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Ce Ce Ce Ce You are titrating 100.0 mL of 0.0400 M Fe2 in 1 M HCIO4 with 0.100 M Ce to give Fe and Ce E E-0.767-0.05916log 0.24I using Pt and calomel electrodes to find the endpoint. (a) Write the balanced titration reaction. Fe aq)Ce (aq)Fe (b) Complete the two half reactions for (d) Calculate the values of E for the cell when the following volumes of the Ce titrant have been added: (Activity coefficients may be ignored as they tend to cancel when calculating concentration ratios.) FE-0.767-0.05916 log 0.241 the Pt electrode G) E-1.70-0.05916 logta t 0.241 Fe +e E = 0.767 V 1.00 mL 0.241V Ce (c) From the list in the column at the right, select the Number H E 1.70-0.05916log 0.241 two correct Nernst equations for the cell (Each applying at different points in the titration.) E-0.767-0.05916 log 0.241 Ce 20.0 mL Fe Eo of the calomel electrode is 0.241 V (B) E-0.767-0.059 1 6 log| 2+1|-0.24 1 39.0 mL O □ (B) (C) 6-1.70-0.059 16 log- -0.241 40.0 mL 0.993 □ (E) I (F) Fe Number (D) E= 1.70-0.059 16 log 210.241 5.0 mL 80.0 mL 1.459

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