1	1^st you need the BALANCED EQUATION Pb²⁺ + 2 e- ÚPb_(s)E⁰_red = -0.13 V Cu²⁺ + 2 e- ÚCu_(s) E⁰_red = 0.34 V You need an OXIDATION and a REDUCTION. E⁰_cell >0 or nothing happens!
2	1^st you need the BALANCED EQUATION Pb_(s)ÚPb²⁺ + 2 e- E⁰_ox = +0.13 V Cu²⁺ + 2 e- ÚCu_(s) E⁰_red = 0.34 V Pb_(s) + Cu²⁺ ÚPb²⁺ + Cu_(s) E⁰_cell = 0.13+0.34V That assumes 1 M concentrations. Since they aren’t 1 M: NERNST EQUATION
3	1^st you need the BALANCED EQUATION Pb_(s) + Cu²⁺ ÚPb²⁺ + Cu_(s) E⁰_cell = 0.47 v E_cell = E⁰_cell – 0.0592/n log Q E_cell = E⁰_cell – 0.0592 log Pb²⁺ n Cu²⁺ E_cell = 0.47V – 0.0592 log 0.050 2 1.50 E_cell = 0.51 V
4	What is the cell potential when the concentration of Cu²⁺ has fallen to 0.200 M? E_cell = E⁰_cell – 0.0592 log Pb²⁺ n Cu²⁺ E_cell = 0.47V – 0.0592 log 1.35 2 0.2 E_cell = 0.45 V
5	What are the concentrations of Pb²⁺ and Cu²⁺ when the cell potential falls to 0.35 V? E_cell = E⁰_cell – 0.0592 log Pb²⁺ n Cu²⁺ 0.35 = 0.47V – 0.0592 log 0.05+x 2 1.5-x -0.12 = -0.0296 log (0.05+x/1.5-x) 4.054 = log (0.05+x/1.5-x) 10^4.054= 0.05+x/1.5-x 1.1324x10⁴ = 0.05+x/1.5-x 1.6986x10⁴ – 1.1324x10⁴ x = 0.05 +x 1.6986x10⁴ = 1.1325x10⁴ x X = 1.4999
6	What are the concentrations of Pb²⁺ and Cu²⁺ when the cell potential falls to 0.35 V? E_cell = E⁰_cell – 0.0592 log Pb²⁺ n Cu²⁺ 0.35 = 0.47V – 0.0592 log 0.05+x 2 1.5-x -0.12 = -0.0296 log (0.05+x/1.5-x) 4.054 = log (0.05+x/1.5-x) 10^4.054= 0.05+x/1.5-x 1.1324x10⁴ = 0.05+x/1.5-x 1.6986x10⁴ – 1.1324x10⁴ x = 0.05 +x 1.6986x10⁴ = 1.1325x10⁴ x X = 1.4999
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