For the reaction $F_2 + 2e^{-} \to 2F^{-}$,$E^{\circ} = 2.8 \, V$. What is the $E^{\circ}$ for the reaction $\frac{1}{2} F_2 + e^{-} \to F^{-}$?

Give the symbolic representation of the following half-cells (electrodes):
$(i)$ $2H^{+}_{(aq)} + 2e^- \to H_{2_{(g)}}$
$(ii)$ $Br_{2_{(aq)}} + 2e^- \to 2Br^{-}_{(aq)}$
$(iii)$ $2Br^{-}_{(aq)} \to Br_{2_{(aq)}} + 2e^-$

The $E^0_{cell}$ of the cell $Al(s)|Al^{3+}(1M)||Pb^{2+}(1M)|Pb(s)$ is $1.5 \text{ V}$. If $E^0_{Pb^{2+}/Pb}$ is $-0.14 \text{ V}$, then the standard electrode potential $E^0_{Al^{3+}/Al}$ will be: (in $\text{ V}$)

If $Cu^{2+} + 2e^{-} \rightarrow Cu, E^{0} = 0.337 \ V$ and $Cu^{2+} + e^{-} \rightarrow Cu^{+}, E^{0} = 0.153 \ V$,then for the reaction $Cu^{+} + e^{-} \rightarrow Cu$,$E^{0}_{cell} =$ .............. $V$.

For the half-reactions $Zn \rightarrow Zn^{2+} + 2e^{-}$ and $Fe \rightarrow Fe^{2+} + 2e^{-}$,the standard oxidation potentials are given as $E^{0}_{Oxi} = +0.76 \ V$ and $E^{0}_{Oxi} = +0.41 \ V$ respectively. Calculate the cell potential for the reaction $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$ in $V$.

The standard Gibbs energy for the given cell reaction in $kJ \, mol^{-1}$ at $298 \, K$ is $Zn_{(s)} + Cu^{2+}_{(aq)} \to Zn^{2+}_{(aq)} + Cu_{(s)}$,given $E^o = 2 \, V$ at $298 \, K$ [Faraday's constant $F = 96500 \, C \, mol^{-1}$].