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

Consider the following relations for $EMF$ of an electrochemical cell:
$(i)$ $EMF$ of cell = (Oxidation potential of anode) $-$ (Reduction potential of cathode)
$(ii)$ $EMF$ of cell = (Oxidation potential of anode) $+$ (Reduction potential of cathode)
$(iii)$ $EMF$ of cell = (Reduction potential of anode) $+$ (Reduction potential of cathode)
$(iv)$ $EMF$ of cell = (Oxidation potential of anode) $-$ (Oxidation potential of cathode)
Which of the above relations are correct?

For the half-cell $Zn^{2+} | Zn$,the standard electrode potential $E^{\circ}$ is $-0.76 \ V$. What is the $e.m.f.$ of the cell $Zn_{(s)} | Zn^{2+}_{(aq)} (1 \ M) || 2H^{+}_{(aq)} (1 \ M) | H_{2(g)} (1 \ atm)$ in $V$?

Ferrous ion $(Fe^{2+})$ can be oxidised by which of the following ions?

The reaction $Zn^{2+} + 2e^{-} \to Zn$ has a standard electrode potential of $-0.76 \ V$. This means:

The standard Gibbs energy change for the Daniel cell reaction is $Zn_{(s)} + Cu^{2+}_{(aq)} \longrightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$ where $E_{\text{cell}}^{\circ} = 1.1 \ V$.