Q12. Standard electrode potential of \( Ni ^{2+} / \) Ni electrode, if the cell potential of the cell \( Ni \mid Ni ^{2+} \) \( Ni \mid Ni ^{2+}(0.01 M ) \| Cu ^{2+}(0.1 M ) \) | Cu is \( 0.59 v \) given : \( E ^{0} cu ^{2+} / cu =0.34 V \) is (a) \( -0.2205 V \) (b) \( 0.3205 V \) (c) \( +0.014 V \) (d) \( -0.34 V \)

Question

 Standard electrode potential of \( Ni ^{2+} / \) Ni electrode, if the cell potential of the cell \( Ni \mid Ni ^{2+} \) \( Ni \mid Ni ^{2+}(0.01 M ) \| Cu ^{2+}(0.1 M ) \) | Cu is \( 0.59 v \) given : \( E ^{0} cu ^{2+} / cu =0.34 V \) is (a) \( -0.2205 V \) (b) \( 0.3205 V \) (c) \( +0.014 V \) (d) \( -0.34 V \)

Answer 1

Fort given cell

Ni(B) → Ni²⁺ + 2e^-

Cu²⁺ + 2e^- → Cu(B)

Ni(B) + Cu²⁺ → Cu(B) + Ni²⁺

E_Cell = E⁰_cell - \(\frac{0.0591}{2}\) log \(\Big[\frac{Ni^{2+}}{cu^{2+}}\Big]\) 

0.059 = E⁰_cell  - \(\frac{0.0591}{2}\) log \(\Big(\frac{0.01}{0.1}\Big)\)

0.059 = E⁰_cell + 0.02955

E_cell = 0.059 - 0.02955

E⁰_cell = 0.0295V