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Ion Electron Method

Calculate EMF...

Question

Calculate EMF of the cell given: Zn(s) / Zn²⁺ (0.1M) // Pb²⁺ (0.02M) / Pb(s). Also write the chemical reaction at the anode & cathode. (E⁰ Zn²⁺/Zn = -0.76V, E⁰ Pb²⁺/Pb = -0.13V)

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Solution

$T h e g i v e n c e l l i s Z n / Z n^{2 +} / / P b^{2 +} / P b A n o d e : Z n \to Z n^{2 +} + 2 e^{-} C a t h o d e P b^{2 +} + 2 e^{-} \to P b T h e o v e r a l l r e a c t i o n i s Z n + P b^{2 +} \to P b + Z n^{2 +} E^{0} = {E^{0}}_{o x} + {E^{0}}_{r e d} = E_{Z n / Z n^{2 +}}^{0} + E_{P b^{2 +} / P b}^{0} = 0.76 - 0.13 = 0.63 V$

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Similar questions

Zn {(N O_{3})}_{2}

Pb {(N O_{3})}_{2}

E_{Z n^{2 +}, Z n}^{\circ} = - 0.76 V, E_{P b^{2 +}, P b}^{\circ} = 0.13 V

E_{(P b^{2 +} | P b)}^{⊖} = - 0.126 V, E_{(Z n^{2 +} | Z n)}^{⊖} = - 0.763 V

25^{\circ} C

Z n | Z n^{2 +} (1 M) | | P b^{2 +}

P b^{2 +}

P b^{2 +}

2.5 \times 10^{- x} M

Z n | {Z n}^{2 +} (0.1 M) ∥ {P b}^{2 +} (1 M) | P b

Z n \to Z n^{2 +} + 2 e^{-} (E_{(Z n^{2 +} / Z n)}^{0} = - 0.76 V)

F e \to F e^{2 +} + 2 e^{-} E^{0} = 0.41 V

F e^{2 +} + Z n \to Z n^{2 +} + F e

E^{0}

E

S n | {S n}^{2 +} (1 M) ∥ {P b}^{2 +} (10^{- 3} M) | P b, E^{0} ({S n}^{2 +} / S n) = - 0.14 V, E^{0} ({P b}^{2 +} / P b) = - 0.13 V

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