Calculate $E_{cell}^o$ of the following galvanic cell at $298 \ K$:
$Ca_{(s)} | Ca^{2+}_{(aq)} || Fe^{2+}_{(aq)} | Fe_{(s)}$
Given: $E_{Ca^{2+}/Ca}^o = -2.87 \ V$; $E_{Fe/Fe^{2+}}^o = 0.41 \ V$

$FeO_4^{2-}$ $\xrightarrow{2.0 \ V} Fe^{3+}$ $\xrightarrow{0.8 \ V} Fe^{2+}$ $\xrightarrow{-0.44 \ V} Fe^0$
In the above diagram,the standard electrode potentials are given in volts. The value of $E_{FeO_4^{2-}/Fe^{2+}}^{\Theta}$ is: (in $V$)

The $E^{0}_{Red}$ values for $P, Q, R$ and $S$ are $-2.90 \, V, +0.34 \, V, +1.20 \, V$ and $-0.76 \, V$ respectively. The decreasing order of their reactivity is:

The standard reduction potential at $298 \ K$ for the following half-cell reactions is given as:
$Zn^{2+}_{(aq)} + 2e^{-} \rightarrow Zn_{(s)} ; \quad E^{\circ} = -0.762 \ V$
$Cr^{3+}_{(aq)} + 3e^{-} \rightarrow Cr_{(s)} ; \quad E^{\circ} = -0.740 \ V$
$2H^{+}_{(aq)} + 2e^{-} \rightarrow H_{2(g)} ; \quad E^{\circ} = 0.0 \ V$
$F_{2(g)} + 2e^{-} \rightarrow 2F^{-}_{(aq)} ; \quad E^{\circ} = 2.87 \ V$
Which of the following is the strongest reducing agent?

Given:
$Co^{3+} + e^- \longrightarrow Co^{2+}; E^o = 1.81 \ V$
$Pb^{4+} + 2e^- \longrightarrow Pb^{2+}; E^o = + 1.67 \ V$
$Ce^{4+} + e^- \longrightarrow Ce^{3+}; E^o = + 1.61 \ V$
$Bi^{3+} + 3e^- \longrightarrow Bi; E^o = + 0.20 \ V$
Oxidizing power of the species will increase in the order:

If the electrode potential of $Zn^{2+} | Zn$ is $-0.76 \, V$ and the potential of $Cu^{2+} | Cu$ is $0.34 \, V$,what will be the $EMF$ of the cell formed by both electrodes in $V$?