Calculate the emf of the cell. $Z n | Z n^{2 +} (0.001 M) | | C u^{2 +} (0.1 M) | C u$
The standard potential of $C u / C u^{2 +}$ half-cell is +0.34 and $Z n / Z n^{2 +}$ is -0.76 V.

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Solution

Given $Z n | Z n^{2 +} (0.001 M) | | C u^{2 +} (0.1 M) | C u$
Overall cell reaction:
$Z n ⟶ {Z n}^{2 +} + 2 e^{-} {C u}^{2 +} + 2 e^{-} ⟶ C u - --------------------------- - Z n + {C u}^{2 +} ⟶ {Z n}^{2 +} + C u - ---------------------------- -$
$E_{c e l l}^{o} =$ standard reduction potential of cathode + standard oxidation potential of anode
$E_{c e l l}^{o} =$ 0.34 to 0.76 V
$E_{c e l l}^{o} =$ 1.1 V
$K_{C} = \frac{[{Z n}^{2 +}]}{[{C u}^{2 +}]} = \frac{10^{- 3}}{10^{- 1}} = 10^{- 2}$
EMF of the cell at any electrode concentration is:
$E = E^{o} - \frac{0.059}{n} log (K_{C}) = 1.1 - \frac{0.059}{2} log (10^{- 2}) = 1.1 - \frac{0.059}{2} \times (2) = 1.1 - 0.059 = 1.041 V$

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