Question

What is the value of the reaction quotient Q for the cell?
${\text{Ni }}\left( s \right){\text{|}}\,{\text{Ni}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}\,\left( {{\text{0}}{\text{.190 M}}} \right)\,{\text{||}}\,{\text{KCl}}\,\left( {0.40\,\,{\text{M}}} \right)|\,{\text{C}}{{\text{l}}_2}\left( g \right)\,,\,0.10\,{\text{atm}}\,{\text{|}}\,{\text{Pt}}\,\left( s \right)$
A.$3 \times {10^{ - 1}}$
B.$1.3 \times {10^{ - 1}}$
C.$8.0 \times {10^{ - 2}}$
D.$3.0 \times {10^{ - 2}}$

Answer 1

Hint: Electrochemistry is an important branch in physical chemistry, it deals with the study of electricity and related chemical reactions. In which electricity is generated in a reaction by the movements of electrons from one element to another, the reaction is known as redox or oxidation -reduction reaction. An electrochemical cell is a device which produces electric current from energy released by a spontaneous redox reaction. Examples of such cells are galvanic cells, voltaic cells etc.
Reaction quotient (${\text{Q}}$) measures the relative amount of products and reactant present during a reaction at a particular point in time
Formula used: Nernst equation is the most important equation used in electrochemistry at standard temperature ${25^o}{\text{C}}$ or $298\,{\text{K}}$ the equation is given below
$
  {\text{E}}\, = \,{{\text{E}}^0}\,{\text{ - }}\,\left( {\dfrac{{0.0591}}{{\text{n}}}} \right)\,{\text{log}}\,{\text{Q}} \\
  {{\text{E}}^0} \to {\text{standard reduction potential for the reaction in volts}}{\text{.}} \\
  {\text{Q}} \to \,{\text{reaction quotient}} \\
  {\text{n}} \to {\text{no of electrons exchanged}} \\
  {\text{where Q = }}\dfrac{{{{\left[ {{\text{products}}} \right]}^x}}}{{{{\left[ {{\text{reactants}}} \right]}^y}}} \\
 $

Complete step by step answer:
We already discussed that the reaction taking place in an electrochemical cell is known as redox reaction. Where the oxidation reaction occurs at one electrode and reduction occurs at another electrode. The electrode at which reduction takes place is called cathode half cell whereas the oxidation takes place at anode half cell. And the term half cell consists of electrodes and the species to be oxidized or reduced. Also if a material conducts electricity, it can be used as an electrode.
Here we have given an electrochemical reaction, for an electrochemical reaction certain rules have to be followed to write the cell notation for a reaction. The rules are given below
Here the half cell which is described first is anode followed by the cathode half cell. Within the half cell, reactants are written first then the products. A single vertical line which is drawn between two chemical species used to describe that chemical species are in two different phases but are in physical contact with each other. The double vertical line is used to represent a salt bridge or a porous membrane which is used to separate the individual half cell. The phase of each chemical is given in the cell notation. If the electrolytes in the cells are not in standard conditions then concentrations and pressure will be included with the phase notation.
Here we have given a electrochemical reaction, ${\text{Ni }}\left( s \right){\text{|}}\,{\text{Ni}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}\,\left( {{\text{0}}{\text{.190 M}}} \right)\,{\text{||}}\,{\text{KCl}}\,\left( {0.40\,\,{\text{M}}} \right)|\,{\text{C}}{{\text{l}}_2}\left( g \right)\,,\,0.10\,{\text{atm}}\,{\text{|}}\,{\text{Pt}}\,\left( s \right)$
Reaction at anode half cell
Here ${\text{Ni}}\,\left( s \right)$ give up $2{e^ - }$ per atom and goes into the solution ${\text{Ni}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}$ as ${\text{N}}{{\text{i}}^{2 + }}$. Oxidation reactions take place at anode.
${\text{Ni}}\left( s \right) \to {\text{N}}{{\text{i}}^{2 + }}\left( {aq} \right)\, + \,2{e^ - }$
Reaction at cathode half cell
In cathode ${\text{C}}{{\text{l}}_2}\left( g \right)$ takes the electrons from ${\text{Ni}}\,\left( s \right)$ and get reduced to ${\text{C}}{{\text{l}}^ - }$. Reduction reactions take place at cathode.
${\text{C}}{{\text{l}}_2}\left( g \right)\, + \,2{e^ - } \to 2{\text{C}}{{\text{l}}^ - }$
Overall cell reaction can written as
${\text{Ni}}\left( s \right)\, + \,{\text{C}}{{\text{l}}_2}\left( g \right) \to {\text{N}}{{\text{i}}^{2 + }}\left( {aq} \right)\, + \,2{\text{C}}{{\text{l}}^ - }\left( {aq} \right)$
Here we have to calculate the reaction quotient of the above reaction. Reaction quotient is the same as the equilibrium constant expression ${\text{K}}$, but the reaction quotient can be calculated at any conditions.
Reaction quotient ${\text{Q}}$ of reaction is given by
${\text{Q = }}\,\dfrac{{\left[ {{\text{N}}{{\text{i}}^{2 + }}} \right]\,{{\left[ {{\text{C}}{{\text{l}}^ - }} \right]}^2}}}{{\left[ {{\text{Ni}}\left( s \right)} \right]\,\left[ {{\text{C}}{{\text{l}}_2}\left( g \right)} \right]}}$
Since here in the cell notation both the concentration and pressure is noted hence the cells are not in the standard conditions. But for ${\text{Ni}}\,\left( s \right)$ Concentration value is not mentioned hence we can take it as $1.00\,{\text{M}}$.
$\therefore \,\,\,\,{\text{Q}}\,{\text{ = }}\,\dfrac{{0.190\, \times {{\left( {0.40} \right)}^2}}}{{1 \times 0.10}} = 3.0 \times {10^{ - 1}}$
So the reaction quotient of the reaction is $3 \times {10^{ - 1}}$.

So, the correct answer is option A.

Note:
${\text{Ni }}\left( s \right){\text{|}}\,{\text{Ni}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}\,\left( {{\text{0}}{\text{.190 M}}} \right)\,{\text{||}}\,{\text{KCl}}\,\left( {0.40\,\,{\text{M}}} \right)|\,{\text{C}}{{\text{l}}_2}\left( g \right)\,,\,0.10\,{\text{atm}}\,{\text{|}}\,{\text{Pt}}\,\left( s \right)$
In an electrochemical cell reduction and oxidation reactions take place in electrodes, here ${\text{Pt }}$is used as an electrode, which allows electron movement. ${\text{Pt }}$is non reactive hence it will not take part in redox reactions in electrochemical cells.