Calculate the emf of the following cell at $25^{o} C$ $A g (s) ∣ ∣ {A g}^{+} (10^{- 3} M) ∣ ∣ ∣ ∣ {C u}^{2 +} (10^{- 1} M) ∣ ∣ C u (s)$
Given: $E_{c e l l}^{o} = + 0.46 V$ and $log 10^{n} = n$

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Solution

Cathode $- C C l^{2 +} + 2 e^{-} \to C C l$
Anode $- 2 A g \to 2 A g^{+} + 2 e^{-}$
$- - - - - - - - -$
Cell reaction $C u^{2 +} + 2 A g \to 2 A g^{+} + C u$
$E_{c e l l}^{0} = + 0.46 v$
$E_{c e l l} = 0.46 V - \frac{0.0591}{2} l o g \frac{[A g^{+}]^{2}}{[C u^{2 +}]} E_{c e l l} = 0.46 V - \frac{0.0591}{2} l o g \frac{[10^{- 3}]^{2}}{[10^{- 1}]} E_{c e l l} = + 0.61 V$
Thus the Emf of the reaction is $+ 0.61 V$

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Similar questions

25^{\circ} C

A g (s) | A g^{+} (10^{- 3} M) | | C u^{2 +} (10^{- 1} M) | C u (s)

E_{c e l l}^{0} = + 0.46 V

log 10^{n} = n

25^{\circ} C

A g (s) | A g^{+} (10^{- 3} M) | | C u^{2 +} (10^{- 1} M) | C u (s)

E_{c e l l}^{\circ} = + 0.46 V

l o g 10^{n} = n .

C u | C u^{2 +} (1 M) | | A g^{+} (1 M) | A g, E^{o} = 0.46 V

Z n | Z n^{2 +} (1 M) | | C u^{2 +} (1 M) | C u, E^{o} = 1.10 V

Z n | Z n^{2 +} | | A g^{+} (1 M) | A g

C u | C u^{2 +} (1 M) | | A g^{+} (1 M) | A g, E^{0} = 0.46 V

Z n | Z n^{2 +} (1 M) | | C u^{2 +} (1 M) | C u, E^{0} = 1.10 V

Z n | Z n^{2 +} (1 M) | | A g^{+} (1 M) | A g

(E^{\circ})

A g / A g^{+} (1 M) ∥ C u^{2 +} (1 M) | C u; E^{\circ} = - 0.46 v o l t

Z n / Z n^{2 +} (1 M) ∥ C u^{2 +} (1 M) | C u; E^{\circ} = + 1.10 v o l t

Z n | Z n^{2 +} (1 M) ∥ A g^{+} (1 M) | A g

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