Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App