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?

In the cell

Pt$$\left| {\left( s \right)} \right|$$H₂(g, 1 bar)$$\left| {HCl\left( {aq} \right)} \right|$$AgCl$$\left| {\left( s \right)} \right|$$Ag(s)|Pt(s)

the cell potential is 0.92 V when a 10^–6 molal HCl solution is used. The standard electrode potential of (AgCl/ AgCl^– ) electrode is :
$$\left\{ {} \right.$$Given,  $${{2.303RT} \over F} = 0.06V$$  at  $$\left. {298} \right\}$$

Consider the following reduction processes :
Zn²⁺ + 2e^– $$ \to $$ Zn(s) ; E^o = – 0.76 V
Ca²⁺ + 2e^– $$ \to $$ Ca(s); E^o = –2.87 V
Mg²⁺ + 2e^– $$ \to $$ Mg(s) ; E^o = – 2.36 V
Ni² + 2e^– $$ \to $$ Ni(s) ; E^o = – 0.25
The reducing power of the metals increases in the order :