The concentrations of NaCl and KI are at 0.160M and 0.120M respectively, in the

Question

The concentrations of NaCl and KI are at 0.160M and 0.120M respectively, in the following electrochemical cell: Cu(s) | CuI(s) | I-(aq) || Cl-(aq) | AgCl(s) | Ag(s) a) using the following half-reactions, calculate cell voltage: CuI(s) + e- <--> Cu(s) + I- Eo = -0.185V AgCl(s) + e- <--> Ag(s) + Cl- Eo = 0.222V

b) now calculate the cell voltage using the following half-reactions. The Ksp for AgCl and CuI are 1.8E-10 and 1.0E-12 respectively. Cu+ + e- <--> Cu(s) Eo = 0.518V Ag+ + e- <--> Ag(s) Eo = 0.7993V

Explanation / Answer

Oxidation half-cell: Cu(s) + I-(aq) ----- > CuI(s) + e- , E0(oxi) = + 0.185 V

Reduction half-cell: AgCl(s) + e- ----> Ag(s) + Cl-(aq), E0(red) = + 0.222V

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Cell reaction:  Cu(s) + I-(aq) + AgCl(s) ---- > CuI(s) + Ag(s) + Cl-(aq),

E0(cell) = E0(oxi) + E0(red) = 0.185 V + 0.222 V = 0.407 V

n = 1

Now applying Nernst equation

E(cell) = E0(cell) - (0.0591 /n) x log[Cl-(aq)] / [I-(aq)] = 0.407 V - 0.0591 x log( 0.160 M / 0.120M)

=> E(cell) = 0.400 V (answer)

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