Question

A piece of zinc metal is dipped in a 0.1 M solution of zinc salt. The salt is dissociated to the extent 0f 20%. Calculated the electrode potential $Z{{n}^{2+}}/Zn$. (Given : $E_{Z{{n}^{2+}}/Zn}^{{}^\circ }=-0.76\,volt$).

Answer 1

Hint: To solve this question we have to use the formula of Nernst equation to find the electrode potential that is given as \[{{E}_{{{M}^{+}}/M}}=E_{{{M}^{+}}/M}^{{}^\circ }-\dfrac{0.0591}{2}\log \left[ {{M}^{+}} \right]\] where, $E_{{{M}^{+}}/M}^{{}^\circ }$ is the standard electrode potential of the elements and n represents the no. of electrons of the element used in the reaction.

Complete step by step solution:
From your chemistry lesson you have learned about the electrode potential and the standard electrode potential and the formula of Nernst equation to find the values of electrode potential for elements. Electrode potential is defined as the tendency of the element when it is placed in contact with its own ions to either gain or lose electrons and in turn become negatively or positively charged. Standard electrode potential is defined as the electrode potential of an element which is determined in relative to standard hydrogen electrode under the standard conditions. As we know that the Nernst equation is used to relate the electrode potential with the concentrations of the ion and it is given as:
\[{{E}_{{{M}^{+}}/M}}=E_{{{M}^{+}}/M}^{{}^\circ }-\dfrac{0.0591}{n}\log \left[ {{M}^{+}} \right]\]
Where, $E_{{{M}^{+}}/M}^{{}^\circ }$= Standard electrode potential
n = no. of electrons
$\left[ {{M}^{+}} \right]$ = concentration of the ion.
In this question we have to find the electrode potential of $Z{{n}^{2+}}/Zn$. Therefore the Nernst equation for $Z{{n}^{2+}}/Zn$ will be:
\[{{E}_{Z{{n}^{2+}}/Zn}}=E_{Z{{n}^{2+}}/Zn}^{{}^\circ }-\dfrac{0.0591}{n}\log \left[ Z{{n}^{2+}} \right]\]
Now, the value for standard electrode potential is given as $E_{Z{{n}^{2+}}/Zn}^{{}^\circ }=-0.76\,volt$ and for concentration it is given as 0.1 M in which 20 % salt is dissociated. So the value of concentration will be $0.1\times \dfrac{20}{100}=0.02M$ and the value of n will be 2 because the reaction that will take place is:
\[Z{{n}^{2+}}+2{{e}^{-}}\to Zn\]
Now, put the values in the equation (1) we will get:
\[{{E}_{Z{{n}^{2+}}/Zn}}=-0.76-\dfrac{0.0591}{2}\log \left[ 2\times {{10}^{-2}} \right]\]
\[\therefore {{E}_{Z{{n}^{2+}}/Zn}}=-0.81\,V\]

Thus the electrode potential will be -0.81 V.

Note: The unit of electrode potential is Volt. The name of the electrode potential is given as reduction electrode potential or oxidation electrode potential on the basis of whether reduction or oxidation is taking place. Standard reduction potentials are taken on standard conditions that are 1 M concentration, at temperature 289 K and at 1 bar pressure.