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Nernst Equation

The emf of a ...

Question

The emf of a galvanic cell, with electrode potentials of $Z n = + 0.76 V$ , is:

- 1.1 V

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+ 1.1 V

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+ 0.34 V

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+ 0.76 V

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Solution

The correct option is D $+ 1.1 V$
Left electrode : $Z n (s) \to Z n^{2 +} + 2 e^{-}$ .
Right electrode : $C u^{2 +} (a q) + 2 e^{-} \to C u (s)$

The overall reaction of the cell is the sum of above two reactions:

$Z n (s) + C u^{2 +} (a q) ⟶ Z n^{2 +} (a q) + C u (s)$

Emf of the cell $, E^{o} c e l l = E^{o} (c a t h o d e) - E^{o} (a n o d e)$

Emf of the cell $= 0.34 V - (- 0.76) V = + 1.1 V$

Hence, option B is correct.

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Similar questions

{Z n}^{2 +} | Z n (- 0.76 V)

{F e}^{2 +} | F e (- 0.41 V)

{Z n}^{2 +} | Z n

- 0.76 V

{C u}^{2 +} | C u

0.34 V

(V)

(k J {m o l}^{- 1})

Z n^{2 +} | Z n

C u^{2 +} | C u

Z n | Z n^{2 +} (0.05 M) | | C u^{2 +} (0.005 M) | C u

\frac{R T}{F} = 0.059

Z n^{2 +} + 2 e^{-} \to Z n; E = - 0.76 V

F e^{2 +} + 2 e^{-} \to F e; E = - 0.44 V

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

Z n ∣ ∣ {Z n}^{2 +} (1 M) ∣ ∣ ∣ ∣ {C u}^{2 +} (1 M) ∣ ∣ C u

E_{{C u}^{2 +} / C u}^{o} = 0.34 V

E_{{z n}^{2 +} / Z n}^{o} = - 0.76 V

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