Which is the increasing order of reducing power of the following metals on the basis of standard electrode potential? $Ag^{+}/Ag = 0.80 \ V$,$Mg^{2+}/Mg = -2.37 \ V$,$Hg^{2+}/Hg = 0.79 \ V$,$Cr^{3+}/Cr = -0.74 \ V$
- A$Mg < Cr < Hg < Ag$
- B$Cr < Mg < Ag < Hg$
- C$Hg < Ag < Mg < Cr$
- D$Ag < Hg < Cr < Mg$
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Calculate the $emf$ of the cell $Cu_{(s)} | Cu^{2+}_{(aq)} || Ag^+_{(aq)} | Ag_{(s)}$. Given: $E^0_{Cu^{2+}/Cu} = +0.34 \ V$,$E^0_{Ag^+/Ag} = +0.80 \ V$.
The standard electrode potential $(E^\circ)$ for the half-cell reaction $Fe^{3+} + e^- \rightarrow Fe^{2+}$ at $298 K$ is (Given: $E^\circ(Fe^{3+}/Fe) = -0.04 V$ and $E^\circ(Fe^{2+}/Fe) = -0.44 V$ at $298 K$)
DifficultNEET 2026
View SolutionIf the cell potential value is negative for a cell constructed by attaching a standard hydrogen electrode to another half-cell,then the other half-cell will be:
$(i)$ Anode or cathode?
$(ii)$ Positive or negative?
$(iii)$ On left side or right side?
$(i)$ Anode or cathode?
$(ii)$ Positive or negative?
$(iii)$ On left side or right side?
Easy
View SolutionIf $E^0_{Fe^{2+} / Fe} = -0.441 \ V$ and $E^0_{Fe^{3+} / Fe^{2+}} = 0.771 \ V$,the standard emf of the cell reaction $Fe_{(s)} + 2 Fe^{3+}_{(aq)} \longrightarrow 3 Fe^{2+}_{(aq)}$ is
The $EMF$ of a cell whose half-cell reactions are given below is .......... $V$.
$Mg^{2+} + 2e^- \to Mg_{(s)}; E^o = -2.37 \ V$
$Cu^{2+} + 2e^- \to Cu_{(s)}; E^o = +0.34 \ V$ (in $V$)
$Mg^{2+} + 2e^- \to Mg_{(s)}; E^o = -2.37 \ V$
$Cu^{2+} + 2e^- \to Cu_{(s)}; E^o = +0.34 \ V$ (in $V$)
Medium
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