Calculate $E_{\text{cell}}^{\circ}$ for $Cd_{(s)}|Cd^{2+}_{(1M)}||Ag^{+}_{(1M)}|Ag_{(s)}$. Given: $E^{\circ}_{Cd^{2+}/Cd} = -0.403 \ V$ and $E^{\circ}_{Ag^{+}/Ag} = 0.799 \ V$. (in $V$)

Calculate the standard cell potential (in $V$) of the cell in which the following reaction takes place:
$Fe^{2+}_{(aq)} + Ag^{+}_{(aq)} \to Fe^{3+}_{(aq)} + Ag_{(s)}$
Given that:
$E^o_{Ag^{+}/Ag} = x \ V$
$E^o_{Fe^{2+}/Fe} = y \ V$
$E^o_{Fe^{3+}/Fe} = z \ V$

The standard electrode potentials of the half-cells are given below:
$Zn^{2+} + 2e^- \to Zn$; $E^{\circ} = -0.76 \, V$
$Fe^{2+} + 2e^- \to Fe$; $E^{\circ} = -0.44 \, V$
The $EMF$ of the cell $Fe^{2+} + Zn \to Zn^{2+} + Fe$ is ............ $V$.

Give the cell reaction and $E_{cell}^o$ value for the cell constructed using the given standard electrode potentials: $E_{(H^+|O_2|H_2O)}^o = 1.23 \ V$ and $E_{(Fe^{2+}|Fe)}^o = -0.44 \ V$.

What is denoted by a negative $E^o$ and a positive $E^o$ value in electrochemistry?

If $E^0_{Fe^{3+}/Fe} = x \ V$ and $E^0_{Fe^{2+}/Fe} = y \ V$,then what will be the value of $E^0_{Fe^{3+}/Fe^{2+}}$?