Consider the following cell: Ni(s)|Ni2+(aq,0.100 M) || Cl–(aq, ? M)|Cl2(g, 0.800

Question

Consider the following cell:
         Ni(s)|Ni2+(aq,0.100 M) || Cl–(aq, ? M)|Cl2(g, 0.800 atm)|Pt
For this cell, E° = 1.59 V and E = 1.79 V at 25°C.Calculate the Cl–(aq) concentration in the cathodecompartment.
a) 4.2 × 10–4 M d) 3.4 × 10–3 M
b) 6.5 × 10–2 M e) 1.2 × 10–3 M
c) 5.8 × 10–2 M

Please show work.

Explanation / Answer

Ni => Ni2+ + 2e- Cl2 + 2e- => 2Cl- E = E0 - RT/nF *ln{[Cl-]^2[Ni2+]/PCl2} n = 2 (# electrons moved), F = 96485 C/mol (Faraday's Constant) 1.79 = 1.59 - 8.314 *298/2/96485*ln([Cl-]^2*(0.1)/0.8) -15.58 = ln([Cl-]^2*(0.1)/0.8) [Cl-] = 0.0012 M = 1.2 x 10^-3 M (pick e)

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