From the following values of ${\mathrm{E}}^0$ drawn from the emf series, calculate standard emf and the equilibrium constant for the reaction,

${\mathrm{Hg}}^{2+}+\mathrm{Hg}\rightleftharpoons {\mathrm{Hg}}_2^{2+}$

${\mathrm{E}}_{{\mathrm{Hg}}_2^{2+},\mathrm{Hg}}=0.788 \mathrm{V}; {\mathrm{E}}_{{\mathrm{Hg}}^{2+},{\mathrm{Hg}}_2^{2+}}^0=0.92 \mathrm{V}$

The standard electrode potential corresponding to the reduction ${\mathrm{Cr}}^{3+}+{\mathrm{e}}^{-}\to {\mathrm{Cr}}^{2+}$ is ${\mathrm{E}}^{°}=-0.407 \mathrm{V}$. If excess $\mathrm{Fe} \left(\mathrm{s}\right)$ is added to a solution in which $\left[{\mathrm{Cr}}^{3+}\right]=1 \mathrm{M},$ what will be $\left[{\mathrm{Fe}}^{2+}\right]$ when equilibrium is established at $25°\mathrm{C}?$

${\mathrm{E}}_{{\mathrm{Fe}}^{2+}|\mathrm{Fe}}^{°} = -0.44\mathrm{V}$

$\{\mathrm{Fe} (\mathrm{s})+2{\mathrm{Cr}}^{3+}\rightleftharpoons {\mathrm{Fe}}^{2+}+2{\mathrm{Cr}}^{2+}\}$

The EMF of a cell consisting of a copper and a lead electrode, immersed in $1 \mathrm{M}$ solution of salts of these metals, is $0.47 \mathrm{V}$. Will the EMF change if $0.001 \mathrm{M}$ solution is taken?

What is the potential of a hydrogen electrode at $\mathrm{pH}=10$ ? 

${{\mathrm{E}}^0}_{{\mathrm{H}}^{+}|{\mathrm{H}}_2} = 0$

Report your answer in volts correct up to three decimal places.

We have an oxidation-reduction system:

${\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{3-}+{\mathrm{e}}^{-}\underset{}{\rightleftarrows } {\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{4-}; {\mathrm{E}}^0=+0.36 \mathrm{V}$

At what ratio of the concentrations of the oxidised and reduced forms will the potential of the system be $0.28 \mathrm{V}?$

Answer correct up to three decimal places with proper rounding off.

Calculate the EMF of the following cell at $25°\mathrm{C}$:

$\begin{array}{c}\mathrm{Fe}\left|{\mathrm{FeSO}}_4\Vert {\mathrm{CuSO}}_4\right|\mathrm{Cu}\\ (0.1 \mathrm{M}) (0.1 \mathrm{M})\end{array}$

Given that ${\mathrm{E}}_{\mathrm{oxidation}}^{°}$ of $\mathrm{Fe}$ and $\mathrm{Cu}$ are $0.44 \mathrm{V}$ and  $-0.34 \mathrm{V}$, respectively, report your answer in Volts.

Calculate the $\mathrm{emf}$ of the following cells and find their cell reactions.

$\mathrm{Ag}\left|{\mathrm{Ag}}^{+}(0.01 \mathrm{M})\right|\left|{\mathrm{Zn}}^{2+}(0·1 \mathrm{M})\right|\mathrm{Zn}$

In this case, is the reaction as written spontaneous or not?

$$\begin{gathered} {\mathrm{E}}_{{\mathrm{Zn}}^{2+}|\mathrm{Zn}}^0 = -0.76\mathrm{V} \\ {\mathrm{E}}_{{\mathrm{Ag}}^{+}|\mathrm{Ag}}^0 = +0.8\mathrm{V} \end{gathered}$$

Calculate the EMF of the following cells, find their cell reactions

$\mathrm{Pt} \left| {\mathrm{Fe}}^{2+}(1 \mathrm{M}), {\mathrm{Fe}}^{3+}(0·1 \mathrm{M})\right| \left|{\mathrm{Cl}}^{-}(0·001 \mathrm{M}) \right| \mathrm{AgCl}\mid \mathrm{Ag}$

In each case, is the reaction as written spontaneous or not?

$$\begin{gathered} {{\mathrm{E}}^0}_{{\mathrm{Fe}}^{3+},{\mathrm{Fe}}^{2+}} = 0.22\mathrm{V} \\ {{\mathrm{E}}^0}_{{\mathrm{Cl}}^{-}\left|\mathrm{AgCl}\right|\mathrm{Ag}} = 0.77\mathrm{V} \end{gathered}$$