Identify which of the following cell reactions is spontaneous under standard state conditions.
- A$Ca(s) + Cd^{2+}(aq) \rightarrow Ca^{2+}(aq) + Cd(s)$ $[E^0_{Ca^{2+}/Ca} = -2.866 \text{ V}, E^0_{Cd^{2+}/Cd} = -0.403 \text{ V}]$
- B$2Br^-(aq) + Sn^{2+}(aq) \rightarrow Br_2(l) + Sn(s)$ $[E^0_{Br_2/Br^-} = 1.08 \text{ V}, E^0_{Sn^{2+}/Sn} = -0.136 \text{ V}]$
- C$2Ag(s) + Ni^{2+}(aq) \rightarrow 2Ag^+(aq) + Ni(s)$ $[E^0_{Ag^+/Ag} = 0.799 \text{ V}, E^0_{Ni^{2+}/Ni} = -0.257 \text{ V}]$
- D$2Au(s) + Zn^{2+}(aq) \rightarrow 2Au^+(aq) + Zn(s)$ $[E^0_{Au^+/Au} = 1.68 \text{ V}, E^0_{Zn^{2+}/Zn} = -0.763 \text{ V}]$
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Similar Questions
Which metal does not give the following reaction? $M + \text{water} \rightarrow \text{oxide or hydroxide} + H_2$
Easy
View SolutionFor the reduction of $NO_3^{-}$ ion in an aqueous solution,$E^0$ is $+0.96 \ V$. Values of $E^0$ for some metal ions are given below:
$V^{2+}_{(aq)} + 2e^{-} \rightarrow V \ \ \ \ E^0 = -1.19 \ V$
$Fe^{3+}_{(aq)} + 3e^{-} \rightarrow Fe \ \ \ \ E^0 = -0.04 \ V$
$Au^{3+}_{(aq)} + 3e^{-} \rightarrow Au \ \ \ \ E^0 = +1.40 \ V$
$Hg^{2+}_{(aq)} + 2e^{-} \rightarrow Hg \ \ \ \ E^0 = +0.86 \ V$
The pair$(s)$ of metals that is(are) oxidized by $NO_3^{-}$ in aqueous solution is(are):
$(A) V$ and $Hg$
$(B) Hg$ and $Fe$
$(C) Fe$ and $Au$
$(D) Fe$ and $V$
$V^{2+}_{(aq)} + 2e^{-} \rightarrow V \ \ \ \ E^0 = -1.19 \ V$
$Fe^{3+}_{(aq)} + 3e^{-} \rightarrow Fe \ \ \ \ E^0 = -0.04 \ V$
$Au^{3+}_{(aq)} + 3e^{-} \rightarrow Au \ \ \ \ E^0 = +1.40 \ V$
$Hg^{2+}_{(aq)} + 2e^{-} \rightarrow Hg \ \ \ \ E^0 = +0.86 \ V$
The pair$(s)$ of metals that is(are) oxidized by $NO_3^{-}$ in aqueous solution is(are):
$(A) V$ and $Hg$
$(B) Hg$ and $Fe$
$(C) Fe$ and $Au$
$(D) Fe$ and $V$
DifficultIIT 2009
View SolutionStandard reduction electrode potentials of three metals $A$,$B$,and $C$ are respectively $+0.5 \ V$,$-3.0 \ V$,and $-1.2 \ V$. The reducing powers of these metals are
MediumIIT 1998
View SolutionThe reduction potential ($E^{\circ}$,in $V$) of $MnO_4^-{_{\text{(aq)}}} / Mn_{\text{(s)}}$ is. . . . . $[Given : E^{\circ}_{(MnO_4^{-(aq)} / MnO_{2(s)})} = 1.68 \ V ; E^{\circ}_{(MnO_{2(s)} / Mn^{2+}_{(aq)})} = 1.21 \ V ; E^{\circ}_{(Mn^{2+(aq)} / Mn_{(s)})} = -1.03 \ V]$
MediumIIT 2022
View SolutionThe standard electrode potential for the Daniell cell is $1.1 \ V$. What will be the value of standard Gibbs energy for the reaction?
$Zn_{(s)} + Cu_{(aq)}^{2+} \rightarrow Zn_{(aq)}^{2+} + Cu_{(s)}$
$(1 \ F = 96487 \ C \ mol^{-1})$
$Zn_{(s)} + Cu_{(aq)}^{2+} \rightarrow Zn_{(aq)}^{2+} + Cu_{(s)}$
$(1 \ F = 96487 \ C \ mol^{-1})$
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