\({ \Lambda }_{ m }=\frac { 1000\times k }{ M } =\frac { 1000\times 0.025\times { 10 }^{ -2 } }{ 0.20 } =125 \ S{ cm }^{ 2 }{ mol }^{ -1 }\)

It means zinc is more reactive than hydrogen, therefore, acts as anode when coupled to S.H.E and gets oxidised to Zn²⁺ (aq) and H⁺ will get reduced to H₂ (g).

At the stage of chemical equilibrium in the cell.

No but \(\mathrm{E}_{\text {cell }} \text { or } \triangle_r G^o\) of reaction can be zero at equilibrium.

\(\mathrm{Cr}(\mathrm{s})\left|\mathrm{Cr}^{3+}(\mathrm{aq}) \| \mathrm{Cd}^{2+}(\mathrm{aq})\right| \mathrm{Cd}(\mathrm{s})\)

\( \Lambda_m^{\infty} N a_2 S O_4=2 \lambda_m^{\infty} N a^{+}+\lambda_m^{\infty} S O_4^{2-} or\\ \Lambda^{\circ} \mathrm{Na}_2 \mathrm{SO}_4=2 \lambda^{\circ} \mathrm{Na}^{+}+\lambda^{\circ} \mathrm{SO}_4^{2-} \)

\(\mathrm{CuSO}_4 \rightarrow \mathrm{Cu}^{2+}+\mathrm{SO}_4^{2-}\)
\( \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}^{+}+\mathrm{OH}^{-} \)
\( \mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu}(\mathrm{s}) , At\ anode 2 \mathrm{OH}^{-} \rightarrow \mathrm{O}_2+4 \mathrm{H}^{+}+4 e^{-} or 2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{O}_2+4 \mathrm{H}^{+}+4 e^{-}\)
Copper is formed at cathode and oxygen gas is liberated at anode

\(\mathrm{E}^{\circ} \mathrm{Zn}^{2+} / \mathrm{Zn}=-0.76 \mathrm{~V}\) because it acts as anode when coupled with SHE, i.e. it undergoes oxidation, therefore, oxidation potential is + ve and sign of reduction potential will be negative.

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

\(K=1.29\times { 10 }^{ -2 }{ ohm }^{ -1 }{ cm }^{ -1 }\)
\(K=\frac { 1 }{ R } \times \frac { 1 }{ a } \Rightarrow \frac { 1 }{ a } =K\times R=1.29\times { 10 }^{ -2 }\times 85=109.65\times { 10 }^{ -2 }=1.0965{ cm }^{ -1 }\)
\({ \wedge }_{ m }=\frac { 1000K }{ M } =\frac { 1000 }{ M } \times \frac { 1 }{ R } \times \frac { 1 }{ a } \)
\({ \wedge }_{ m }=\frac { 1000\times 1\times 1.0965 }{ 0.052\times 96 } =\frac { 1096.50 }{ 4.992 } =219.65 \ S \ { cm }^{ 2 }{ mol }^{ -1 }\)

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
\(E_1=E^0-{2.303\ RT\over 2F}log{c\over [Cu^{2+}]}\)
\(E_2=E^0-{2.303\ RT\over 2F}log{c\over 0.5}\)
\(E_2-E_1={2.303\ RT\over 2F}\left( log{c\over [Cu^{2+}]}-log{c\over 0.5}\right)=0.03V\)
\({0.06\over 2}log{0.5\over [Cu^{2+}]}=0.03\ or\ log{0.5\over [Cu^{2+}]}=1\ or\ {0.5\over {[Cu^{2+}]}}=10\ or\ [Cu^{2+}]={0.5\over 10}=0.05M\)