For the cell :
$Z n | Z n^{2 +} | | C u^{2 +} | C u$
$E_{z n}^{⊖} = 0.736 V; E_{(c u)}^{⊖} = - 0.35 V$
Calculate the cell EMF.

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Solution

Anode: $Z n ⟶ Z n^{2 +} + 2 e^{-}$ $E^{⊖} = - 0.736 V$
Cathode: $C u^{2 +} + 2 e^{-} ⟶ C u$ $E^{⊖} = 0.35 V$
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ Z n + C u^{2 +} ⟶ Z n^{2 +} + C u - ---------------------------- -$

$E_{c e l l} = E_{c}^{⊖} - E_{a}^{⊖} = 0.35 - (- 0.736) = 1.086 V$
It is a Daniell cell, it is reversible in the sense that it reaches equilibrium and then it does not deliver any energy.

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