Question

Given 2Ce4+ + Co → 2Ce3+ + Co2+, E° - 1.89 V.
The standard reduction potential for Co2+ is
-0.28 V. The standard reduction potential of Ce* to
Ce3+ is
(1) -1.61 V +2 1.61 v
(3) 2.17 V
(4) -2.17 V​

Answer 1

Answer : The correct option is, (2) 1.61 V

Solution :

The balanced cell reaction will be,

$Co(s)+2Ce^{4+}(aq)\rightarrow 2Ce^{3+}(aq)+Co^{2+}(s)$

Here cobalt (Co) undergoes oxidation by loss of electrons, thus act as anode. Cerium (Ce) undergoes reduction by gain of electrons and thus act as cathode.

Given :

$E^o_{[Co^{2+}/Co]}=-0.28V$

$E^o=1.89V$

$E^o_{[Ce^{4+}/Ce^{3+}]}=?$

$E^o=E^o_{[cathode]}-E^o_{[anode]}$

$E^o=E^o_{[Ce^{4+}/Ce^{3+}]}-E^o_{[Co^{2+}/Co]}$

Now put all the values in this expression, we get:

$1.89V=E^o_{[Ce^{4+}/Ce^{3+}]}-(-0.28V)$

$E^o_{[Ce^{4+}/Ce^{3+}]}=1.89-0.28=1.61V$

Therefore, the standard reduction electrode potential of $Ce^{4+}/Ce^{3+}$ is 1.61 V