For the following concentration cell at $298 K$ :

$Z n (s) | Z n^{2 +} (a q, C_{1}) | | Z n^{2 +} (a q, C_{2}) | Z n (s)$

$C_{1} & C_{2} are concentrations of Z n^{2 +} ions of each half-cell.$
Choose the correct option:

E_{c e l l}^{0} = 0

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E_{c e l l}^{0} > 1

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0 < E_{c e l l}^{0} < 1

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E_{c e l l}^{0} < 0

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Solution

The correct option is A $E_{c e l l}^{0} = 0$
$Z n (s) | Z n^{2 +} (a q, C_{1}) | | Z n^{2 +} (a q, C_{2}) | Z n (s)$

$C_{1} & C_{2} are concentrations of Z n^{2 +} ions of each half-cell.$
Anode:
$Z n (s) ⇌ Z n^{2 +} (a q) + 2 e^{-}$

Cathode:
$Z n^{2 +} (a q) + 2 e^{-} ⇌ Z n (s)$

Net cell reaction:
$Z n^{2 +} (C_{2}) ⇌ Z n^{2 +} (C_{1})$

Since in concentration cells, same metal and its solution is present.
Cathode and anode consist of same electrode.
$E_{c a t h o d e}^{0} = E_{a n o d e}^{0}$
For all concentrationc cells
$E_{c e l l}^{0} = E_{c a t h o d e}^{0} - E_{a n o d e}^{0}$
$E_{c e l l}^{0} = 0$

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Similar questions

298 K

Z n (s) | Z n^{2 +} (a q, C_{1}) | | Z n^{2 +} (a q, C_{2}) | Z n (s)

C_{1} & C_{2} are concentrations of Z n^{2 +} ions of each half-cell.

C u^{2 +} (C_{1}) (a q .) + Z n (s) \to Z n^{2 +} (C_{2}) (a q .) + C u (s)

(△ G)

C d^{2 +} (a q) + Z n (s) \to Z n^{2 +} (a q) + C d (s)

Z n_{(s)} + C u_{(a q)}^{2 +} ⇌ Z n_{(a q)}^{2 +} + C u_{(s)}

= 1.1 v o l t

F = 96500

E_{c e l l} = 0.2905 V

Z n (s) + F e^{2 +} (a q) 0.01 M \to Z n^{2 +} 0.1 M (a q) + F e (s)

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