Question

EMF of the given cell
$A _{( s )}\left| A _{( aq )}^{2+} \| B _{( aq )}^{2+}\right| B _{( s )}$
Given $E_{ A / A ^{2+}}^{\circ}:+1.4 V$ and $E _{ B / B ^{2+}}^{\circ}:-1.4 V$

• A. $2.8 V$
• B. $1.8 V$
• C. $0 V$
• D. $-1.8 V$

Answer 1

Answer: (A)

$E _{\text {cell }}^{0} =\left( E _{\text {cathode }}^{0}\right)_{ SRP }-\left( E _{\text {anode }}^{0}\right)_{ SRP } $
$=\left( E _{\text {anode }}^{0}\right)_{ SOP }-\left( E _{\text {cathode }}^{0}\right)_{ SOP }$
Where SRP is standard reduction potential and SOP is standard oxidation potential
$\therefore E_{ Cell } =[1.4-(-1.4)] V $
$=+2.8 \,V$