Give the symbolic representation of the following half-cells (electrodes):
$(i)$ $2H^{+}_{(aq)} + 2e^- \to H_{2_{(g)}}$
$(ii)$ $Br_{2_{(aq)}} + 2e^- \to 2Br^{-}_{(aq)}$
$(iii)$ $2Br^{-}_{(aq)} \to Br_{2_{(aq)}} + 2e^-$
$(i)$ $2H^{+}_{(aq)} + 2e^- \to H_{2_{(g)}}$
$(ii)$ $Br_{2_{(aq)}} + 2e^- \to 2Br^{-}_{(aq)}$
$(iii)$ $2Br^{-}_{(aq)} \to Br_{2_{(aq)}} + 2e^-$
The symbolic representation of a half-cell is written as $\text{Electrode} | \text{Electrolyte}$ for oxidation and $\text{Electrolyte} | \text{Electrode}$ for reduction.
$(i)$ $2H^{+}_{(aq)} + 2e^- \to H_{2_{(g)}}$ represents a reduction half-cell involving a gas electrode: $H^{+}_{(aq)} | H_{2_{(g)}} | Pt_{(s)}$.
$(ii)$ $Br_{2_{(aq)}} + 2e^- \to 2Br^{-}_{(aq)}$ represents a reduction half-cell involving a non-metal electrode: $Br_{2_{(aq)}} | Br^{-}_{(aq)} | Pt_{(s)}$.
$(iii)$ $2Br^{-}_{(aq)} \to Br_{2_{(aq)}} + 2e^-$ represents an oxidation half-cell involving a non-metal electrode: $Pt_{(s)} | Br^{-}_{(aq)} | Br_{2_{(aq)}}$.
$(i)$ $2H^{+}_{(aq)} + 2e^- \to H_{2_{(g)}}$ represents a reduction half-cell involving a gas electrode: $H^{+}_{(aq)} | H_{2_{(g)}} | Pt_{(s)}$.
$(ii)$ $Br_{2_{(aq)}} + 2e^- \to 2Br^{-}_{(aq)}$ represents a reduction half-cell involving a non-metal electrode: $Br_{2_{(aq)}} | Br^{-}_{(aq)} | Pt_{(s)}$.
$(iii)$ $2Br^{-}_{(aq)} \to Br_{2_{(aq)}} + 2e^-$ represents an oxidation half-cell involving a non-metal electrode: $Pt_{(s)} | Br^{-}_{(aq)} | Br_{2_{(aq)}}$.
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The values of $E^0$ for metals $A$,$B$,and $C$ are $0.34 \ V$,$-0.80 \ V$,and $-0.46 \ V$ respectively. State the correct order for their ability to act as reducing agents.
What is the standard potential of the cell $Ni|Ni^{2+}_{(1 \ M)} || Cu^{2+}_{(1 \ M)}| Cu$ (in $V$)? Given $E^{\circ}_{Cu^{2+}/Cu} = 0.337 \ V$ and $E^{\circ}_{Ni^{2+}/Ni} = -0.236 \ V$.
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Easy
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.
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