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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$A$ gas $X$ at $1 \, atm$ is bubbled through a solution containing a mixture of $1 \, M \, Y^{-}$ and $1 \, M \, Z^{-}$ at $25 \, ^oC$. If the reduction potential of $Z > Y > X$,then
MediumIIT 1999
View SolutionThe standard Gibbs energy change for the Daniel cell reaction is $Zn_{(s)} + Cu^{2+}_{(aq)} \longrightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$ where $E_{\text{cell}}^{\circ} = 1.1 \ V$.
Given:
$Fe^{3+}_{(aq)} + e^- \to Fe^{2+}_{(aq)}; E^o = +0.77 \ V$
$Al^{3+}_{(aq)} + 3e^- \to Al_{(s)}; E^o = -1.66 \ V$
$Br_{2(aq)} + 2e^- \to 2Br^-_{(aq)}; E^o = +1.09 \ V$
Considering the electrode potentials,which of the following represents the correct order of reducing power?
$Fe^{3+}_{(aq)} + e^- \to Fe^{2+}_{(aq)}; E^o = +0.77 \ V$
$Al^{3+}_{(aq)} + 3e^- \to Al_{(s)}; E^o = -1.66 \ V$
$Br_{2(aq)} + 2e^- \to 2Br^-_{(aq)}; E^o = +1.09 \ V$
Considering the electrode potentials,which of the following represents the correct order of reducing power?
DifficultJEE MAIN 2014
View SolutionGiven electrode potentials:
$Fe^{3+} + e^- \to Fe^{2+}$; $E^o = 0.771 \, V$
$I_2 + 2e^- \to 2I^-$; $E^o = 0.536 \, V$
$E^o_{\text{cell}}$ for the cell reaction
$2Fe^{3+} + 2I^- \to 2Fe^{2+} + I_2$ is:
$Fe^{3+} + e^- \to Fe^{2+}$; $E^o = 0.771 \, V$
$I_2 + 2e^- \to 2I^-$; $E^o = 0.536 \, V$
$E^o_{\text{cell}}$ for the cell reaction
$2Fe^{3+} + 2I^- \to 2Fe^{2+} + I_2$ is:
Medium
View SolutionThe correct order of reduction potentials of the following pairs is
$A.$ $Cl_2 / Cl^{-}$
$B.$ $I_2 / I^{-}$
$C.$ $Ag^{+} / Ag$
$D.$ $Na^{+} / Na$
$E.$ $Li^{+} / Li$
Choose the correct answer from the options given below.
$A.$ $Cl_2 / Cl^{-}$
$B.$ $I_2 / I^{-}$
$C.$ $Ag^{+} / Ag$
$D.$ $Na^{+} / Na$
$E.$ $Li^{+} / Li$
Choose the correct answer from the options given below.
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