The potential of hydrogen electrode is $-118 \mathrm{mV}$. The concentration of ${\mathrm{H}}^{+}$ in the solution is

The standard reduction potentials at $298 \mathrm{K}$ for the following half-reactions are given against each.

$$\begin{gathered} {\mathrm{Zn}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Zn}; {\mathrm{E}}^0=-0.762 \mathrm{V} \\ {\mathrm{Cr}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{Cr}; {\mathrm{E}}^0=-0.74 \mathrm{V} \\ 2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2; {\mathrm{E}}^0=0.0 \mathrm{V} \\ {\mathrm{Fe}}^{3+}+{\mathrm{e}}^{-}\to {\mathrm{Fe}}^{2+}; {\mathrm{E}}^0=+0.77 \mathrm{V}. \end{gathered}$$

Which is the strongest reducing agent?

The standard reduction potentials, ${\mathrm{E}}^0$, for the half-reactions are as

$\begin{array}{l}\mathrm{Zn}\to {\mathrm{Zn}}^{2+}+2{\mathrm{e}}^{-}; {\mathrm{E}}^0=+0.76 \mathrm{V}\\ \mathrm{Fe}\to {\mathrm{Fe}}^{2+}+2{\mathrm{e}}^{-}; {\mathrm{E}}^0=+0.41 \mathrm{V}\end{array}$

the $\mathrm{emf}$ for the cell reaction,

${\mathrm{Fe}}^{2+}+\mathrm{Zn}\to {\mathrm{Zn}}^{2+}+\mathrm{Fe}$ is

From the following ${\mathrm{E}}^0$ values of half cells,

 $\left(\mathrm{i}\right) \mathrm{A}+{\mathrm{e}}^{-}\to {\mathrm{A}}^{-}; {\mathrm{E}}^0=-0·24 \mathrm{V}$
$\left(\mathrm{ii}\right) {\mathrm{B}}^{-}+{\mathrm{e}}^{-}\to {\mathrm{B}}^{2-}; {\mathrm{E}}^0=+1·25 \mathrm{V}$
 $\left(\mathrm{iii}\right) {\mathrm{C}}^{-}+2{\mathrm{e}}^{-}\to {\mathrm{C}}^{3-}; {\mathrm{E}}^0=-1·25 \mathrm{V}$
 $\left(\mathrm{iv}\right) \mathrm{D}+2{\mathrm{e}}^{-}\to {\mathrm{D}}^{2-}; {\mathrm{E}}^0=+0.68 \mathrm{V}$

What combination of two half cells would result in a cell with the largest potential?

From the following ${\mathrm{E}}^{\circ }$ values of half cells

 $\left(\mathrm{i}\right) \mathrm{A}\to {\mathrm{A}}^{+}+{\mathrm{e}}^{-} ; {\mathrm{E}}^0=+1·2 \mathrm{V}$
 $\left(\mathrm{ii}\right) {\mathrm{B}}^{-}\to \mathrm{B}+{\mathrm{e}}^{-} ; {\mathrm{E}}^0=-2·1 \mathrm{V}$
$\left(\mathrm{iii}\right) \mathrm{C}\to {\mathrm{C}}^{2+}+2{\mathrm{e}}^{-} ; {\mathrm{E}}^0=-0.38 \mathrm{V}$
 $\left(\mathrm{iv}\right) {\mathrm{D}}^{-2}\to {\mathrm{D}}^{-}+{\mathrm{e}}^{- }; {\mathrm{E}}^0=-0.59 \mathrm{V}$

What combination of two half cells would result in a cell with the largest potential?

From the following ${\mathrm{E}}^0$ values of half cells

${\mathrm{A}}^{3-}\to {\mathrm{A}}^{2-}+{\mathrm{e}}^{-}; {\mathrm{E}}^0=1.5 \mathrm{V}$

${\mathrm{B}}^{+}+{\mathrm{e}}^{-}\to \mathrm{B}; {\mathrm{E}}^0=-0.5 \mathrm{V}$

${\mathrm{C}}^{2+}+{\mathrm{e}}^{-}\to {\mathrm{C}}^{+}; {\mathrm{E}}^0=+0.5 \mathrm{V}$

$\mathrm{D}\to {\mathrm{D}}^{2+}+2{\mathrm{e}}^{-}; {\mathrm{E}}^0=-1·5 \mathrm{V}$

What combination of two half cells would result in a cell with the largest potential?

If, the following half cells have the ${\mathrm{E}}^0$ values as ${\mathrm{Fe}}^{3+}+\mathrm{e}\to {\mathrm{Fe}}^{2+}; {\mathrm{E}}^0=+0.77 \mathrm{V}$ and ${\mathrm{Fe}}^{2+}+2\mathrm{e}\to \mathrm{Fe}; {\mathrm{E}}^0=-0.44 \mathrm{V};$ the ${\mathrm{E}}^0$ of the half cell ${\mathrm{Fe}}^{3+}+3\mathrm{e}\to \mathrm{Fe}$ will be

${\mathrm{E}}^0$ (red.) values of the half cells ${\mathrm{Mg}}^{2+}/\mathrm{Mg}$ and ${\mathrm{Cl}}_2/{\mathrm{Cl}}^{-}$ are respectively $-2·36 \mathrm{V}$ and $+1·36 \mathrm{V}.$ The ${\mathrm{E}}^0$ value of the cell $\mathrm{Mg}\left|{\mathrm{Mg}}^{2+}\right|\left|{\mathrm{Cl}}_2\right|{\mathrm{Cl}}^{-}$ is