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 Ce4+ to
Ce 3+ is
(1) -1.61 V
(2) 1.61 V
(3) 2.17 V
(4) -2.17 V​

Answer 1

Answer: (2) 1.61 V

Explanation:

$2Ce^{4+}+Co\rightarrow 2Ce^{3+}+Co^{2+}$

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

$E^0=E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials.

$E^0_{[Ce^{4+}/Ce]}= ?$

$E^0_{[Co^{2+}/Co]}=-0.28V$

$E^0=1.89V$

Putting in the values we get:

$1.89=E^0_{[Ce^{4+}/Ce^{3+}]}- E^0_{[Co^{2+}/Co]}$

$1.89=E^0_{[Ce^{4+}/Ce^{3+}]}-(-0.28V)$

$E^0_{[Ce^{4+}/Ce^{3+}]}=1.61V$

Thus The standard reduction potential of $Ce^{4+}$ to  $Ce^{3+}$ is 1.61 V