(a) Two half-reactions of an electrochemical cell are given below :
$\begin{array}{l}{\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+8{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+5{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{I}\right),{\mathrm{E}}^{\circ }=1.51\mathrm{V}\\ {\mathrm{Sn}}^{2+}\left(\mathrm{aq}\right)\to {\mathrm{Sn}}^{4+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-},{\mathrm{E}}^{\circ }=+0.15\mathrm{V}\end{array}$
Construct the redox equation from the standard potential of the cell and predict if the reaction is reactant favored or product favored.
(b) The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm² mol^-1. Calculate the conductivity of this solution. 

(c) The standard electrode potential (E°) for Daniel cell is +1.1 V. Calculate the ΔG° for the reaction
$\begin{array}{r}\mathrm{Zn}\left(\mathrm{s}\right)+{\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)\to {\mathrm{Zn}}^{2+}\left(\mathrm{aq}\right)+\mathrm{Cu}\left(\mathrm{s}\right)\\ \left(1\mathrm{F}=96500\mathrm{C}{\mathrm{mol}}^{-1}\right)\end{array}$
OR 

Define the following terms : (a) (i) Limiting molar conductivity (ii) Fuel cell (b) Resistance of a conductivity cell filled with 0.1 mol L^-1 KC1 solution is 100 Q. If the resistance of the same cell when filled with 0.02 mol L^-1 KC1 solution is 520 Q, calculate the conductivity and molar conductivity of 0.02 mol L KC1 solution. The conductivity of 0.1 mol L 1.29 X 10^-2 CT1 cm^-1

(a)

EMF of the cell is positive reaction is product favoring.

${\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cathode}}^{°}-{\mathrm{E}}_{\mathrm{anode}}^{°}$

=1.51-(0.15)=1.36 v .Spontaneous and product favored.

(b)${\mathrm{\lambda }}_{\mathrm{M}}=\frac{\mathrm{\kappa }\times 1000}{\mathrm{M}}$

$\mathrm{\kappa }=\frac{{\mathrm{\lambda }}_{\mathrm{M}}\times \mathrm{M}}{1000}=\frac{138.9\times 1.5}{1000}=208.35\times {10}^{-3}$ S/cm

(c)$∆{\mathrm{G}}^{\circ }=-{\mathrm{nFE}}^0=-2\times 96500\times 1.1 \mathrm{J}$ =212.3 kJ

OR

i) When a concentration of an electrolyte approaches zero, then its molar conductivity is known as limiting molar conductivity.

(ii) Fuel cells are the galvanic cells in which the energy of combustion of the fuels likes hydrogen, methanol. etc is directly convert into electrical energy.

(B) Given that:

Concentration of the KCl solution = 0.1 mol L⁻¹

Resistance of cell filled with 0.1 mol L⁻¹ KCl solution = 100 ohm

Cell constant  = G* = conductivity × resistance

1.29×10⁻²ohm−1cm−1×100ohm=1.29cm⁻¹=129m⁻¹

Cell constant for a particular conductivity cell is constant.

Conductivity of 0.02 mol L⁻¹ KCl solution = Cell constant G*/ Resistance R=129m⁻¹ /520ohm​=0.248Sm⁻¹

Concentration = 0.02mol l⁻¹

=1000×0.02molm⁻³=20molm⁻³

Now,

Molar conductivity =Λ_m​=k​ /c=248×10⁻³Sm⁻¹​/ 20molm⁻³=124×10⁻⁴Sm²mol⁻¹

Therefore the molar conductivity of 0.02 molL−1KCl solution is 124×10⁻⁴Sm²mol⁻¹