$$ \wedge _m^ \circ $$ for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm² mol^–1, respectively. If the conductivity of 0.001 M HA is-

5 $$ \times $$ 10^–5 S cm^–1, degree of dissociation of HA is -

The standard electrode potential $${E^o }$$ and its temperature coefficient $$\left( {{{d{E^o }} \over {dT}}} \right)$$ for a cell are 2V and $$-$$ 5 $$ \times $$ 10^$$-$$4 VK^$$-$$1 at 300 K respectively.
The cell reaction is
Zn(s) + Cu²⁺ (aq) $$\buildrel \, \over \longrightarrow $$ Zn²⁺ (aq) + Cu(s)

The standard reaction enthalpy ($$\Delta $$_rH$${^o }$$) at 300 K in kJ mol^–1 is, [Use R = 8 JK^–1 mol^–1 and F = 96,000C mol^–1]

Given the equilibrium constant:

K_C of the reaction :

Cu(s) + 2Ag⁺ (aq) $$ \to $$ Cu²⁺ (aq) + 2Ag(s) is

10 $$ \times $$ 10¹⁵, calculate the E$$_{cell}^0$$ of this reaciton at 298 K

[2.303 $${{RT} \over F}$$ at 298 K = 0.059V]

For the cell Zn(s) |Zn²⁺ (aq)| |M^x+ (aq)| M(s), different half cells and their standard electrode potentials are given below :

Mx+ (aq)/M(s)	Au3+(aq)/Au(s)	Ag+(aq)/Ag(s)	Fe3+(aq)/Fe2+ (aq)	Fe2+(aq)/Fe(s)
E0Mx+/M/(V)	1.40	0.80	0.77	$$-$$0.44

If $$E_{z{n^{2 + }}/zn}^0$$ = $$-$$ 0.76 V, which cathode will give maximum value of E^o_cell per electron transferred?