$5.1 \times 10^{-9}=\left[\mathrm{Ba}^{2+}\right] \times 1 \times 10^{-4}$
$5.1 \times 10^{-5}=\left[\mathrm{Ba}^{2+}\right]$
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$(I)\,\,[Fe(CN)_6]^{4-}$ $(II)\,\,[Fe(CN)_6]^{3-}$
$(III) [Cr(NH_3)_6]^{3+}$ $(IV)\,\,[Ni(H_2O)_4]^{2+}$
$(1)$ $C{H_3}C{H_2}N{H_2}$ $(2)$ $\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_2}C{H_3}}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,}\\
{\,C{H_3}C{H_2}NH\,\,\,\,\,\,\,}
\end{array}$
$(3)$ $\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,C{H_3}}\\
{\,|}\\
{{H_3}C - N - C{H_3}}
\end{array}$ $(4)$ $\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,C{H_3}}\\
{\,|}\\
{Ph - N - H}
\end{array}$
If $E_{op}^o$ for this electrode is $1.30\,volt$ then what will be the oxidation electrode potential at $pH = 3$ ? .............. $\mathrm{volt}$
