Given the standard electrode potentials, K+/K=−2.93V,Ag+/Ag=0.80V,Hg2+/Hg=0.79V,Mg2+/Mg=−2.37V,Cr3+/Cr=−0.74V
Arrange these metals in their increasing order of reducing power.
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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=−2.93V,Ag+/Ag=0.80V,Hg2+/Hg=0.79V,Mg2+/Mg=−2.37V,Cr3+/Cr=−0.74V
Arrange these metals in their increasing order of reducing power.
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Q2
Given the standard reduction potentials, E∘K+/K=−2.93V,E∘Ag+/Ag=+0.80V,E∘Hg2+/Hg=0.79V E∘Mg2+/Mg=−2.37V,E∘Cr3+/Cr=−0.74V. 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.
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Q4
Given the standard electrode potentials,
K+K=−2.93V,
Ag+Ag=0.80V
Hg2+Hg=0.79V,
Mg2+Mg=−2.37V
Cr3+Cr=−0.74V
Arrange these metals in their increasing order of reducing power.
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Q5
Arrange Ag,Cr and Hg metals in the increasing order of reducing power. Given: E∘Ag+/Ag=+0.80V E∘Cr3/Cr=−0.74V E∘Hg3+/Hg=+0.79V