Question

Find the EMF of $Zn|Z{{n}^{2+}}$, if the zinc rod is dipped in $0.1M$$ZnS{{O}_{4}}$solution assuming $80%$dissociation. Given $E_{Z{{n}^{2+}}/Zn}^{\circ }=0.76V$.

Answer 1

Hint:The standard electrode potential is used for measuring the potential for equilibrium. The potential of the electrode is defined as the potential difference present between the electrode and electrolyte. When the concentration of all the species is equal to the unity involved in a semi-cell the potential of electrodes is known as standard electrode potential.

Complete step-by-step answer:The concentration of the Zinc (II) ion in aqueous solution will be $0.1+$$80%$dissociated ion.
$\Rightarrow 0.1+\dfrac{80}{100}=0.9$
Now, look at the following equation:
$Z{{n}^{2+}}\left( aq \right)+2{{e}^{-}}\to Zn$
We know that the Nernst equation is given by the formula
${{E}_{cell}}=E_{cell}^{\circ }+\dfrac{0.059}{n}\log \dfrac{\text{ }\!\![\!\!\text{ reactants }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ products }\!\!]\!\!\text{ }}$
So, the formula for the Nernst equation, for this reaction will be as follows
${{E}_{cell}}=E_{cell}^{\circ }+\dfrac{0.059}{n}\log \dfrac{[Z{{n}^{2+}}]}{[Zn]}$
Now, in the question the value of $E_{Cell}^{\circ }=-0.76V$
The value of n= number of electrons transferred is 2 as two electrons are needed to convert the given zinc cation to zinc.
We have calculated the value of $[Z{{n}^{2+}}]=0.9$ and the value of $[Zn]=1$
Now substituting these values in the above formulae, we get
$\begin{align}
  & \Rightarrow {{E}_{cell}}=-0.76+\dfrac{0.059}{2}\log \dfrac{\text{ }\!\![\!\!\text{ 0}\text{.9 }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ 1 }\!\!]\!\!\text{ }} \\
 & \text{we know that the value of log}\left( 0.9 \right)=-0.045 \\
 & \Rightarrow {{E}_{cell}}=-0.76+0.0295\left( -0.045 \right) \\
 & \Rightarrow {{E}_{cell}}=-0.76-0.0013275 \\
 & \therefore {{E}_{cell}}=-0.7613V \\
\end{align}$
The electrochemical cells are based on the concept of redox reaction which is made up of two half reactions. The one of the half reactions is oxidation which usually occurs at anode and it is done by losing the electrons. The reduction reaction occurs at the cathode which occurs by gaining the electrons.
Hence, the value of electrode potential is $-0.763V$

Note:Students should be careful while determining the value of n in the Nernst equation as this is the most critical point, where mistake happens. Also perform the calculations properly. Sometimes oxidation potential is given and we need to convert it to reduction potential. For this always remember that Oxidation potential of an electrode is considered to be the negative of the reduction potential.