Question

Report Sheet 5 Part D. Calculation of expected voltage, concluded. (4) (e). Calculation of the expected voltage for Agcl. AgCl (5) ► Ag + CIK -[AgICI] Calculation of [C ] in the 50 mL of solution after addition of the NaCl solution to the AgNO, solu tion. Take into account that each AgCl molecule formed took a Clion from the solution. m 1V1-mava 0.01x25 + Gasxas=30* Y-0.05 +.105 -0.11n [CI] - 0.2 M 2 significant digits) Calculation of [Ag] after addition of NaCl solution Using the researched value of K for AgCl, determine the (Agexpected under the experimental conditions Kf=17 x107 = COJI TA9] [Ag] - 215 x 16 M (2 significant digits) Calculation of E for AgCl -1.157 Use the Nernst equation, (E® vs SHE from your data in volts and the (Ag determined above, calculate E (V vs Ag AgCl) after addition of NaCl in Part D(4c). E- (2 significant digit) Table value: K -2 * 10-10
calculation of E for AgCl

Answer 1

Nesse equation is, Full - Full - 230 3 RI [сховці) Theactant) nF Eulp = full – 2.303 RT log fred nf u cox] For Ag (Ag ce electrode, the reaction is Agt te P ag (s) Here n = np of electron As we are given t = - 1015V = 1 Ecell = É - .303x8.314 x 8898 96500 x n Tlagt] Eccl- & ur 9.0.591 log Tag -4 part) As calculated above, rag*] = 1,5x16'm [Given] and t'iu = -1.15V &mal € = -6.15 - 0.0591 log Tas xlou 2-1.18 – 0.0591 109 (6167x109) =-1115 - 00591 X (-4. = -1.15 + 0.25 = -0.901 E