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Electrochemical Cells, Galvanic Cells

Given:Cu 2++ ...

Question

Given:
Cu2++e−→Cu+;E∘=0.15VCu++e−→Cu;E∘=0.50VZn2++ze−→Zn;E∘=−0.76V
If 9.65 amperes of current is passed, making Cu anode and Zn cathode for 1000 seconds in the cell

Zn|Zn2+||Cu2+|Cu
containing one litre of 0.55 M zinc (II) ions and one litre of 0.05 M copper (II) ions in the half cells, the E.M.F of cell the after passage of current would be (log 2 = 0.3, log 3 = 0.5, log 5 = 0.7)
:

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Solution

$\begin{matrix} Z n \to Z n^{2 +} + 2 e^{-} & E_{a n o d e}^{\circ} = - 0.76 V C u^{2 +} + 2 e \to C u; \end{matrix}$
For the cathode, we need to use both the copper equations.
Remember, $G^{\circ}$ is additive, $E^{\circ}$ is not.
$E_{c a t h o d e}^{\circ} = (0.15 + 0.5) \div 2 = 0.325 V$
$E^{\circ} o f Z n | Z n^{2 +} | | C u^{2 +} | C u = 1.085 V$
$Q = I t = 9.65 \times 1000 = 9650 = 0.1 e q = 0.05 m o l$
After electrolysis concentration will be
$Z n | Z n^{2 +} | | C u^{2 +} | C u$
$0.50 M 0.10 M$
$E M F = 1.085 V - \frac{0.059}{2} l o g \frac{0.50}{0.10} = 1.064$

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