For the Following Voltaic Cell, Determine the [Cl-] When PCl₂

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)

A) 5.48 × 10^-6 M
B) 2.32 × 10^-3 M
C) 0.0296 M
D) 0.0939 M
E) 0.0352 M

Correct Answer:

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