If the cell potential value is negative for a cell constructed by attaching a standard hydrogen electrode to another half-cell,then the other half-cell will be:
$(i)$ Anode or cathode?
$(ii)$ Positive or negative?
$(iii)$ On left side or right side?
$(i)$ Anode or cathode?
$(ii)$ Positive or negative?
$(iii)$ On left side or right side?
(A) The cell potential $E_{cell}$ is given by $E_{cell} = E_{cathode} - E_{anode}$.
Given that the standard hydrogen electrode $(SHE)$ has a potential of $0.00 \ V$,if the cell potential is negative,it implies that the potential of the other half-cell is lower than the $SHE$.
$(i)$ Since the other half-cell has a lower potential,it acts as the $Anode$ (oxidation occurs here).
$(ii)$ The $Anode$ in a galvanic cell is the $Negative$ electrode.
$(iii)$ By convention,the $Anode$ is placed on the $Left$ side of the cell diagram.
Given that the standard hydrogen electrode $(SHE)$ has a potential of $0.00 \ V$,if the cell potential is negative,it implies that the potential of the other half-cell is lower than the $SHE$.
$(i)$ Since the other half-cell has a lower potential,it acts as the $Anode$ (oxidation occurs here).
$(ii)$ The $Anode$ in a galvanic cell is the $Negative$ electrode.
$(iii)$ By convention,the $Anode$ is placed on the $Left$ side of the cell diagram.
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Calculate the cell potential in volts for the cell reaction given below:
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$Zn \rightarrow Zn^{2+} + 2e^{-}, E^{0} = 0.76 \ V$
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$Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$
Given:
$Zn \rightarrow Zn^{2+} + 2e^{-}, E^{0} = 0.76 \ V$
$Fe \rightarrow Fe^{2+} + 2e^{-}, E^{0} = 0.41 \ V$ (in $V$)
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View SolutionAssertion : Copper reacts with $HCl$ and liberates hydrogen.
Reason : Hydrogen is present above $Cu$ in the reactivity series.
Reason : Hydrogen is present above $Cu$ in the reactivity series.
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View SolutionThe $E^o$ for the reduction of $NO_3^-$ ion in aqueous solution is $+0.96 \, V$. The $E^o$ values for several metal ions are given below:
$V^{2+}_{(aq)} + 2e^{-} \rightarrow V, \, E^o = -1.19 \, V;$
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$Au^{3+}_{(aq)} + 3e^{-} \rightarrow Au, \, E^o = +1.40 \, V;$
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Which of the following pairs of metals is $NOT$ oxidized by $NO_3^-$ in aqueous solution?
$V^{2+}_{(aq)} + 2e^{-} \rightarrow V, \, E^o = -1.19 \, V;$
$Fe^{3+}_{(aq)} + 3e^{-} \rightarrow Fe, \, E^o = -0.04 \, V;$
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$Hg^{2+}_{(aq)} + 2e^{-} \rightarrow Hg, \, E^o = +0.86 \, V$
Which of the following pairs of metals is $NOT$ oxidized by $NO_3^-$ in aqueous solution?
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View SolutionIf $Cu^{+} + e^- \to Cu$ ; $E^o = X_1$ and $Cu^{2+} + 2e^- \to Cu$ ; $E^o = X_2$,then the value of $E^o$ for $Cu^{2+} + e^- \to Cu^{+}$ will be
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View SolutionGiven:
$Co^{3+} + e^- \longrightarrow Co^{2+}; E^o = 1.81 \ V$
$Pb^{4+} + 2e^- \longrightarrow Pb^{2+}; E^o = + 1.67 \ V$
$Ce^{4+} + e^- \longrightarrow Ce^{3+}; E^o = + 1.61 \ V$
$Bi^{3+} + 3e^- \longrightarrow Bi; E^o = + 0.20 \ V$
Oxidizing power of the species will increase in the order:
$Co^{3+} + e^- \longrightarrow Co^{2+}; E^o = 1.81 \ V$
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Oxidizing power of the species will increase in the order:
DifficultJEE MAIN 2019
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