The standard reduction potential at $298 \ K$ for the following half cells are given below :-
$NO_3^{-} + 4H^{+} + 3e^{-} \rightarrow NO_{(g)} + 2H_2O \quad E^0 = 0.97 \ V$
$V^{2+}_{(aq)} + 2e^{-} \rightarrow V_{(s)}$ $E^0 = -1.19 \ V$ $Fe^{3+}_{(aq)} + 3e^{-} \rightarrow Fe_{(s)}$ $E^0 = -0.04 \ V$ $Ag^{+}_{(aq)} + e^{-} \rightarrow Ag_{(s)}$ $E^0 = 0.80 \ V$ $Au^{3+}_{(aq)} + 3e^{-} \rightarrow Au_{(s)}$ $E^0 = 1.40 \ V$
The number of metal$(s)$ which will be oxidized by $NO_3^{-}$ in aqueous solution is $....$.
$NO_3^{-} + 4H^{+} + 3e^{-} \rightarrow NO_{(g)} + 2H_2O \quad E^0 = 0.97 \ V$
| $V^{2+}_{(aq)} + 2e^{-} \rightarrow V_{(s)}$ | $E^0 = -1.19 \ V$ |
| $Fe^{3+}_{(aq)} + 3e^{-} \rightarrow Fe_{(s)}$ | $E^0 = -0.04 \ V$ |
| $Ag^{+}_{(aq)} + e^{-} \rightarrow Ag_{(s)}$ | $E^0 = 0.80 \ V$ |
| $Au^{3+}_{(aq)} + 3e^{-} \rightarrow Au_{(s)}$ | $E^0 = 1.40 \ V$ |
The number of metal$(s)$ which will be oxidized by $NO_3^{-}$ in aqueous solution is $....$.
- A$1$
- B$2$
- C$3$
- D$4$
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View SolutionUsing the standard electrode potentials given below,identify the correct statements from the following.
$Fe^{2+} + 2e^{-} \longrightarrow Fe ; E^{\circ} = -0.44 \ V$
$Cu^{2+} + 2e^{-} \longrightarrow Cu ; E^{\circ} = +0.34 \ V$
$Ag^{+} + e^{-} \longrightarrow Ag ; E^{\circ} = +0.80 \ V$
$(i)$ Copper can displace iron from $FeSO_4$ solution.
$(ii)$ Iron can displace copper from $CuSO_4$ solution.
$(iii)$ Silver can displace copper from $CuSO_4$ solution.
$(iv)$ Iron can displace silver from $AgNO_3$ solution.
$Fe^{2+} + 2e^{-} \longrightarrow Fe ; E^{\circ} = -0.44 \ V$
$Cu^{2+} + 2e^{-} \longrightarrow Cu ; E^{\circ} = +0.34 \ V$
$Ag^{+} + e^{-} \longrightarrow Ag ; E^{\circ} = +0.80 \ V$
$(i)$ Copper can displace iron from $FeSO_4$ solution.
$(ii)$ Iron can displace copper from $CuSO_4$ solution.
$(iii)$ Silver can displace copper from $CuSO_4$ solution.
$(iv)$ Iron can displace silver from $AgNO_3$ solution.
The standard electrode potentials are: $K^{+}/K = -2.93 \ V$,$Ag^{+}/Ag = 0.80 \ V$,$Hg^{2+}/Hg = 0.79 \ V$,$Mg^{2+}/Mg = -2.37 \ V$,$Cr^{3+}/Cr = -0.74 \ V$. Arrange these metals in the increasing order of their reducing power.
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View SolutionThe $EMF$ of a cell in terms of the reduction potential of its left and right electrodes is:
MediumAIEEE 2002
View SolutionThe standard electrode potentials $E^{\circ} (V)$ for $Li^{+} / Li$ and $Na^{+} / Na$ respectively are:
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