1. Tardigrade
  2. Question
  3. Chemistry
  4. Consider the following cell reaction: Cd ( s )+ Hg 2 SO 4( s )+(9/5) H 2 O (l) leftharpoons CdSO 4 ⋅ (9/5) H 2 O (s)+2 Hg (l) The value of E text cell 0 is 4.315 V at 25° C. If Δ H °=-825.2 kJ mol -1, the standard entropy change Δ S ° in J K -1 is (Nearest integer) [. Given: Faraday constant .=96487 C mol -1]

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Q. Consider the following cell reaction:
$Cd _{( s )}+ Hg _{2} SO _{4( s )}+\frac{9}{5} H _{2} O _{(l)} \rightleftharpoons CdSO _{4} \cdot \frac{9}{5} H _{2} O _{(s)}+2 Hg _{(l)}$
The value of $E _{\text {cell }}^{0}$ is $4.315\, V$ at $25^{\circ} C$. If $\Delta H ^{\circ}=-825.2 \, kJ \, mol ^{-1}$, the standard entropy change $\Delta S ^{\circ}$ in $J \, K ^{-1}$ is ______ (Nearest integer) $\left[\right.$ Given : Faraday constant $\left.=96487 \, C\, mol ^{-1}\right]$

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Solution:

$\Delta G ^{\circ}=- nFE ^{\circ}=\Delta H ^{\circ}- T \Delta S ^{\circ}$
$=\frac{\Delta H ^{\circ}+ nFE ^{\circ}}{ T }$
$=\frac{\left(-825.2 \times 10^{3}\right)+(2 \times 96487 \times 4.315)}{298}$
$=\frac{-825.2 \times 10^{3}+832.682 \times 10^{3}}{298}$
$=\frac{7.483 \times 10^{3}}{298}=25.11 \,JK ^{-1}\, mol ^{-1}$
$=\frac{7.483 \times 10^{3}}{298}=25.11 \,JK ^{-1} \,mol ^{-1}$
$\therefore $ Nearest integer answer is $25$