The standard electrode potential of a ${\mathrm{Cl}}_2/{\mathrm{Cl}}^{-}$ half-cell is $+1.36 \mathrm{V}$. This ${\mathrm{Cl}}_2/{\mathrm{Cl}}^{-}$ half-cell was connected to a standard half-cell containing solid iodine in equilibrium with iodide ions. The standard electrode potential of an ${\mathrm{I}}_2/{\mathrm{I}}^{-}$ half-cell is $+0.54 \mathrm{V}$. Calculate the standard cell voltage for this cell.

In the presence of acid, the manganate($\mathrm{VII}$) ion is a powerful oxidising agent. The half-equation for its reduction in acid solution is:

${{\mathrm{MnO}}_4}^{-}\left(\mathrm{aq}\right)+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\stackrel{ }{\rightleftharpoons }{\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right): {\mathrm{E}}^{°}=+1.51 \mathrm{V}$

Explain why the presence of an acid is necessary for the ${{\mathrm{MnO}}_4}^{-}\left(\mathrm{aq}\right)$ to function as an oxidising agent.

In the presence of acid, the manganate($\mathrm{VII}$) ion is a powerful oxidising agent. The half-equation for its reduction in acid solution is:

${{\mathrm{MnO}}_4}^{-}\left(\mathrm{aq}\right)+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\stackrel{ }{\rightleftharpoons }{\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right): {\mathrm{E}}^{°}=+1.51 \mathrm{V}$

Give two reasons for${{\mathrm{MnO}}_4}^{-}\left(\mathrm{aq}\right)$ acting as an oxidising agent in acidic solution.

Liquid bromine is added to an aqueous solution of potassium iodide. The following reaction takes place:

 ${\mathrm{Br}}_2\left(\mathrm{l}\right)+2{\mathrm{I}}^{-}\left(\mathrm{aq}\right)\stackrel{ }{\stackrel{ }{\rightleftharpoons }}2{\mathrm{Br}}^{-}\left(\mathrm{aq}\right)+{\mathrm{I}}_2\left(\mathrm{aq}\right): {\mathrm{E}}^{°}=+0.53 \mathrm{V}$

Write two half-equations for this reaction.

Liquid bromine is added to an aqueous solution of potassium iodide. The following reaction takes place:

 ${\mathrm{Br}}_2\left(\mathrm{l}\right)+2{\mathrm{I}}^{-}\left(\mathrm{aq}\right)\stackrel{ }{\stackrel{ }{\rightleftharpoons }}2{\mathrm{Br}}^{-}\left(\mathrm{aq}\right)+{\mathrm{I}}_2\left(\mathrm{aq}\right): {\mathrm{E}}^{°}=+0.53 \mathrm{V}$

Draw a labelled diagram to show two linked half-cells that could be used to measure the standard cell potential for this reaction. 

Liquid bromine is added to an aqueous solution of potassium iodide. The following reaction takes place:

 ${\mathrm{Br}}_2\left(\mathrm{l}\right)+2{\mathrm{I}}^{-}\left(\mathrm{aq}\right)\stackrel{ }{\stackrel{ }{\rightleftharpoons }}2{\mathrm{Br}}^{-}\left(\mathrm{aq}\right)+{\mathrm{I}}_2\left(\mathrm{aq}\right): {\mathrm{E}}^{°}=+0.53 \mathrm{V}$

The standard cell potential for this reaction is $+0.53 \mathrm{V}$. State whether the position of equilibrium favours the reactants or the products. Explain your answer.