Calculate $E_{cell}^{\circ}$ for the following cell: $Zn_{(s)} | Zn^{2+}_{(1 \ M)} || Pb^{2+}_{(1 \ M)} | Pb_{(s)}$ given that $E^{\circ}_{Zn^{2+}/Zn} = -0.763 \ V$ and $E^{\circ}_{Pb^{2+}/Pb} = -0.126 \ V$. (in $V$)

Given 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:

$Li$ occupies a higher position in the electrochemical series of metals as compared to $Cu$ since:

The standard electrode potentials $(E^o)$ for $OCl^{-}/Cl^{-}$ and $\frac{1}{2}Cl_2/Cl^{-}$ are $0.94 \ V$ and $+1.36 \ V$ respectively,the $E^o$ value for $OCl^{-}/\frac{1}{2}Cl_2$ will be ........... $V$.

Explain: Reduction reaction is possible with higher $E^{\theta}$ value.

Calculate the standard cell potential $(E_{cell}^{o})$ for the following electrochemical cells:
$(i)$ $Al_{(s)}|Al_{(1M)}^{3+}||Cu_{(1M)}^{2+}|Cu_{(s)}$
$(ii)$ $Al_{(s)}|Al_{(1M)}^{3+}||Zn_{(1M)}^{2+}|Zn_{(s)}$
$(iii)$ $Al_{(s)}|Al_{(1M)}^{3+}||Ag_{(1M)}^{+}|Ag_{(s)}$
$($ Given: $E_{Al^{3+}|Al}^{o} = -1.66 \ V$,$E_{Zn^{2+}|Zn}^{o} = -0.76 \ V$,$E_{Cu^{2+}|Cu}^{o} = 0.34 \ V$,$E_{Ag^{+}|Ag}^{o} = 0.80 \ V$ $)$