Which one of the following metals cannot evol | vedclass

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(A) The ability of a metal to displace $H_2$ from acids or $H_2O$ depends on its standard reduction potential $(E^{\circ}_{red})$. Metals with a negat

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$Hg$

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(A) The ability of a metal to displace $H_2$ from acids or $H_2O$ depends on its standard reduction potential $(E^{\circ}_{red})$. Metals with a negative reduction potential (placed above $H_2$ in the Electrochemical Series) can displace $H_2$. $Hg$ has a positive standard reduction potential $(E^{\circ}_{Hg^{2+}/Hg} = +0.85 \ V)$,which is greater than that of $H^+/H_2$ $(0.00 \ V)$. Therefore,$Hg$ cannot displace hydrogen from acids or water. Thus,the correct option is $(A)$.

Which one of the following metals cannot evolve $H_2$ from acids,$H_2O$,or from its compounds?

Similar Questions

The reaction $Zn^{2+} + 2e^{-} \to Zn$ has a standard electrode potential of $-0.76 \ V$. This means:

The standard electrode potentials are: $K^{+}/K = -2.93 \ V$,$Ag^{+}/Ag = 0.80 \ V$,$Hg^{2+}/Hg = 0.79 \ V$,$Mg^{2+}/Mg = -2.37 \ V$,$Cr^{3+}/Cr = -0.74 \ V$. Arrange these metals in the increasing order of their reducing power.

$E^{\circ}$ of $Cu$ is $+0.34 \ V$ while that of $Zn$ is $-0.76 \ V$. Explain.

The standard $E^{\circ}_{red}$ values of $A, B$ and $C$ are $+0.68 \ V, -2.54 \ V$ and $-0.50 \ V$ respectively. The order of their reducing power is:

Which electrode is used to measure the electrode potential of any electrode?

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