JEE Main 2022 · 26 June · Evening Shift · Chemistry · Electrochemistry
Cu(s)+Sn2+(0.001M)→Cu2+(0.01M)+Sn(s)Cu\left(s\right) + {Sn}^{2 +}\left(0 . 001M\right) \to {Cu}^{2 +}\left(0 . 01M\right) + Sn\left(s\right)Cu(s)+Sn2+(0.001M)→Cu2+(0.01M)+Sn(s)
The Gibbs free energy change for the above reaction at 298K298K298K is x×10−1kJmol−1x \times {10}^{- 1}kJ{mol}^{- 1}x×10−1kJmol−1. The value of xxx is____[nearest integer]
[Given : ECu2+/Cu⊖=0.34V;ESn2+/Sn⊖=−0.14V;F=96500Cmol−1{E}_{{Cu}^{2 +} / Cu}^{⊖} = 0 . 34V; {E}_{{Sn}^{2 +} / Sn}^{⊖} = - 0 . 14V; F = 96500{Cmol}^{- 1}ECu2+/Cu⊖=0.34V;ESn2+/Sn⊖=−0.14V;F=96500Cmol−1]
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Ecell°=ESn2+Sn°−ECu2+Cu°=−0.14−0.34=−0.48{E}_{cell}^{°} = {E}_{{Sn}^{2 +}Sn}^{°} - {E}_{{Cu}^{2 +} Cu}^{°} = - 0 . 14 - 0 . 34 = - 0 . 48Ecell°=ESn2+Sn°−ECu2+Cu°=−0.14−0.34=−0.48
Ecell=Ecell°−0.0592log[Cu2+][Sn2+]{E}_{cell} = {E}_{cell}^{°} - \dfrac{0 . 059}{2}log\dfrac{\left[{Cu}^{2 +}\right]}{\left[{Sn}^{2 +}\right]}Ecell=Ecell°−20.059log[Sn2+][Cu2+]
=−0.48−0.0295[log10−210−3]= - 0 . 48 - 0 . 0295\left[log\dfrac{{10}^{- 2}}{{10}^{- 3}}\right]=−0.48−0.0295[log10−310−2]
=−0.48−0.0295= - 0 . 48 - 0 . 0295=−0.48−0.0295
=−0.5095= - 0 . 5095=−0.5095
ΔG=−nFEcell\Delta G = - {nFE}_{cell}ΔG=−nFEcell
=−2×96500(−0.5095)= - 2 \times 96500\left(- 0 . 5095\right)=−2×96500(−0.5095)
=98333.5=98.33kJ=983.3×10−1kJ= 98333 . 5 = 98 . 33kJ = 983 . 3 \times {10}^{- 1}kJ=98333.5=98.33kJ=983.3×10−1kJ
x=983.3Nearestinteger=983x = 983 . 3Nearestinteger = 983x=983.3Nearestinteger=983
