Calculate the standard cell potential in (V) of the cell in which following reaction takes place :

Fe²⁺(aq) + Ag⁺(aq) $$ \to $$ Fe³⁺(aq) + Ag (s)

Given that

$$E_{A{g^ + }/Ag}^o = xV$$

$$E_{Fe^{2+ }/Fe}^o = yV$$

$$E_{Fe^{3+ }/Fe}^o = zV$$

Given that $${E^\Theta }_{{O_2}/{H_2}O} = 1.23\,V$$ ;

$${E^\Theta }_{{S_2}O_8^{2 - }/SO_4^{2 - }} = 2.05\,V$$

$${E^\Theta }_{B{r_2}/B{r^ - }} = 1.09\,V$$

$${E^\Theta }_{A{u^{3 + }}/Au} = 1.4\,V$$

The strongest oxidizing agent is :

$$ \wedge _m^ \circ $$ 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]