93V,Ag+/Ag=0.80V,Mg2+/Mg=–2.37V,Cr3+/Cr=–0.74V. What will be the correct order of these metals in increasing reducing power?

Given below are the standard electrode potentials of a few half-cells. $K^{+} / K = - 2.93 V, {Ag}^{+} / Ag = 0.80 V, {Mg}^{2 +} / Mg = - 2.37 V, {Cr}^{3 +} / Cr = - 0.74 V$ . What will be the correct order of these metals in increasing reducing power?