The E^o values of Mg^{2+}/Mg is -2.37 V,Zn^{ | vedclass

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Question

(A) The standard reduction potentials are given as: $E^o_{Mg^{2+}/Mg} = -2.37 \ V$,$E^o_{Zn^{2+}/Zn} = -0.76 \ V$,and $E^o_{Fe^{2+}/Fe} = -0.44 \ V$.

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$Zn$ will reduce $Fe^{2+}$

Vedclass · Answer

(A) The standard reduction potentials are given as: $E^o_{Mg^{2+}/Mg} = -2.37 \ V$,$E^o_{Zn^{2+}/Zn} = -0.76 \ V$,and $E^o_{Fe^{2+}/Fe} = -0.44 \ V$. $A$ metal with a lower (more negative) reduction potential can reduce the ion of a metal with a higher (less negative) reduction potential. Since $E^o_{Zn^{2+}/Zn} (-0.76 \ V) $Zn$ cannot reduce $Mg^{2+}$ because $E^o_{Zn^{2+}/Zn} > E^o_{Mg^{2+}/Mg}$. Similarly,$Mg$ and $Zn$ act as reducing agents,not oxidizing agents for $Fe$. Therefore,the correct statement is $A$.

The $E^o$ values of $Mg^{2+}/Mg$ is $-2.37 \ V$,$Zn^{2+}/Zn$ is $-0.76 \ V$,and $Fe^{2+}/Fe$ is $-0.44 \ V$. Which of the following statements is correct?

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