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Question: At \(300K\), \(\Delta H\) for the reaction, \[Z{n_{\left( s \right)}} \oplus 2AgC{l_{\left( s \right...

At 300K300K, ΔH\Delta H for the reaction, Zn(s)⊕2AgCl(s)→ZnCl2(s)⊕2Ag(s)Z{n_{\left( s \right)}} \oplus 2AgC{l_{\left( s \right)}} \to ZnC{l_2}_{\left( s \right)} \oplus 2A{g_{\left( s \right)}} is −218KJ/mol - 218KJ/mol while the emf of the cell was 1.015V1.015V. (dEdT)p{\left( {\dfrac{{dE}}{{dT}}} \right)_p} of the cell is:
A.−4.2×10−4VK−1 - 4.2 \times {10^{ - 4}}V{K^{ - 1}}
B.−3.81×10−4VK−1 - 3.81 \times {10^{ - 4}}V{K^{ - 1}}
C.0.11VK−10.11V{K^{ - 1}}
D.7.6×10−4VK−17.6 \times {10^{ - 4}}V{K^{ - 1}}

Explanation

Solution

Thermodynamic principles can be employed to derive a relation between electrical energy and the maximum amount of work. The maximum amount of work obtainable from the cell is the product of charge flowing per mole and maximum potential difference E, through which the charge is transferred.
Formula used: ΔH=nF(dEdT)pT−nFEcell\Delta H = nF{\left( {\dfrac{{dE}}{{dT}}} \right)_p}T - nF{E_{cell}}
FF is faraday constant having value 96500Cmol−196500Cmo{l^{ - 1}}
nn is no of moles of electrons gained by one mole of oxidized state to get changed into the reduced state in the process of reduction occurring at the electrode
TT is the temperature in kelvin
Ecell{E_{cell}} is the emf of the cell
ΔH\Delta H is the net enthalpy change that occurs when the cell reaction gets completed

Complete step by step answer:
For the given question as we can see that the value of nn is 22 since there is net transfer of two moles of electrons
Also the temperature is given as 300K300K while emf of cell Ecell{E_{cell}} is also given as 1.015V1.015V
Finally the change in enthalpy (ΔH)\left( {\Delta H} \right) of the following reaction is needed which is also provided in the question as −218KJmol−1 - 218KJmo{l^{ - 1}} which we need to convert in joules by multiplying it by 10001000 which is done below
Now putting all the values in the above mentioned formula we get
−218×1000=2×96500×300(dEdT)p−2×96500×1.015- 218 \times 1000 = 2 \times 96500 \times 300{\left( {\dfrac{{dE}}{{dT}}} \right)_p} - 2 \times 96500 \times 1.015
On solving the above equation we get the result as
(dEdT)p=−3.81×10−4VK−1{\left( {\dfrac{{dE}}{{dT}}} \right)_p} = - 3.81 \times {10^{ - 4}}V{K^{ - 1}}
And hence option “B” is the correct option.

Additional Information: A cell is characterized by its voltage. A particular kind of cell generates the same voltage irrespective of the size of the cell. The only thing that depends on the cell voltage is the chemical composition of the cell, given the cell is operated at ideal conditions.Normally, the cell voltage may be different from this ideal value, due to several factors like temperature difference, change in concentration, etc. Nernst equation formulated by Walther Nernst can be used to calculate the EMF value of a given cell, provided the standard cell potential of the cell.

Note: In the given question as we see the cell emf was positive hence the cell reaction is feasible. Also the free energy change would have been negative for this reaction making the conversion spontaneous.