${\mathrm{E}}_{\mathrm{cell}}^0$  is 1.89V. For the cell reaction $2 {\mathrm{Ce}}^{4+}+\mathrm{Co}\to 2{\mathrm{Ce}}^{3+ }+ {\mathrm{Co}}^{2+}$. If ${\mathrm{E}}_{{\mathrm{Co}}^{2+}/\mathrm{Co}}$  is – 0.28V. The value of ${\mathrm{E}}_{{\mathrm{Cu}}^{4+}/{\mathrm{Ce}}^{3+}}^0$ is

For the given reaction $2{\mathrm{Ce}}^{4+}+\mathrm{Co}\to 2{\mathrm{Ce}}^{3+}+{\mathrm{Co}}^{2+}$
We know that,
Oxidation takes place at anode i.e. $\mathrm{Co}\to {\mathrm{Co}}^{2+}+2{\mathrm{e}}^{-}$
Reduction takes place at cathode i.e. $2{\mathrm{Ce}}^{4+}+2{\mathrm{e}}^{-}\to {\mathrm{Ce}}^{3+}$
We have to calculate Standard reduction potential for ${\mathrm{Ce}}^{4+}/{\mathrm{Ce}}^{3+} \mathrm{i}.\mathrm{e}. {\mathrm{E}}_{\mathrm{RP}}^0=?$
Now for ${\mathrm{E}}_{\mathrm{RP}}^0$ we have to add oxidation potential with reduction potential
It means  ${\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{ox}}+{\mathrm{E}}_{\mathrm{red}}$

By putting values in equation we get,
$1.89=-(-0.28)+{\mathrm{E}}_{{\mathrm{Ce}}^{4+}/{\mathrm{Ce}}^{3+}}$
${\mathrm{E}}_{{\mathrm{Ce}}^{4+}/{\mathrm{Ce}}^{3+}}=1.89-0.28=1.61\mathrm{V}$

${\mathrm{E}}_{{\mathrm{Ce}}^{4+}/{\mathrm{Ce}}^{3+}}=1.61\mathrm{V}$
Hence option B is correct.