Question

Iit question mate solve .......​

Answer 1

Answer:

See attachment mate .........

Answer 2

${\bold{\huge{\mathcal{\red{SOLUTION}}}}}$

The above cell reaction is as follows,

Zn(s) + Ni2+(aq) ⇌ Zn2+(aq)+ Ni(s)

Cell can be represented as,

Zn(s) | Zn2+(aq) || Ni2+(aq) | Ni(s)

Given, Standard Cell Potentials,

E0Zn | Zn2+ = -0.75V

& E0Ni2+ | Ni = -0.24V

E0cell = E0cathode - E0anode

= 0.75 - 0.24 = 0.51V

${\large{\bold{\tt{\green{Nerst \:equation}}}}}$

can be written as,

From the cell reaction depicted above,

Zn(s) + Ni2+(aq) ⇌ Zn2+(aq)+ Ni(s)

1 - x x

keq = [Zn2+] / [Ni2+]

= x / (1-x)

Hence, concentration of Ni2+ in solution at equilibrium, [Ni2+] = 1 / 1.78 x 1017 = 5.6 x 10-8 M