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Question

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

Solution
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The increasing order of reducing power is as follows:

Ag+|Ag<Hg2+|Hg<Cr3+|Cr<Mg2+|Mg<K+|K
Metal with most positive standard electrode potential has the least reducing power.
Metal with most negative standard electrode potential has maximum reducing power.

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Similar Questions
Q1
Given the standard electrode potentials, K+/K=2.93 V,Ag+/Ag=0.80 V, Hg2+/Hg=0.79 V, Mg2+/Mg=2.37V,Cr3+/Cr=0.74 V
Arrange these metals in their increasing order of reducing power.
View Solution
Q2
Given the standard reduction potentials,
EK+/K=2.93 V,EAg+/Ag=+0.80 V,EHg2+/Hg=0.79 V
EMg2+/Mg=2.37 V,ECr3+/Cr=0.74 V.
The correct increasing order of reducing power is:
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Q3
Given are the standard electrode potentials of few half-cells. The correct order of these metals in increasing reducing power will be:
K+/K=2.93V,Ag+/Ag=0.80V,
Mg2+/Mg=2.37V,Cr3+/Cr=0.74V.
View Solution
Q4

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.

View Solution
Q5
Arrange Ag,Cr and Hg metals in the increasing order of reducing power. Given:
EAg+/Ag=+0.80 V
ECr3/Cr=0.74 V
EHg3+/Hg=+0.79 V
View Solution