For 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 $V$
- A$(A), (B), (D)$
- B$(D), (B), (C)$
- C$(B), (C), (D)$
- D$(C), (A), (B)$
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In a cell,the following reactions take place:
$Fe^{2+} \rightarrow Fe^{3+} + e^{-}$ $\quad$ $E^{\circ}_{Fe^{3+} / Fe^{2+}} = 0.77 \, V$
$2I^{-} \rightarrow I_{2} + 2e^{-}$ $\quad$ $E^{\circ}_{I_{2} / I^{-}} = 0.54 \, V$
The standard electrode potential for the spontaneous reaction in the cell is $x \times 10^{-2} \, V$ at $298 \, K$. The value of $x$ is .... (Nearest Integer)
$Fe^{2+} \rightarrow Fe^{3+} + e^{-}$ $\quad$ $E^{\circ}_{Fe^{3+} / Fe^{2+}} = 0.77 \, V$
$2I^{-} \rightarrow I_{2} + 2e^{-}$ $\quad$ $E^{\circ}_{I_{2} / I^{-}} = 0.54 \, V$
The standard electrode potential for the spontaneous reaction in the cell is $x \times 10^{-2} \, V$ at $298 \, K$. The value of $x$ is .... (Nearest Integer)
Bubbles of gas $z$ are evolved from a solution containing $x$ and $y$. If the order of reduction potential is $x > y > z$,then:
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View SolutionWhat is the procedure to determine the cell potential?
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View SolutionBased on the given data,which is the strongest oxidizing agent? $E_{Li^+|Li}^o = -3.03 \ V$,$E_{Ba^{2+}|Ba}^o = -2.73 \ V$,$E_{Na^+|Na}^o = -2.71 \ V$,and $E_{Mg^{2+}|Mg}^o = -2.37 \ V$.
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View SolutionThe $E^o$ values of half-cells are given below. Which combination of two half-cells will form a cell with the maximum potential?
$(i)$ $A + e^- \rightarrow A^- ; E^o = -0.24 \ V$
$(ii)$ $B^- + e^- \rightarrow B^{2-} ; E^o = +1.25 \ V$
$(iii)$ $C^- + 2e^- \rightarrow C^{3-} ; E^o = -1.25 \ V$
$(iv)$ $D + 2e^- \rightarrow D^{2-} ; E^o = +0.68 \ V$
$(i)$ $A + e^- \rightarrow A^- ; E^o = -0.24 \ V$
$(ii)$ $B^- + e^- \rightarrow B^{2-} ; E^o = +1.25 \ V$
$(iii)$ $C^- + 2e^- \rightarrow C^{3-} ; E^o = -1.25 \ V$
$(iv)$ $D + 2e^- \rightarrow D^{2-} ; E^o = +0.68 \ V$
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