$E^*{ }_{\mathrm{Ag}^{+} / \mathrm{Ag}}=0.80 \mathrm{~V}$ and $E^{\circ}{ }_{\mathrm{NI}^{+} / \math

Question

$E^*{ }_{\mathrm{Ag}^{+} / \mathrm{Ag}}=0.80 \mathrm{~V}$ and $E^{\circ}{ }_{\mathrm{NI}^{+} / \mathrm{Ni}}=-0.25 \mathrm{~V}$. The EMF of the cell $\mathrm{Ni}-\mathrm{Ag}$ is

Professor PC Thomas Classes · Answer

$E^\circ_{	ext{cell}} = E^\circ_{	ext{cathode}} - E^\circ_{	ext{anode}} = 0.80 - (-0.25) = 1.05	ext{ V}$.