Question

Ive attacthed all my results. I just need help calculating question 4 and 5

(5 pts) Part II. Your experimental data: how many sig figs should your temperature data have? Show work for one calculation of temp in Kelvin. Show work for one AG calculation: Approximate Measured Cell Calculated Temp (°C) Calculated Temp (°C) potential Temp (K) AG Ecell (V)* (kJ/mol)* * measured value 70.21 0.581 343 1-112.1 65.12 0.680338 - 111.9 Show work: 60.02 0.5715333 -1110 °C -> 5.05+278.15 66.11 0.610 328 - 110.0 - 278.K 50.23 0.565 323 -109.0 AGonFEcele 45.14 0.563 318 - 108. 1 2mole") (96500 40.00 0.656313 -101.3 (0.615V) = -99.4 KJ 35.15 0.541008 -105.60 30.31 0.5351 303 -103.3 25.13 0.531 298 -102.5 120.10 10.520 293 -100.4. 15.25 0.518 288 -100.0 110.35 0.5161284 -99.6 6.05 0.5151278 -99.4 ) ~15 POST-LAB QUESTIONS (5) Include your graph of Gibbs Free Energy vs Absolute Temperature. Graph (12) 1. In Part I of the procedure, you measured the cell potential for a variety of combinations of half cells. Use a standard reduction potential table in your textbook to calculate the expected E cell using the formula E cell-Ecathode - Eanode. Calculate E cell for each combination of cells. Show work for one calculation here, but you can enter the values in the table on page 6. E'cell - Eºcuze / co- Eºznatlan -0.342-(-0.762) = 1.104

Answer 1

Question - 4 ) given equation of straight line y = —0.2283x —34.856

compare with $\Delta$ G = —T$\Delta$S + $\Delta$ H

slope = —$\Delta$S = —0.2283

$\Delta$S = 0.2283 KJ / K

Intercept $\Delta$ H = —34.856 KJ / mol

Question -5 ) at equilibrium E_cell =0

E^o_cell = 0.531 volt at 298 K

log K = nE^o_cell / 0.0591 = 2 x 0.531 / 0.0591 =17.96

K_eq = 10^17.96 = 9.12 x 10¹⁷