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\[Zn(s)+C{{u}^{2+}}(aq)\to Z{{n}^{2+}}(aq)+Cu(s)\]

The standard reaction enthalpy $({{\Delta }_{r}}{{H}^{\Theta }})$ at 300k in $KJmo{{l}^{-1}}$ is:

[Use R = 8 $J{{K}^{-1}}mo{{l}^{-1}}$ and F = 96,000 $Cmo{{l}^{-1}}$]

A. -412.8

B. -384.0

C. 206.4

D. 192.0

Answer

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From your chemistry lessons you have learned about the electrode potential and their relation to Gibbs free energy and also about the entropy, enthalpy in thermodynamics. Standard electrode potential is the measurement of potential under standard conditions of half-reaction against the standard hydrogen potential which is taken as reference electrode.

The electrochemical cell has a relation with the thermodynamic quantity like entropy, Gibbs free energy which we are going to see here. Gibbs free energy or change in free energy is used to measure the maximum amount of work which is performed in a chemical reaction. There is relation between Standard change in free and standard potential of electrochemical cell which is shown as,

\[\Delta {{G}^{{}^\circ }}=-nFE_{cell}^{\Theta }\]………………… (1)

Where, $\Delta {{G}^{{}^\circ }}$= Standard Change in free energy,

n = overall charge in a reaction

F = Faraday

\[{{E}^{\Theta }}\]= standard potential of electrochemical cells.

-Now, the change in free energy and entropy are also related with each other as,

\[\Delta {{S}^{{}^\circ }}=nF\left( \dfrac{d{{E}^{\Theta }}}{dT} \right)\]……………… (2)

-In this question we have to find the standard reaction enthalpy, from your thermodynamic lessons you are have learned about the relation between Gibbs free energy, entropy and enthalpy which is expressed as , $\Delta G=\Delta H-T\Delta S$ (T is the temperature)

-Which can also be written as, $\Delta {{H}^{{}^\circ }}=\Delta {{G}^{{}^\circ }}+T\Delta {{S}^{{}^\circ }}$…………. (3)

-So, in the question the value of ${{E}^{\Theta }}$ is given as 2 V and the value of $\left( \dfrac{d{{E}^{\Theta }}}{dT} \right)$ as $-5\times {{10}^{-4}}V{{K}^{-1}}$, the value of F= 96,000 and n = 2

-Now put the values of (1) and (2) in equation three, you will get

\[\Delta {{H}^{{}^\circ }}=-nFE_{cell}^{\Theta }+T\left[ nF\dfrac{d{{E}^{\Theta }}}{dT} \right]\]

\[\therefore \Delta H=-2\times 96,000\times 2+300\left[ 2\times 96,000\times (-5\times {{10}^{-4}}) \right]\]

= -412.8 $KJmo{{l}^{-1}}$