Question

In which metal container, the aqueous solution of $CuS{{O}_{4}}$can be stored?
$\begin{align}
  & E_{C{{u}^{2+}}/Cu}^{\circ }=0.34V \\
 & E_{Fe/F{{e}^{2+}}}^{\circ }=0.44V \\
 & E_{Al/A{{l}^{3+}}}^{\circ }=1.66V \\
 & E_{Ni/N{{i}^{2+}}}^{\circ }=0.25V \\
 & E_{A{{g}^{+}}/Ag}^{\circ }=0.80 \\
\end{align}$
A. Ag
B. Ni
C. Fe
D. Al

Answer 1

Hint: In such questions, we will find out the electrochemical potential $E_{cell}^{\circ }$by placing the copper sulphate solution in every container. The reaction for which the $E_{cell}^{\circ }$will be negative, will not be spontaneous and thus copper sulphate stored in the vessel will not react with the vessel and can be easily stored in it.

Complete step-by-step answer:We know that the electrochemical potential of a reaction is calculated by subtracting the reduction electrode potential anode from the reduction potential of cathode. This can be expressed in the form of a formula as $E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }$. But we should be careful that we are subtracting the reduction potential and not the oxidation potential.
In the above question only Silver’s electrode potential is given as reduction potential ($A{{g}^{+}}\to Ag$). Rest all the electrode potentials are given as oxidation potential (For example $Al\to A{{l}^{+3}}$).
So, firstly we need to express them in reduction potential.
Reduction potential for all the elements can be written as
$\begin{align}
  & E_{C{{u}^{2+}}/Cu}^{\circ }=0.34V \\
 & E_{F{{e}^{2+}}/Fe}^{\circ }=-0.44V \\
 & E_{A{{l}^{3+}}/Al}^{\circ }=-1.66V \\
 & E_{N{{i}^{2+}}/Ni}^{\circ }=-0.25V \\
 & E_{A{{g}^{+}}/Ag}^{\circ }=0.80 \\
\end{align}$
Now we will see the redox reaction of copper with all the above elements and will calculate the electrode potential. This is mentioned as follows :

Element	Reaction	Cathode	Anode	$E_{cathode}^{\circ }(red)$	$E_{anode}^{\circ }(red)$	$E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }$
Fe	$C{{u}^{2+}}+Fe\to Cu+ F{{e}^{2+}}$	Cu	Fe	0.34	-0.44	$0.34-\left( -0.44 \right)=0.78$
Al	$C{{u}^{2+}}+Al\to A{{l}^{3+}}+Cu$	Cu	Al	0.34	-1.66	$0.34-\left( -1.66 \right)=2.00$
Ni	$C{{u}^{2+}}+Ni\to Cu+ N{{i}^{2+}}$	Cu	Ni	0.34	-0.25	$0.34-\left( -0.25 \right)=0.59$
Ag	$Cu+ A{{g}^{+}}\to C{{u}^{2+}}+Ag$	Ag	Cu	0.80	0.34	$0.34-0.80=-0.46$

In the above table, we can see that the electrode potential of reaction of copper with iron, aluminum, Nickel are positive. Only Ag has negative electrode potential. We should always remember the fact that whenever the electrode potential is negative, it means that the reaction is nonspontaneous and as a result it will not occur on its own.
So, even if we store copper sulphate in a silver vessel, there would be no reaction and copper will not be affected. So, we can store the copper in silver vessels.

Hence, the correct option is A.

Note:Always remember while calculating the electrode potential of a cell that electrode potential of cation and anion are reduction potential and if oxidation potential is given, it can be converted into reduction potential by just placing a negative sign in front of it. This is the most crucial step and students commit a lot of mistakes here.