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Given that $E^{\large\circ}$ values of $Ag^+|Ag,K^+|K,Mg^{2+}|Mg$ and $Cr^{3+}|Cr$ are 0.80V,-2.93V,-2.37V,and -0.74V,respectively.Which of the following orders regarding the reducing power of metals is correct?

$\begin{array}{1 1}Ag > Cr > Mg > K\\Ag < Cr < Mg < K\\Ag > Cr > K > Mg\\Cr > Ag > Mg > K\end{array} $

1 Answer

Answer : $Ag < Cr < Mg < K$
More negative the $E^{\large\circ}$ value,larger the reducing power of the metal.In present case,$K > Mg > Cr > Ag$
answered Jul 21, 2014 by sreemathi.v
 

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