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Question: The temperature coefficient of a cell reaction is $Pb(s) + HgCl_2(aq) \rightarrow PbCl_2(aq) + Hg(l)...

The temperature coefficient of a cell reaction is Pb(s)+HgCl2(aq)→PbCl2(aq)+Hg(l)Pb(s) + HgCl_2(aq) \rightarrow PbCl_2(aq) + Hg(l) (dEdT)P=1.5×10−4VK−1(\frac{dE}{dT})_P = 1.5 \times 10^{-4}VK^{-1} at 298 K The change in entropy in JK−1mol−1JK^{-1} mol^{-1} for given reaction will be

A

14.475

B

57.9

C

28.95

D

86.82

Answer

28.95

Explanation

Solution

The temperature coefficient of a cell reaction is related to the change in entropy (ΔS\Delta S) by the following thermodynamic relationship:

ΔS=nF(dEdT)P\Delta S = nF\left(\frac{dE}{dT}\right)_P

Where:

  • ΔS\Delta S is the change in entropy of the reaction (in JK−1mol−1JK^{-1} mol^{-1}).
  • nn is the number of moles of electrons transferred in the balanced cell reaction.
  • FF is Faraday's constant (96485 C mol−196485 \text{ C mol}^{-1}, commonly approximated as 96500 C mol−196500 \text{ C mol}^{-1} for calculations).
  • (dEdT)P\left(\frac{dE}{dT}\right)_P is the temperature coefficient of the cell potential at constant pressure (in VK−1VK^{-1}).

1. Determine the number of electrons (n) transferred: The given cell reaction is: Pb(s)+HgCl2(aq)→PbCl2(aq)+Hg(l)Pb(s) + HgCl_2(aq) \rightarrow PbCl_2(aq) + Hg(l)

Let's write the oxidation and reduction half-reactions: Oxidation: Pb(s)→Pb2+(aq)+2e−Pb(s) \rightarrow Pb^{2+}(aq) + 2e^- Reduction: Hg2+(aq)+2e−→Hg(l)Hg^{2+}(aq) + 2e^- \rightarrow Hg(l) From the half-reactions, it is clear that 2 moles of electrons are transferred for every mole of the reaction. So, n=2n = 2.

2. Identify the given values:

  • Temperature coefficient, (dEdT)P=1.5×10−4VK−1\left(\frac{dE}{dT}\right)_P = 1.5 \times 10^{-4} VK^{-1}
  • Faraday's constant, F=96500 C mol−1F = 96500 \text{ C mol}^{-1}

3. Calculate the change in entropy (ΔS\Delta S): Substitute the values into the formula: ΔS=2×96500 C mol−1×1.5×10−4 VK−1\Delta S = 2 \times 96500 \text{ C mol}^{-1} \times 1.5 \times 10^{-4} \text{ VK}^{-1} ΔS=193000×1.5×10−4 J K−1 mol−1\Delta S = 193000 \times 1.5 \times 10^{-4} \text{ J K}^{-1} \text{ mol}^{-1} (Since 1 C⋅V=1 J1 \text{ C} \cdot \text{V} = 1 \text{ J}) ΔS=19.3×104×1.5×10−4 J K−1 mol−1\Delta S = 19.3 \times 10^4 \times 1.5 \times 10^{-4} \text{ J K}^{-1} \text{ mol}^{-1} ΔS=19.3×1.5 J K−1 mol−1\Delta S = 19.3 \times 1.5 \text{ J K}^{-1} \text{ mol}^{-1} ΔS=28.95 J K−1 mol−1\Delta S = 28.95 \text{ J K}^{-1} \text{ mol}^{-1}

The change in entropy for the given reaction is 28.95 J K−1 mol−128.95 \text{ J K}^{-1} \text{ mol}^{-1}.