Question

If for a cell reaction, \[{\text{Zn + C}}{{\text{u}}^{{\text{ + 2}}}} \to {\text{Cu + Z}}{{\text{n}}^{{\text{ + 2}}}}\], entropy change $\Delta{\text{ }}{{\text{S}}^{\text{0}}}$ is $96.5$ ${\text{Jmo}}{{\text{l}}^{{\text{ - 1}}}}{{\text{K}}^{{\text{ - 1}}}}$, then temperature coefficient of the emf of the cell is:
A. ${{5 \times 1}}{{\text{0}}^{{\text{ - 4}}}}$${\text{V}}{{\text{K}}^{{\text{ - 1}}}}$
B. ${{1 \times 1}}{{\text{0}}^{{\text{ - 3}}}}$${\text{V}}{{\text{K}}^{{\text{ - 1}}}}$
C. ${{2 \times 1}}{{\text{0}}^{{\text{ - 3}}}}$${\text{V}}{{\text{K}}^{{\text{ - 1}}}}$
D. \[{{9}}{{.65 \times 1}}{{\text{0}}^{{\text{ - 4}}}}\]${\text{V}}{{\text{K}}^{{\text{ - 1}}}}$

Answer 1

Hint: This question can be answered from knowledge of the Gibbs free energy released in the reaction occurring in the electrochemical cells, where the Gibbs free energy is the measure of the work that is obtained as a result of the reaction taking place in the system.

Formula used: \[{{\Delta G = - nFE}}\]
${{\Delta G = \Delta H - T\Delta S}}$
${{\Delta H = - nFE + nFT}}\left( {\dfrac{{{\text{dE}}}}{{{\text{dT}}}}} \right)$
where, T is the absolute temperature, $\Delta H$ and $\Delta S$ is the change in entropy, $\Delta G$ is the energy change for the reaction, n is the no. of moles, F is faraday constant, and ${\text{E}}$ is the cell potential.

Complete step by step answer:
The basis of any electrochemical cell is a redox reaction that can be broken down into two half-cell reactions, where oxidation occurs at the anode and electrons are lost and reduced at the cathode, resulting in the gain of electrons. The Gibbs free energy of the system is negative due to the release in energy in a spontaneous reaction.
Hence, as per the equation,
$ - {{nFE = \Delta H}} - {{T\Delta S}}$
$ \Rightarrow {{T\Delta S = \Delta H + nFE}}$
Substituting the value of E, we get:
$ \Rightarrow {{T\Delta S = nFT}}\left( {\dfrac{{{\text{dE}}}}{{{\text{dT}}}}} \right)$
Rearranging:
$ \Rightarrow {{\Delta S = nF}}\left( {\dfrac{{{\text{dE}}}}{{{\text{dT}}}}} \right)$
Therefore,
$ \Rightarrow \left( {\dfrac{{{\text{dE}}}}{{{\text{dT}}}}} \right){\text{ = }}\dfrac{{{{\Delta S}}}}{{{\text{nF}}}}$
$ \Rightarrow \left( {\dfrac{{{\text{dE}}}}{{{\text{dT}}}}} \right) = \dfrac{{96.5}}{{2 \times 96500}}$,
where n is the number of electrons involved in the reaction and F = 96500 coulombs, and $\left( {\dfrac{{{\text{dE}}}}{{{\text{dT}}}}} \right)$ is the temperature coefficient.
Therefore, $\left( {\dfrac{{{\text{dE}}}}{{{\text{dT}}}}} \right) = 0.000495{\text{V}}{{\text{K}}^{{\text{ - 1}}}} = 5 \times {10^{ - 4}}{\text{V}}{{\text{K}}^{{\text{ - 1}}}}$

Hence the correct option is A.

Note:
The cell potential is the measure of the potential difference between the two half cells in an electrochemical cell.
The potential difference is created due to the flow of the electrons from one half of the cell to the other.
Electrons are able to move between the electrodes due to the redox reaction occurring on the cell.