First slide

Nernst Equation

Question

Cell notation of Daniel cell is $Zn (s) / {Zn}^{2 +} (1 M) / / {Cu}^{2 +} (1 M) / Cu (s)$ . Standard free energy change for the reaction occurring in Daniel cell is $(E_{{Zn}^{2 +} / Zn}^{0} = - 0.76 V, E_{{Cu}^{2 +} / Cu}^{0} = + 0.34 V)$

Moderate

A

-234.9 kJ
B

-212.3 kJ
C

-251.1 kJ
D

-205.2 kJ

Solution

$E_{cell}^{0} = E_{cathode}^{0} - E_{anode}^{0}; E_{cell}^{0} = + 0.34 V - (- 0.76 V) = 1.1 V; Zn (s) + {Cu}^{2 +} (1 M) \to {Zn}^{2 +} (1 M) + Cu (s); ∆^{0} G = - {nFE}_{cell}^{0}; ∆^{0} G = - 2 (96500) (1.1) J; ∆^{0} G = - 2 (96.5) (1.1) kJ; ∆^{0} G = - 212.3 kJ;$

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