Question

Zn\left| {\begin{array}{*{20}{c}} {Zn{{\left( {N{O_3}} \right)}_2}\left( {aq} \right)} \\\ {100mL,1M} \end{array}} \right|\;\left| {\begin{array}{*{20}{c}} {Cu{{\left( {N{O_3}} \right)}_2}\left( {aq} \right)} \\\ {100mL,1M} \end{array}} \right|\;{\mathbf{Cu}}
The following galvanic cell was operated as an electrolytic cell using $Cu$ as anode and $Zn$ as cathode. A current of 0.48 ampere was passed for 10 hours and then the cell was allowed to function as a galvanic cell. The e.m.f. of the cell at ${25^0}C$ is (write the nearest integer value) :
Assume that the only electrode reactions occurring were those involving ${\text{Cu}}/{\text{C}}{{\text{u}}^{2 + }}$and $Zn/Z{n^{(2 + )}},$
Given: $E_{C{u^{2 + }},Cu}^0 = 0.34V$ and $E_{(Zn,Z{n^{2 + }})}^0 = - 0.76V$

Answer 1

So, here we can use the Nernst equation to find the emf of the cell. It is often used to find the cell potential of an electrochemical cell at any given pressure, temperature and concentration.

Formula Used:
Nernst equation for finding single electrode potential.
${E_{cell}} = E_{cell}^0 - \left[ {\dfrac{{RT}}{{nF}}} \right]ln\left( Q \right)$
Where,
${E_{cell}}$ is the cell potential of the cell
$E_{cell}^0$ is the cell potential under standard conditions
$R$ is the universal gas potential
$T$ is the temperature
$n$ is the number of electrons transferred in the reaction
$F$ is the Faraday constant
$Q$ is the reaction constant

Complete step by step answer:
We have given the representation of electrochemical cell, we understand that,
At anode $Zn$ undergoes oxidation, and the half-reaction is
$Z{n_{\left( s \right)}} \to Z{n^{2 + }} + 2{e^ - }$
At cathode $Cu$ undergoes reduction and the half-reaction is
$C{u^{2 + }} + 2{e^ - } \to Cu$
Therefore, the total reaction becomes:
$Zn + C{u^{2 + }} \to Z{n^{2 + }} + Cu$
In question, it is given that the current of 0.48A is passed through it for 10 hours.
i.e., $I = 0.48A$
$t\; = \;10hours\; = \;10 \times 60 \times 60s$
We know that
$Current\;\left( I \right) = \dfrac{{Charge\left( q \right)}}{{time\left( t \right)}}$
$q = It$
Substituting the values, we get
$charge\;\left( q \right)\; = \;0.48 \times 10 \times 60 \times 60\; = \;17280C$
From faraday's first law of electrolysis we know that If n electrons are involved in electrode reaction, the passage of n faraday's $\left( {n \times 96500C} \right)$ of electricity will deposit or liberate 1 mole of the substance.
Applying this here,
We have 2 electrons involved in the reaction, therefore $\left( {2 \times 96500C} \right)$ of electricity will deposit or liberate 1 mole of the substance.
Therefore, the number of moles of substance deposited/liberated equals to
Number of moles, $n\; = \dfrac{{17280}}{{2\times96500}}$
$\Rightarrow \;n = 0.09$
Here in question, it is given that the molarity of an aqueous solution of $Zn{\left( {N{O_3}} \right)_2}$ and $Cu{\left( {N{O_3}} \right)_2}$ is $1M$.
We know the molarity of a solution is the number of moles in 1-litre solution
i.e., Molarity of solution after 10 hours will be
$Molarity,\;M = \dfrac{{number\;of\;moles}}{{Volume\left( {in\;litre} \right)}}$
Here, we know that number of moles in 0.09 and the volume given is $100ml = 0.1l$
Therefore, $Molarity,\;M\; = \dfrac{{\;0.09}}{{0.1}} = 0.9M$
i.e., the concentration of $C{u^{2 + }}$ deposited and concentration of $Z{n^{2 + }}$ liberated is 0.9
So, the concentration $C{u^{2 + }}$ will be increased by 0.9, since $C{u^{2 + }}$ is getting deposited therefore the concentration will be increased by 0.09 from their initial concentration of 1.
$\Rightarrow Molarity\left( M \right) = 1 + 0.9\; = 1.9M$ --(1)
Also, the concentration $Z{n^{2 + }}$ will be decreased by 0.9, since $Z{n^{2 + }}$ is getting liberated therefore the concentration will be decreased by 0.09 from their initial concentration of 1.
$\Rightarrow Molarity\left( M \right) = 1 - 0.9\; = 0.1M$ --(2)
Now we know that,
The Nernst equation for the standard condition is,
${E_{cell}} = E_{cell}^0 - \dfrac{{0.059}}{n}ln\left( Q \right)$
where, ${E_{cell}}$ is the cell potential of the cell
$E_{cell}^0$ is the cell potential under standard conditions
$n$ is the number of electrons transferred in the reaction
$Q$ is the reaction constant
For the given reaction, the equation becomes,
${E_{cell}} = E_{cell}^0 - \dfrac{{0.059}}{2}log\left[ {\dfrac{{Z{n^{2 + }}}}{{C{u^{2 + }}}}} \right]$
Since $n = 2$ from the reactions.
Also, we know that,
$E_{cell}^0\; = \;E_{cathode}^0 - E_{anode}^0$
$\Rightarrow E_{cell}^0\; = \;E_{C{u^{2 + }},Cu}^0 - E_{Zn,Z{n^{2 + }}}^0$
$\Rightarrow E_{cell}^0 = 0.34 - ( - 0.76) = 1.1V$-- (3)
From equations (1), (2) and (3), we get
$\Rightarrow {E_{cell}}\; = 1.1 - \dfrac{{0.059}}{2}{\text{lo}}{{\text{g}}_{10}}\left( {\dfrac{{0.1}}{{1.9}}} \right)$
$\Rightarrow {E_{cell}}=1.1-0.02645\log(100.1)-\log(101.9)$
Now, substituting the values of log
$\Rightarrow {E_{cell}}=1.1-(0.02645)\times(-1.27)$
$\therefore {E_{cell}} = 1.1 + 0.038V = 1.138V$

**Therefore, the emf of the cell at ${25^0}C$ is $1.138V$

Note: **
Here, we used the molarity for measuring the concentration of ions present in it. It can be molality as well. Both have the same property while using Nernst equation.
Also at $25^\circ C$ Nernst equation behaves as a standard condition that’s why we used direct equation
${E_{cell}} = E_{cell}^0 - \dfrac{{0.059}}{n}ln\left( Q \right)$