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For the following voltaic cell, determine the [Cl-] when PCl2 = 0.500 atm, [Zn2+] = 1.77 × 10-2 M, and Ecell = 2.250 V. The half-reactions at 25°C are: Cl2(g) + 2 e- → 2 Cl-(aq) E° = +1.358 V
Zn2+(aq) + 2 e- → Zn(s) E° = -0.763 V
Zn(s) ∣ Zn2+(aq) || Cl-(aq) , Cl2(g) ∣ Pt(s)
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