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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Redox Reactions
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Which of the following is wrong statement?

$\begin{array}{1 1}(a)\;\text{Metals with negative SRP give }H_2\;\text{gas with normal acids}\\(b)\;Cu,Ag,Hg\;\overset{\overset{HCl}{\rightarrow}}{H_2SO_4}\;No\;rx_n\\(c)\;Cu,Ag,Hg\;\overset{\overset{HNO_3}{\rightarrow}}{oxidative}\;metal\;ion+NO\;or\;NO_2\\(d)\;Al,Zn,Fe,Sn\;\overset{\overset{HCl}{\rightarrow}}{H_2SO_4}\;No\;rx_n\end{array}$

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1 Answer

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$E^0_{m^{+x}/m}=-ve$
$E^0_{m^{H^+}/H_2}=0$
So this reaction is possible.
$M+x^{H+}\rightarrow M^{x+}+\large\frac{1}{2}$$H_2$
So (a) is right
$E^o_{Cu^{+2}/Cu},E^o_{Hg^{+2}/Hg},E^o_{Ag^+/Ag} > E^o_{H^+/H_2}$
So $Cu/Ag/Hg +HCl/H_2SO_4\rightarrow No \;reaction$
So $E^o_{cell}< 0$
$E^o_{NO_3^-/NO\;or\;NO_2} > E^o_{Cu},E^o_{Cu},E^o_{Ag},E^o_{Hg}$
So $E^0_{cell} > 0$
So reaction possible.
$E^o_{Al},E^o_{Zn},E^o_{Fe},E^o_{Sn} < E^o_{H^+/H_2}$
So $E^o_{cell} > 0$
So this reaction is possible
Hence (d) is the correct answer.
answered Dec 11, 2013 by sreemathi.v
 

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