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Question: The standard electrode potentials of the two half calls are given below \(Ni^{2 +} + 2e^{-}\)**⇌**\...

The standard electrode potentials of the two half calls are given below

Ni2++2e−Ni^{2 +} + 2e^{-}⇌Ni;E0=−0.25voltNi;E^{0} = - 0.25volt;

Zn2++2e−Zn^{2 +} + 2e^{-}⇌Zn;E0=−0.77voltZn;E^{0} = - 0.77volt

The voltage of cell formed by combining the two half - cells would be

A

– 1.02 V

B
  • 0.52 V
C
  • 1.02 V
D

– 0.52 V

Answer
  • 0.52 V
Explanation

Solution

Ni2++2e−Ni^{2 +} + 2e^{-}⇌ Ni;Eo=−0.25volt\mathbf{Ni;}\mathbf{E}^{\mathbf{o}}\mathbf{=}\mathbf{-}\mathbf{0.25volt}

Zn2++2e−\mathbf{Z}\mathbf{n}^{\mathbf{2 +}}\mathbf{+ 2}\mathbf{e}^{\mathbf{-}}⇌ Zn;E0=−0.77volt\mathbf{Zn;}\mathbf{E}^{\mathbf{0}}\mathbf{=}\mathbf{-}\mathbf{0.77volt}

Ecell=\mathbf{E}_{\mathbf{cell}}\mathbf{=}Reduction potential of cathode – Reduction potential of anode =−0.25−(−0.77)= - 0.25 - ( - 0.77)

=−0.25+0.77=0.52- 0.25 + 0.77 = 0.52V