Given the standard reduction potentials- E-K- K - -2-93 V- E-Ag- Ag - - 0-80 V- E-Hg2- Hg - 0-79 V E-Mg2- Mg - -2-37 V- E-Cr3- Cr - -0-74 V-The correct increasing order of reducing power is-

The correct option is A K > Mg > Cr > Hg > Ag Reducing power of ions is α E^∘ reduction.So, in the question we can see that, E^∘_K^+/K l ...

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Electrochemical Series

Given the sta...

Question

Given the standard reduction potentials,
$E_{K^{+} / K}^{\circ} = - 2.93 V, E_{A g^{+} / A g}^{\circ} = + 0.80 V, E_{H g^{2 +} / H g}^{\circ} = 0.79 V$
$E_{M g^{2 +} / M g}^{\circ} = - 2.37 V, E_{C r^{3 +} / C r}^{\circ} = - 0.74 V$ .
The correct increasing order of reducing power is:

K > M g > C r > H g > A g

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A g > H g > C r > M g > K

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M g > K > C r > H g > A g

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C r > H g > K > M g > A g

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Solution

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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

Given the standard electrode potentials,

$\frac{K^{+}}{K} = - 2.93 V,$

$\frac{A g^{+}}{A g} = 0.80 V$

$\frac{H g^{2 +}}{H g} = 0.79 V,$

$\frac{M g^{2 +}}{M g} = - 2.37 V$

$\frac{C r^{3 +}}{C r} = - 0.74 V$

Arrange these metals in their increasing order of reducing power.

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

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