# Can we store $CuS{{O}_{4}}$ solution in an iron vessel? Explain.

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Hint: If the standard EMF or the cell potential of the reaction between copper and iron is positive then the reaction will occur, if it is negative then the reaction will not occur. Reactivity of the element is based on the oxidation potential or reduction potential.

The standard electrode potential of a large number of electrodes has been determined using a standard hydrogen electrode as the reference electrode.
The electrode at which reduction takes place concerning the standard hydrogen electrode has reduction potential and has a positive sign.
The electrode at which oxidation takes place concerning the standard hydrogen electrode has oxidation potential and has a negative sign.
The greater the oxidation potential or lower the reduction potential of metal, the more easily it can lose an electron and hence have greater reactivity. As a result, a metal with greater oxidation potential will displace the metal with lower oxidation potential from their salt.
Copper has a reduction potential of +0.34 V.
Iron has a reduction potential of -0.44 V.
So, the iron has lower reduction potential, it will be more reactive than iron. Therefore it will lose an electron. Copper having higher reduction potential will accept the electron and will change into solid copper.

The reaction will be:
$Fe(s)+CuS{{O}_{4}}(aq)\to FeS{{O}_{4}}+Cu$
Thus, iron will react with $CuS{{O}_{4}}$. Hence, we cannot store copper sulfate in an iron vessel.

Note: We can also find the standard EMF of the cell:
${{E}^{\circ }}_{F{{e}^{2+}},Fe}=-0.44V,\text{ }{{E}^{\circ }}_{C{{u}^{2+}},Cu}=+0.34V$
The reaction is:
$Fe(s)+CuS{{O}_{4}}(aq)\to FeS{{O}_{4}}+Cu$
The cell will be represented as: $Zn|Z{{n}^{2+}}||C{{u}^{2+}}|Cu$
${{E}^{\circ }}_{cell}={{E}^{\circ }}_{C{{u}^{2+}},Cu}-{{E}^{\circ }}_{F{{e}^{2+}},Fe}=+0.34V-(-0.44V)=0.78V$
As the EMF is positive the reaction will take place and we cannot store copper sulfate in an iron vessel.