Given the standard electrode potentials,what is the standard electrode potential for the reaction $Fe^{3+}_{(aq)} + e^{-} \rightarrow Fe^{2+}_{(aq)}$ in $V$?
$Fe^{3+}_{(aq)} + 3e^{-} \rightarrow Fe_{(s)}$ ; $E^o = -0.036 \ V$
$Fe^{2+}_{(aq)} + 2e^{-} \rightarrow Fe_{(s)}$ ; $E^o = -0.440 \ V$

The standard hydrogen electrode has a zero electrode potential because .....

The standard reduction potentials for $Zn^{2+}/Zn$,$Ni^{2+}/Ni$ and $Fe^{2+}/Fe$ are $-0.76 \ V$,$-0.23 \ V$ and $-0.44 \ V$ respectively.
The reaction $X + Y^{2+} \rightarrow X^{2+} + Y$ will be spontaneous when:

Calculate the standard cell potential for the galvanic cell $Zn_{(s)} | Zn^{2+}_{(aq)} || Ag^{+}_{(aq)} | Ag_{(s)}$,given that $E^o_{Zn^{2+}/Zn} = -0.76 \ V$ and $E^o_{Ag^{+}/Ag} = +0.80 \ V$. (in $V$)

Electrode potential data are given below :
$Fe^{3+}_{(aq)} + e^- \to Fe^{2+}_{(aq)}; \, E^o = +0.77 \, V$
$Al^{3+}_{(aq)} + 3e^- \to Al_{(s)}; \, E^o = -1.66 \, V$
$Br_{2(aq)} + 2e^- \to 2Br^{-}_{(aq)}; \, E^o = +1.08 \, V$
Based on the data given above,the reducing power of $Fe^{2+}$,$Al$ and $Br^{-}$ will increase in the order:

What will be the $EMF$ of the cell formed by the following half-cells in $V$?
$Mg^{2+} + 2e^- \to Mg_{(s)}; E = -2.37 \ V$
$Cu^{2+} + 2e^- \to Cu_{(s)}; E = +0.33 \ V$