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Which of the following facts is not true?

$\begin{array}{1 1}\text{If }E^{\large\circ}(M^{n+}|M)\text{ is negative,}H^+\text{will be reduced to }H_2\text{ by the metal M}\\\text{If }E^{\large\circ}(M^{n+}|M)\text{ is positive},M^{n+}\text{ will be reduced to M by }H_2\\\text{In a cell,}M^{n+}|M\text{ assembly is attached to hydrogen-half cell.To produce spontaneous cell reaction,metal M will act as negative electrode if the potential }M^{n+}|M\text{ is negative.It will serve as positive electrode,if }M^{n+}|M\text{ has a positive cell potential.}\\\text{Compounds of active metals (Zn,Na,Mg) are reducible by }H_2\text{ whereas those of noble metals (Cu,Ag,Au) are not reducible.}\end{array} $

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Answer : $\text{Compounds of active metals (Zn,Na,Mg) are reducible by }H_2\text{ whereas those of noble metals (Cu,Ag,Au) are not reducible.}$
If $E(M^{n+}|M)$ is negative,it implies that $M^{n+}$ has a lesser tendency to reduce as compared to $H^+$.In this case $H^+$ is reducible and M is oxidisable.Thus this is correct
If $E(M^{n+}|M)$ is positive,it implies that $M^{n+}$ has a larger tendency to reduce and $H_2$ is oxidizable.This is correct
For a galvanic cell,the electrode with more negative potential serves as anode and the electrode with less negative potential or positive potential acts as cathode.Thus this is correct
Active metals have negative potentials and thus are not reducible by $H_2$.Noble metals have positive potentials and thus are reducible by $H_2$.Thus this is incorrect.
answered Jul 21, 2014 by sreemathi.v

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