Given $E_{Cu^{2+}/Cu}^{\circ} = 0.34 \ V$ and $E_{Cu^{2+}/Cu^{+}}^{\circ} = 0.15 \ V$. Calculate the standard electrode potential $E_{Cu^{+}/Cu}^{\circ}$ in $V$.

Consider the following standard electrode potentials ($E^0$ in volts) in aqueous solution. Based on this data,which of the following statements is correct?

Element	$M^{3+}/M$	$M^{+}/M$
$Al$	$-1.66$	$+0.55$
$Tl$	$+1.26$	$-0.34$

For the cell $Zn_{(s)} | Zn^{2+}_{(aq)} || M^{x+}_{(aq)} | M_{(s)}$,different half cells and their standard electrode potentials are given below:

$M^{x+}_{(aq)} / M_{(s)}$	$Au^{3+}_{(aq)} / Au_{(s)}$	$Ag^{+}_{(aq)} / Ag_{(s)}$	$Fe^{3+}_{(aq)} / Fe^{2+}_{(aq)}$	$Fe^{2+}_{(aq)} / Fe_{(s)}$
$E^o M^{x+} / M (V)$	$1.40$	$0.80$	$0.77$	$-0.44$

If $E^o Zn^{2+}/Zn = -0.76 \ V$,which cathode will give a maximum value of $E^o_{cell}$ per electron transferred?

For the reduction of $NO_3^{-}$ ion in an aqueous solution,$E^0$ is $+0.96 \ V$. Values of $E^0$ for some metal ions are given below:
$V^{2+}_{(aq)} + 2e^{-} \rightarrow V \ \ \ \ E^0 = -1.19 \ V$
$Fe^{3+}_{(aq)} + 3e^{-} \rightarrow Fe \ \ \ \ E^0 = -0.04 \ V$
$Au^{3+}_{(aq)} + 3e^{-} \rightarrow Au \ \ \ \ E^0 = +1.40 \ V$
$Hg^{2+}_{(aq)} + 2e^{-} \rightarrow Hg \ \ \ \ E^0 = +0.86 \ V$
The pair$(s)$ of metals that is(are) oxidized by $NO_3^{-}$ in aqueous solution is(are):
$(A) V$ and $Hg$
$(B) Hg$ and $Fe$
$(C) Fe$ and $Au$
$(D) Fe$ and $V$

Which metal cannot produce $H_2$ gas by reaction with $HCl$ solution?
$E^0_{Fe^{2+}/Fe} = -0.44 \ V$
$E^0_{Cu^{2+}/Cu} = +0.34 \ V$
$E^0_{Ni^{2+}/Ni} = -0.25 \ V$
$E^0_{Zn^{2+}/Zn} = -0.76 \ V$

Standard electrode potential values,$E^{\Theta}$ for $Al^{3+}/Al$ is $-1.66 \ V$ and that of $Tl^{3+}/Tl$ is $+1.26 \ V$. Predict the formation of $M^{3+}$ ion in solution and compare the electropositive character of the two metals.