Given $E_{Cu^{2+}/Cu}^{\circ} = 0.34 \ V$ and $E_{Cu^{2+}/Cu^{+}}^{\circ} = 0.15 \ V$. Calculate the standard electrode potential $E_{Cu^{+}/Cu}^{\circ}$ in $V$.

What is the $e.m.f.$ of the cell in $V$?
$Cr | Cr^{3+} (1.0 \ M) || Co^{2+} (1.0 \ M) | Co$
Given: $E^{o}$ for $Cr^{3+} | Cr = -0.74 \ V$ and $Co^{2+} | Co = -0.28 \ V$

Based on the given standard electrode potentials,identify the metal that can be displaced from its salt solution by all other metals listed.
$E^o Zn^{2+}/Zn = -0.76 \ V, E^o Cu^{2+}/Cu = +0.34 \ V$
$E^o Ag^+/Ag = +0.80 \ V, E^o Co^{2+}/Co = -0.28 \ V$

Given $E^{\circ} Mn^{7+} / Mn^{2+} = 1.51 \ V$ and $E^{\circ} Mn^{4+} / Mn^{2+} = 1.23 \ V$. Calculate $E^{\circ} Mn^{7+} / Mn^{4+}$. (in $V$)

The two half-cell reactions of an electrochemical cell are given as: $Ag^{+} + e^{-} \rightarrow Ag$; $E^{\circ}_{Ag^{+}/Ag} = 0.7995 \ V$ and $Fe^{2+} \rightarrow Fe^{3+} + e^{-}$; $E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.7710 \ V$. The value of cell $EMF$ will be: (in $V$)

Given:
$E^o_{Cr^{3+}/Cr} = -0.74 \ V$,$E^o_{MnO_4^-/Mn^{2+}} = 1.51 \ V$
$E^o_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \ V$,$E^o_{Cl_2/Cl^{-}} = 1.36 \ V$
Based on the data given above,the strongest oxidising agent will be: