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Home » Calculate the standard cell potentials of galvanic cell in which the following reactions take place: (i) 2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd (ii) Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s) Calculate the ∆rGJ and equilibrium constant of the reactions.
Calculate the standard cell potentials of galvanic cell in which the following reactions take place: (i) 2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd (ii) Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s) Calculate the ∆rGJ and equilibrium constant of the reactions.
CBSE | Chemistry | Class 12 | Electrochemistry | NCERT
Calculate the standard cell potentials of galvanic cell in which the following reactions take place: (i) 2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd (ii) Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s) Calculate the ∆rGJ and equilibrium constant of the reactions.

Solution:

(i) Given: E_{C r^{3+} / C r}^{\Theta}=0.74 V

    \[E_{C d^{2+} / C d}^{\Theta}=-0.40 V\]


The galvanic cell of the given reaction is represented as :

    \[C r_{(s)}\left|C r_{(a q)}^{3+} | C d_{a q}^{2+}\right| C d_{(s)}\]


Now, the standard cell potential is

    \[\begin{aligned}&E_{c e l l}^{\Theta}=E_{g}^{\Theta}-E_{L}^{\Theta} \&=-0.40-(-0.74) \&=+0.34 V\end{aligned}\]

    \[E_{c e l l}^{\Theta}=+0.34 V\]


In the given equation, n=6

    \[F =96487 C mol ^{-1}\]



Then substuiting the value of parameters in the formula we get, \Delta_{r} G^{\Theta}=-6 \times 96487 C mol ^{-1} \times 0.34 V

    \[\begin{aligned}&=-196833.48 CV mol -1 \&=-196833.48 J mol -1 \&=-196.83 kJ mol -1\end{aligned}\]

    \[\begin{aligned}&\Delta_{r} G^{\Theta}=-R T l n K \quad \Delta_{r} G^{\Theta}=-2.303 R T l n K \log k=\frac{\Delta_{r} G}{2.303 R T}=\frac{-196.83 \times 10^{3}}{2.303 \times 8.314 \times 298} \&=34.496 \& K =\operatorname{antilog}(34.496)=3.13 \times 10^{34}\end{aligned}\]


ii) The galvanic cell of the given reaction is represented as follows:

    \[F e_{(a q)}^{2+}\left|F e_{(a q)}^{3+} | A g_{(a q)}^{+}\right| A g_{(s)}\]


Now, the standard cell potential is

    \[E_{c e l l}^{\Theta}=E_{g}^{\Theta}-E_{L}^{\Theta}\]


Here, n=1
Then substuiting we have, \Delta_{t} G^{0}=-n F E_{\text {cell }}^{0}

    \[\begin{aligned}&=-1 \times 96487 C mol ^{-1} \times 0.03 V \&=-2894.61 J mol ^{-1} \&=-2.89 kJ mol ^{-1}\end{aligned}\]


Again, \Delta_{t} G^{0}=-2.303 R T \ln K \ln K=\frac{\Delta_{t} G}{2.303 R T}=\frac{-2894.61}{2.303 \times 8.314 \times 298}
=0.5073
K =\operatorname{antilog}(0.5073)
=3.2 (approximately)

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