Electrode potentials of five elements $A, B, C, D$ and $E$ are respectively $-1.36 \ V, -0.32 \ V, 0 \ V, -1.26 \ V$ and $-0.42 \ V$. The reactivity order of these elements is:
- A$A > D > E > B > C$
- B$C > B > E > D > A$
- C$B > D > E > A > C$
- D$C > A > E > D > B$
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Similar Questions
Consider the following ${E^0}$ values:
${E^0}_{Fe^{3+}/Fe^{2+}} = + 0.77 \ V$
${E^0}_{Sn^{2+}/Sn} = - 0.14 \ V$
Under standard conditions,the potential for the reaction $Sn_{(s)} + 2Fe^{3+}_{(aq)} \to 2Fe^{2+}_{(aq)} + Sn^{2+}_{(aq)}$ is ............ $V$.
${E^0}_{Fe^{3+}/Fe^{2+}} = + 0.77 \ V$
${E^0}_{Sn^{2+}/Sn} = - 0.14 \ V$
Under standard conditions,the potential for the reaction $Sn_{(s)} + 2Fe^{3+}_{(aq)} \to 2Fe^{2+}_{(aq)} + Sn^{2+}_{(aq)}$ is ............ $V$.
MediumAIEEE 2004
View SolutionThe standard electrode potential $(E^{\circ}_{M^{3+}/M^{2+}})$ for $V$,$Cr$,$Mn$,and $Co$ are $-0.26 \ V$,$-0.41 \ V$,$+1.57 \ V$,and $+1.97 \ V$,respectively. The metal ions which can liberate $H_2$ from a dilute acid are
DifficultJEE MAIN 2023
View SolutionGiven,the standard potentials $E_{(Cu^{2+}/Cu)}^{\circ}$ and $E_{(Cu^{+}/Cu)}^{\circ}$ as $0.340 \ V$ and $0.522 \ V$ respectively,the value of $E_{(Cu^{2+}/Cu^{+})}^{\circ}$ is $.... \ V$
MediumKVPY 2016
View SolutionIf $E^{\circ}_{Fe^{2+}/Fe} = -0.440 \ V$ and $E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.770 \ V$,then $E^{\circ}_{Fe^{3+}/Fe}$ is (in $V$)
Which is the increasing order of reducing power of the following metals on the basis of standard electrode potential? $Ag^{+}/Ag = 0.80 \ V$,$Mg^{2+}/Mg = -2.37 \ V$,$Hg^{2+}/Hg = 0.79 \ V$,$Cr^{3+}/Cr = -0.74 \ V$
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