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Question

Arrange the following metals in increasing order of their reducing power.
[ Given : E0K+/K = -2.93 V , E0Ag+/Ag = +0.80V,
E0Al3+/Al = -1.66 V , E0Au3+/Au = +1.40 V,
E0Li+/Li = -3.05 V ]

A
Li < K < Al < Ag < Au
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B
Au < Ag < Al < K < Li
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C
K < Al < Au < Ag < Li
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D
Al < Ag < Au < Li < K
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Solution

The correct option is D Au < Ag < Al < K < Li
We know that, lower the standard electrode potential value, greater the reducing power. The standard electrode potential values are increasing in the order Li(3.05V),K(2.93V),Al(1.66V),Ag(+0.80V),Au(+1.40V).
Therefore, the increasing order of reducing power is as follows:
Au<Ag<Al<K<Li.

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