The standard electrode potentials of Ag^{+}/A | vedclass

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(C) The deposition of metals at the cathode during electrolysis depends on their standard reduction potentials $(E^o)$.Metals with higher reduction po

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$Ag, Hg, Cu$ only

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(C) The deposition of metals at the cathode during electrolysis depends on their standard reduction potentials $(E^o)$.Metals with higher reduction potentials are deposited more easily at the cathode.The given standard reduction potentials are: $E^o(Ag^+/Ag) = 0.80 \ V$,$E^o(Hg_2^{2+}/2Hg) = 0.79 \ V$,$E^o(Cu^{2+}/Cu) = 0.34 \ V$,and $E^o(Mg^{2+}/Mg) = -2.37 \ V$.In an aqueous solution,water can also be reduced at the cathode: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$,with $E^o = -0.83 \ V$.Since $Mg^{2+}$ has a reduction potential of $-2.37 \ V$,which is much lower than that of water $(-0.83 \ V)$,$Mg$ will not deposit from an aqueous solution; instead,$H_2$ gas will be evolved.Therefore,the metals that will deposit in the order of increasing voltage are $Ag$,$Hg$,and $Cu$.The correct sequence is $Ag > Hg > Cu$.

$V^{2+}_{(aq)} + 2e^{-} \rightarrow V_{(s)}$	$E^0 = -1.19 \ V$
$Fe^{3+}_{(aq)} + 3e^{-} \rightarrow Fe_{(s)}$	$E^0 = -0.04 \ V$
$Ag^{+}_{(aq)} + e^{-} \rightarrow Ag_{(s)}$	$E^0 = 0.80 \ V$
$Au^{3+}_{(aq)} + 3e^{-} \rightarrow Au_{(s)}$	$E^0 = 1.40 \ V$

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