Question

What will be the ${E_{cell}}$ for the following cell :
$Pt{\text{ | }}{{\text{H}}_2}_{{\text{ }}\left( {1{\text{ atm}}} \right)}{\text{ | }}{{\text{H}}^ + }_{0.001M}{\text{ |}}{{\text{H}}^ + }_{0.1M}{\text{ | }}{{\text{H}}_2}_{\left( {1atm} \right)}{\text{ | Pt }}$
$\left( {1.} \right){\text{ 0}}{\text{.1182 V}}$
$\left( {2.} \right){\text{ - 0}}{\text{.1182 V}}$
$\left( {3.} \right){\text{ 0}}{\text{.0591 V}}$
$\left( {4.} \right){\text{ - 0}}{\text{.0591 V}}$

Answer 1

Hint: The relation between the cell potential , the standard cell potential and reaction quotient is given by the Nernst equation. The standard cell potential of a given cell is taken as zero. Since the partial pressure doesn’t change in a given cell, it can be ignored.
Formula used:${E_{cell}}{\text{ = }}{{\text{E}}_{ \circ {\text{ cell}}}}{\text{ - }}\dfrac{{RT}}{{nF}}{\text{ln}}{{\text{Q}}_c}$

Complete answer:
Firstly we will write the cell reactions taking place at cathode and anode.
At anode: ${H_{2{\text{ }}}}_{\left( {1atm} \right)}{\text{ }}\xrightarrow{{}}{\text{ 2}}{{\text{H}}^ + }_{\left( {0.001M} \right)}{\text{ + 2e}}$
At cathode : ${\text{ 2}}{{\text{H}}^ + }_{\left( {0.1M} \right)}{\text{ + 2e }}\xrightarrow{{}}{\text{ }}{{\text{H}}_{2{\text{ }}\left( {1atm} \right)}}$
On combining the above reactions we can see that the number of electrons is two. Thus , $n{\text{ = 2}}$. Now we will use Nernst equation in finding the ${E_{cell}}$. Therefore,
${E_{cell}}{\text{ = }}{{\text{E}}_{ \circ {\text{ cell}}}}{\text{ - }}\dfrac{{RT}}{{nF}}{\text{ln}}{{\text{Q}}_c}$
Here , ${E_{ \circ {\text{ cell}}}}$ is standard reduction potential of cell $ = {\text{ 0}}$
${Q_c}$ is the reaction quotient of the cell reactions. Here ${Q_c}{\text{ = }}\dfrac{{{{\left( {0.001} \right)}^2}}}{{{{\left( {0.1} \right)}^2}}}{\text{ }} \times {\text{ }}\dfrac{{1{\text{ atm}}}}{{1{\text{ atm}}}}$.
$T{\text{ = 298 K}}$ which is a standard temperature.
$F{\text{ = 96386 C , n = 2 , R = 8}}{\text{.314 J mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}$
Now using the above values ,
${E_{cell}}{\text{ = 0 V - }}\dfrac{{8.314{\text{ }} \times {\text{ 298 }}}}{{96368{\text{ }} \times {\text{ 2}}}}\ln \dfrac{{{{\left( {0.001} \right)}^2}}}{{{{\left( {0.1} \right)}^2}}}{\text{ }} \times {\text{ }}\dfrac{{1atm}}{{1atm}}$
${E_{cell}}{\text{ = - }}\dfrac{{8.314{\text{ }} \times {\text{ 298 }}}}{{96368{\text{ }} \times {\text{ 2}}}}\ln \dfrac{{{{\left( {{{10}^{ - 3}}} \right)}^2}}}{{{{\left( {{{10}^{ - 1}}} \right)}^2}}}{\text{ }}$
Further simplifying the logarithm,
${E_{cell}}{\text{ = - }}\dfrac{{{\text{0}}{\text{.0257 }}}}{{{\text{ 2}}}}\ln {\left( {10} \right)^{ - 4}}{\text{ }}$
To convert $\ln $ to $\log $we need to multiply with $2.303$ .
${E_{cell}}{\text{ = 2}}{\text{.303 }} \times {\text{ }}\dfrac{{{\text{0}}{\text{.0257 }}}}{{{\text{ 2}}}}\log {\left( {10} \right)^{ - 4}}{\text{ }}$
${E_{cell}}{\text{ = 4 }} \times {\text{ 2}}{\text{.303 }} \times {\text{ }}\dfrac{{{\text{0}}{\text{.0257 }}}}{{{\text{ 2}}}}\log {\left( {10} \right)^{}}{\text{ }}$
We know that ${\log _{10}}\left( {10} \right){\text{ = 1}}$
${E_{cell}}{\text{ = 2 }} \times {\text{ 0}}{\text{.0591 }}$
${E_{cell}}{\text{ = 0}}{\text{.1182 V }}$
Therefore the e.m.f of the cell is ${E_{cell}}{\text{ = 0}}{\text{.1182 V }}$ .

Additional information: The standard reduction potential values for hydrogen cells is zero. There is a separate table for the standard reduction potential values for different metals . Higher the value of reduction potential is the best oxidizing agent . Thus it has a great tendency to reduce itself and oxidize others.

Note:
The reaction quotient is the ratio of the concentration of product to reactants . It may also include the partial pressure of the gas exerted during the reaction. We prefer to do calculations in base ten log so always remember to convert it. Use the basic properties of logarithmic efficiently. The number of electrons which take part must be in a balanced reaction.