Why is platinum used in preference to other metals, in half-cells where the reaction does not involve a metallic element?

Show, with the aid of a diagram, how you would measure the ${\mathrm{E}}^{⊝}$value for the half-cell shown by the equation:
${\mathrm{VO}}^{2+}+ 2{\mathrm{H}}^{+}+{\mathrm{e}}^{-}\rightleftharpoons {\mathrm{V}}^{3+}+{\mathrm{H}}_2\mathrm{O}$

(a) Draw a diagram of an electrochemical cell consisting of a ${\mathrm{Cr}}^{3+} / \mathrm{Cr}$ half-cell and a ${\mathrm{Cl}}_2/{\mathrm{Cl}}^{-}$ half-cell.

(b) Use this data to calculate the cell voltage.
$$\begin{gathered} {\mathrm{Cr}}^{3+}+3{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Cr} {\mathrm{E}}^{⊝}=-0.74 \mathrm{V} \\ {\mathrm{Cl}}_2+2{\mathrm{e}}^{-}\rightleftharpoons 2{\mathrm{Cl}}^{-} {\mathrm{E}}^{⊝}=+1.36 \mathrm{V} \end{gathered}$$

(c) Which half-cell is the positive pole?

a Draw a diagram of an electrochemical cell consisting of a ${\mathrm{Mn}}^{2+} / \mathrm{Mn}$ half-cell and a ${\mathrm{Pb}}^{2+}/\mathrm{Pb}$ half-cell.

b Use this data to calculate the cell voltage.
$$\begin{gathered} {\mathrm{Mn}}^{2+}+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Mn} {\mathrm{E}}^{⊝}=-1.18 \mathrm{V} \\ {\mathrm{Pb}}^{2+}+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Pb} {\mathrm{E}}^{⊝}=-0.13 \mathrm{V} \end{gathered}$$

c Which half-cell is the positive pole?

State the direction of the electron flow in the electrochemical cell represented by the following pair of half-equations. Use the given data to help you.
${\mathrm{F}}_2 + 2{\mathrm{e}}^{-} \rightleftharpoons 2{\mathrm{F}}^{-} {\mathrm{E}}^{⊝}=+2.87 \mathrm{V}$  and
${\mathrm{Mn}}^{2+} + 2{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Mn} {\mathrm{E}}^{⊝}=-1.18 \mathrm{V}$

State the direction of the electron flow in the electrochemical cell represented by the following pair of half-equations. Use the given data to help you.

${\mathrm{Sn}}^{4+} + 2{\mathrm{e}}^{-} \rightleftharpoons {\mathrm{Sn}}^{2+} {\mathrm{E}}^{⊝}=+0.15 \mathrm{V}$ and
${\mathrm{I}}_2 + 2{\mathrm{e}}^{-} \rightleftharpoons 2{\mathrm{I}}^{-} {\mathrm{E}}^{⊝}=+0.54 \mathrm{V}$

State the direction of the electron flow in the electrochemical cell represented by the following pair of half-equations. Use the given data to help you.

${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+ 14{\mathrm{H}}^{+}+6{\mathrm{e}}^{-} \rightleftharpoons 2{\mathrm{Cr}}^{3+} +7{\mathrm{H}}_2\mathrm{O} {\mathrm{E}}^{⊝}=+1.33 \mathrm{V}$ and
${\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Cu} {\mathrm{E}}^{⊝}=+0.34 \mathrm{V}$

State the direction of the electron flow in the electrochemical cell represented by the following pair of half-equations. Use the given data to help you.

${\mathrm{Ni}}^{2+} +2{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Ni} {\mathrm{E}}^{⊝}=-0.25 \mathrm{V}$ and
${\mathrm{Fe}}^{3+} + 3{\mathrm{e}}^{-} \rightleftharpoons \mathrm{Fe} {\mathrm{E}}^{⊝}=-0.04 \mathrm{V}$

Use the given data to predict whether the following reaction is feasible or not. If a reaction does occur, write a balanced equation for it. Can ${\mathrm{MnO}}_4^{-}$ ions oxidise ${\mathrm{Cl}}^{-}$ ions to ${\mathrm{Cl}}_2$ in acidic conditions?

$$\begin{gathered} {\mathrm{MnO}}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\rightleftharpoons {\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O} {\mathrm{E}}^{⊝}=+1.52 \mathrm{V} \\ {\mathrm{Cl}}_2+2{\mathrm{e}}^{-}\rightleftharpoons 2{\mathrm{Cl}}^{-} {\mathrm{E}}^{⊝}=+1.36 \mathrm{V} \end{gathered}$$