Question 13 Marks
The equilibrium constant of cell reaction :
$Sn^{4+}(.aq) + Al(s) Al^{3+ }\rightarrow + Sn^{2+} (aq)$ is $4.617 \times 10^{184}, at 25 ^\circ C$
a. Calculate the standard emf of the cell. $($Given: $\log 4.617 \times 10^{184} = 184.6644)$
b. What will be the $E^\circ$ of the half cell $Al^{3+}/Al, if E^\circ$ of half cell $Sn^{4+}/Sn^{2+}$ is $4+ 0.15 V.$
$Sn^{4+}(.aq) + Al(s) Al^{3+ }\rightarrow + Sn^{2+} (aq)$ is $4.617 \times 10^{184}, at 25 ^\circ C$
a. Calculate the standard emf of the cell. $($Given: $\log 4.617 \times 10^{184} = 184.6644)$
b. What will be the $E^\circ$ of the half cell $Al^{3+}/Al, if E^\circ$ of half cell $Sn^{4+}/Sn^{2+}$ is $4+ 0.15 V.$
Answer
View full question & answer→$\mathrm{E}_{\text {Cell }}=\mathrm{E}_{\text {Cell }}^o-\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \cdot \log \mathrm{Kc}$
At $298 K$
|$\mathrm{E}_{\text {Cell }}=\mathrm{E}^{\mathrm{o}}{ }_{\text {Cell }}-\frac{0.0591}{\mathrm{n}} \log \mathrm{Kc}$
At equilibrium Ecell $=0, n=6$
$E_{\text {Cell }}^o =\frac{0.0591}{n} \operatorname{\log~Kc}$
$ =0.059 / 6 \log 4.617 \times 10^{184}$
$ =0.00983 \times 184.6644$
$ =1.8152$
$(ii) \mathrm{E}_{\text {cell }}^0=\mathrm{E}_{\mathrm{Sn}_{n 4+/ S n 2+}}-\mathrm{E}^0{ }_{\mathrm{Al} 3+/ \mathrm{Al}}$
$1.81=-0.15-\mathrm{E}^0{ }^{\mathrm{Al} 3}+\mathrm{Al}$
$ \mathrm{E}_{\mathrm{AA}^0+/ \mathrm{Al}}=-1.66 \mathrm{~V}$
At $298 K$
|$\mathrm{E}_{\text {Cell }}=\mathrm{E}^{\mathrm{o}}{ }_{\text {Cell }}-\frac{0.0591}{\mathrm{n}} \log \mathrm{Kc}$
At equilibrium Ecell $=0, n=6$
$E_{\text {Cell }}^o =\frac{0.0591}{n} \operatorname{\log~Kc}$
$ =0.059 / 6 \log 4.617 \times 10^{184}$
$ =0.00983 \times 184.6644$
$ =1.8152$
$(ii) \mathrm{E}_{\text {cell }}^0=\mathrm{E}_{\mathrm{Sn}_{n 4+/ S n 2+}}-\mathrm{E}^0{ }_{\mathrm{Al} 3+/ \mathrm{Al}}$
$1.81=-0.15-\mathrm{E}^0{ }^{\mathrm{Al} 3}+\mathrm{Al}$
$ \mathrm{E}_{\mathrm{AA}^0+/ \mathrm{Al}}=-1.66 \mathrm{~V}$





