95 %

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

0.296 g.

For EMF to be +ve, oxidation should take place on electrode (ii), i.e., half-cell reactions will b
Cu⁺+e^- ⟶ Cu
Cu⁺ ⟶ Cu²⁺+e^-
Overall cell reaction: 2 Cu⁺ ⟶ Cu + Cu²⁺
Hence, the cell will be represented as : Cu⁺ I Cu²⁺ I I Cu⁺ I Cu
E⁰_cell = E⁰_Red(RHS) - E⁰_Red(LHS) = 0.52-0.16 = 0.36
Δ_rG⁰ = - n FE^o_cell= - 1 x 96500 C mol^-1 x 0·36 V = - 34740 CV mol^-1 = - 34740 J mol^-1

Zn(s) ➝ Zn²⁺ (aq) + 2e^-
Cu²⁺(aq) + 2e^- ➝ Cu(s)
^{______________________________________}
Zn(s) + Cu²⁺(aq) ➝ Zn²+(aq) + Cu(s)
^{______________________________________}
\(E_{cell}=E^0_{cell}-{0.0591\over 2}log{[Zn^{2+}]\over [Cu^{2+}]}\)
\(=\left(E^0_{cu^{2+}/Cu}-E^0_{Zn^{2+}/Zn}\right)-{0.0591V\over 2}log{2\over 0.5}\)
\(=[+0.34V+0.76]-{0.0591V\over v2}log4\)
\(=+1.10V-{0 0591V\over 2}\times0.6021\)
= 1.10 V - 0.018 V = 1.082 V

\(t=50\times 60=3000s\)
\(I=0.1287A\)
\((a) \ { Cu }^{ 2+ }+{ 2e }^{ - }\longrightarrow Cu(s) \ At \ cathode\)
\((b)(i) Q=I\times t=0.1287\times 3000=386.1C\)
\((ii) \ m=Z\times I\times t=\frac { 63.5 }{ 2\times 96500 } \times 386.1=0.127g\)

(i) \(E^0_{cell}=E^0_{cathode}-E^0_{anode}= - 0·40 V - (- 0·74 V) = + 0·34 V\)
\(Δ_r G^0 - n FE^o_{cell}= - 6 mol \times 96500\ C\ mol.^{-1}\times 0·34 V\)
= -196860 CV mol^-1  = -196860 J mol^-1 = -196.86 k.J mol^-1
- Δ_rG^o = 2.303 RT log K
196860 = 2.303 x 8.314 x 298 log K or log K = 34.5014
K = Antilog 34.5014 = 3.192 x 10³⁴
(ii) E⁰_cell = + 0.80 V - 0.77 V = + 0.03 V.
Δ_r G⁰ = - nF E⁰_cell = - (1 mol) x (96500 C mol^-1) x (0·03 V)
= - 2895 CV mol^-1 = - 2895 J mol^-1
= - 2.895 k.J mol^-1
Δ_rG⁰ = - 2.303 RT log K
- 2895 = - 2.303 x 8.314 x 298 x log K
or log K = 0.5074 or K = Antilog (0.5074) = 3.22.

(a) \(\Lambda_{M}=\frac{1000 \times k}{M}=\frac{\left(1000 \mathrm{~cm}^{3}\right) \times\left(2.0 \times 10^{-2} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\right)}{(0.08 \mathrm{~mol})}=250 \mathrm{\ ohm}^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
(b) \(\Lambda_{M}=\frac{1000 \times k}{M}=\frac{1000}{p \times M}=\frac{\left(1000 \mathrm{~cm}^{3}\right)}{(58 \operatorname{ohm} \mathrm{cm}) \times(0.1 \mathrm{~mol})}=172.41 \mathrm{\ ohm}^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}. \)
Solution (a) has larger molar conductance than solution (b).