Question

The emf of the cell,

$\text{Ni }|\text{ N}\text{i}^{\text{2+}}\ (\text{1.0 M})\ ||\text{ A}\text{g}^{+}\ (\text{1.0M})\ |\text{ Ag }\lbrack E^{0}\text{ for N}\text{i}^{\text{2+}}\text{ / Ni = - 0.25}$

volt, E° for Ag+/Ag = 0.80 volt] is given by

• A. –0.25 + 0.80 = 0.55 volt
• B. –0.25 – (+0.80) = –1.05 volt
• C. 0 + 0.80 – (–­0.25) = + 1.05 volt
• D. –0.80 – (–0.25) = – 0.55 vol

Answer 1

Answer: (C)

0 + 0.80 – (–­0.25) = + 1.05 volt

Answer 2

$E_{cell} = E_{Ni/N2 +}^{0} +$= 0.25 + 0.80 = 1.05 Volt