For Daniell cell, Zn + Cu²⁺⇾ Zn²⁺+ Cu
\(E_{cell}=E^0_{cell}-{0.0591\over 2}log{[Zn^{2+}]\over [Cu^{2+}]}\)
It is the equation of straight line (y = C + mx).
Intercept = E^O_cell= 1·10 V (Given)
\(E_{cell}\equiv 1.10-{0.0591\over 2}log{0.1\over 0.01}=1.10-0.0295=1.0705V\)

The two cells may be represented as
Zn I Zn²⁺(cone = c) II Cu²⁺(c = ?) I Cu, EMF = E₁
Zn I Zn²⁺(cone = c) II Cu²⁺(0·5 M) I Cu, EMF = E₂
The cell reaction is : Zn + Cu²⁺ ⇾ Zn²⁺+ Cu
\(E_1=E^0-{2.303\ RT\over 2F}log{c\over [Cu^{2+}]}\)
\(E_2=E^0-{2.303\ RT\over 2F}log{c\over 0.5}\)
\(E_2-E_1={2.303\ RT\over 2F}\left( log{c\over [Cu^{2+}]}-log{c\over 0.5}\right)=0.03V\)
\({0.06\over 2}log{0.5\over [Cu^{2+}]}=0.03\ or\ log{0.5\over [Cu^{2+}]}=1\ or\ {0.5\over {[Cu^{2+}]}}=10\ or\ [Cu^{2+}]={0.5\over 10}=0.05M\)

Quantity of electricity passed = 9·65 x 60 x 60 C = 34740 C
∴ Ag⁺ ions deposited = 34740/96500 mole = 0.36 mole
Cu²⁺ions deposited = 34740/(2 x 96500) = 0.18 mole
[Ag⁺] left = 1 - 0.36 = 0.64 M
[Cu²⁺] left = 1 - 0.18 = 0.82 M
Cell reaction is: Cu + 2 Ag⁺ ➝ Cu²⁺+ 2 Ag
\(ΔE=E^0_{cell}-E_{cell}={0.0591\over 2}log{[Cu^{2+}]\over [Ag^+]^2}={0.0591\over 2}log{0.82\over (0.64)^2}=8.89\times10^{-3}V\)

\(E_{cell}={0.0591\over n}log{c_2\over c_1}\)
\(0.164={0.0591\over 1}log{0.1\over _1}\ or\ log{0.1\over c_1}=2.7750\ or\ 0.1/c_1=5.957\times10^2c_1=1.679\times10^{-4}\)
i.e., [Ag⁺] in saturated Ag₂CrO₄ = 1.679 x 10^-4M
Ag₂CrO₄ ⇌ 2 Ag⁺ +/CrO₄^2-
\([Ag^+]=1.679\times10^{-4}M,\ [CrO^{2-}_4]={[Ag^+]\over 2}={1.679\times10^{-4}\over 2}M\)
\(K_{sp}=[Ag^+]^2[CrO_4^{2-}]=(1.679\times10^{-4})^2\times\left(1.6798\times10^{-4}\over 2\right)=2.37\times10^{-12}\)

Cu⁺ + e^-⇾ Cu; ∆G^o₃=?
(i)------(ii) gives the required result, i.e.,ΔG₃^° = ∆G⁰₁ - ΔG₂^° = [-0.674 - (- 0·153)] F = - 0·521 F
\(-nFE^0_{Cu^+/CU}=-0.5621F\ or\ E^0_{CU^+/Cu}=0.521V\)

Step I
Calculation of the quantity of charge passed:
We know that,
Watt = Ampere x Volt
Ampere =\(\frac { Watt }{ Volt } =\frac { 100 }{ 110 } \)
Now charge = Current x Time
=\(\left( \frac { 100 }{ 110 } amp \right) \times 4\times 60\times 60\)
= 13091 C
Step II
Calculation of mass of Cadmium deposited:
The cathodic reaction is
Cd²⁺_(aq)+2e^-\(\rightarrow\)Cd_(s)
112.2g 2 x 96500C
2 x 96500 C of charge deposited Cd = 112.2 g
13091 C charge will deposite Cd
\(\frac { 112.2g }{ 2\times 96500C } \times 13091C\)
= 7.61g

Cell reaction is
\({ Fe }_{ (s) }+{ 2{ H }^{ + } }_{ (aq) }\rightarrow { { Fe }^{ 2+ } }_{ (aq) }+{ H }_{ 2(g) }\)
\({ { E }^{ ° } }_{ cell }=0.00-(-0.44)=0.44V\)
\({ E }_{ cell }={ { E }^{ ° } }_{ cell }-\frac { 0.0591 }{ 2 } log\frac { \left[ { Fe }^{ 2+ } \right] }{ { \left[ { H }^{ + } \right] }^{ 2 } } \)
\(=0.44-\frac { 0.0591 }{ 2 } log\frac { 0.001 }{ { \left( 0.01 \right) }^{ 2 } } \)
=0.44 - 0.02955 = 0.41045 V

K = 5.25 x 10^-5 S cm^-1, M = 2.5 x 10^-4 M.
\(\lambda_{\mathrm{HCOOH}}^{0}=\lambda_{\left(\mathrm{HCOO}^{-}\right)}^{0}+\lambda_{\mathrm{H}}^{0}+\) 
= 349.5 + 50.5 : 400 S cm² mol^-1.
\(\Lambda_{m}=\frac{1000 \kappa}{\mathrm{M}}=\frac{1000 \times 5.25 \times 10^{-5}}{2.5 \times 10^{-4}}\) 
= \(\frac{1000 \times 525}{10 \times 2.5 \times 100}\) 
\(\Rightarrow \quad \Lambda_{m}=\frac{525}{2.5}=\frac{5250}{25}=210 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) 
\(\alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{0}}=\frac{210}{400}=\frac{21}{40}=0.525\) 
\(\Rightarrow \ \alpha\) =  0.525 x 100% = 52.5%

(a) \(\Delta_{r} G^{\circ}=-n \mathrm{E}^{\circ} \mathrm{F}\) 
= -2 x 2.71 V x 96500 C
= - 523030 J
= - 523.030 kJ mol^-1

(b) H₂ -O₂ Fuel Cell was used in Apollo space programme for providing electrical power.

\(\begin{aligned} \mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \longrightarrow & \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(s) \end{aligned}\)
\(\begin{aligned} E^{\circ} & =+1.1 \mathrm{~V}, \Delta G^{\circ}=? \end{aligned}\)
\(\begin{aligned} 1 \mathrm{~F} & =96500 \space \mathrm{C} \space \mathrm{mol}^{-1}, n=2 \end{aligned}\)
\(\begin{aligned} \Delta G^{\circ} & =-n F E^{\circ} \end{aligned}\)
\(\begin{aligned} =-2 \times 96500 \times 1.1=-212300 \mathrm{~J} \mathrm{~mol}^{-1} \end{aligned}\)