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: $E_{Al^{3+}|Al}^{o} = -1.66 \ V$,$E_{Zn^{2+}|Zn}^{o} = -0.76 \ V$,$E_{Cu^{2+}|Cu}^{o} = 0.34 \ V$,$E_{Ag^{+}|Ag}^{o} = 0.80 \ V$ $)$

The standard cell potential is calculated using the formula: $E_{cell}^{o} = E_{cathode}^{o} - E_{anode}^{o}$.
$(i)$ For $Al|Al^{3+}||Cu^{2+}|Cu$:
$E_{cell}^{o} = E_{Cu^{2+}|Cu}^{o} - E_{Al^{3+}|Al}^{o} = 0.34 \ V - (-1.66 \ V) = 2.00 \ V$.
$(ii)$ For $Al|Al^{3+}||Zn^{2+}|Zn$:
$E_{cell}^{o} = E_{Zn^{2+}|Zn}^{o} - E_{Al^{3+}|Al}^{o} = -0.76 \ V - (-1.66 \ V) = 0.90 \ V$.
$(iii)$ For $Al|Al^{3+}||Ag^{+}|Ag$:
$E_{cell}^{o} = E_{Ag^{+}|Ag}^{o} - E_{Al^{3+}|Al}^{o} = 0.80 \ V - (-1.66 \ V) = 2.46 \ V$.