Question

The emf of the cell:
Ni / Ni2+ (1.0 M) // Au3+ (1.0 M) / Au (E° = -0.25 V for Ni2+/Ni; E° = 1.5 V for Au3+/Au) is
(a) 1.25 V
(b) -1.25 V
(c) 1.75 V
(d) 2.0 V

Answer 1

Answer:

Option (c) is the right answer

Explanation:

Refer attachments for explanation!!!

Answer 2

Answer:

The emf of the given cell is equal to 1.75V.

Therefore, the option (c) is correct.

Explanation:

We have given, the cell representation from which, we can write the half cell reactions:

At anode (oxidation) :

Ni   $\longrightarrow$    Ni²⁺  +  2e⁻              ;    E⁰ (anode) = -0.25V            ..............(1)

At cathode (Reduction):

Au³⁺   +  3e⁻  $\longrightarrow$  Au              ; E⁰ (cathode) = 1.5V                 ...............(2)

Now multiply the equation (1) by 3 and equation (2) by 2, then add:

The overall cell reaction is:

3Ni  +  2Au³⁺ (aq)     $\longrightarrow$   Au  +  Ni²⁺ (aq)

$E^o_{cell}= E^o_{cathode}- E_{anode}^{o}$

$E^o_{cell}= 1.5-(-0.25)$

$E^o_{cell}=1.75V$

From the Nernst equation:

$E_{cell}=E^o_{cell}-\frac{0.059}{n} log\frac{[Ni^{2+}]^3}{[Au^{3+}]^2}$

where 'n' is no. of electrons gained or loss in cell reaction.

Given, [Ni²⁺] = 1.0M and [Au³⁺] = 1.0M and n = 6

$E_{cell}=1.75-\frac{0.059}{6}log\frac{(1.0)^3}{(1.0)^2}$

$E_{cell}=1.75-\frac{0.059}{6}log1$

$E_{cell}=1.75-0$                                          [ ∵  log1 =0]

$E_{cell}= 1.75V$

Therefore, the e.m.f. of cell is 1.75Volt.