Given the standard electrode potentials- K - K -2-93 VAg - Ag -0-80 VHg 2-Hg 2- Hg -0-79 V-Mg 2- Mg -2-37 VCr 3- Cr -0-74 VArrange these metals in their increasing order of reducing power-

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Electrode Potential

Given the sta...

Question

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.

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Solution

Metal which has lower reduction potential (more negative E value) has greater reducing capacity because it itself gets oxidised more easily.

$\begin{matrix} M e t a l s R e d u c t i o n, P o t e n t i a l s (v o l t s) \end{matrix} ∣ ∣ ∣ \begin{matrix} A g < H g < C r < M g < K - ------------------ \to R e d u c i n g p o w e r i n c r e a s e s 0.80 > 0.79 > - 0.74 > - 2.37 > - 2.93 - ------------------ \to R e d u c i n g p o w e r i n c r e a s e s \end{matrix}$

The lower the reduction potential, the higher is the reducing power. The given standard
electrode potentials increase in the order of

$\frac{K^{+}}{K} < \frac{M g^{2 +}}{M g} < \frac{C r^{3 +}}{C r} < \frac{H g^{2 +}}{H g} < \frac{A g^{+}}{A g}$
Hence, the reducing power of the given metals increases in the following order:

$A g < H g < C r < M g < K$

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Similar questions

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. Given the standard electrode potentials of some metals.

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

and

C r^{3 +} / C r = - 0.74 V

Arrange these metals in their increasing order of reducing power.

Q. 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:

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

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Electrode Potential

Standard XII Chemistry

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Given the standard electrode potentials, K+K=−2.93 V, Ag+Ag=0.80 V Hg2+Hg=0.79V, Mg2+Mg=−2.37 V Cr3+Cr=−0.74 V Arrange these metals in their increasing order of reducing power.

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.