Calculate the potential of the indicator electrode at 25o C relative to SCE at the equivalence point of the titration of 25.00 ml of 0.010 Fe2+ with 0.010 M Cr2O72- in a solution that is buffered at pH 3.50.

Calculate the cell potential when 50.00 mL of 0.100 M Ag+ solution is titrated with 50.00 mL of 0.150 M Cl- solution. Silver wire indicator electrode and Ag-AgCl electrode are used as reference electrodes. E0 (Ag/AgCl) = 0,197 V , E0 Ag+/Ag = 0,799 V, Ksp (AgCl) = 1,8 x10-8

Calculate the potential at the equivalence point for titration of 0.100MFe^2+ with 0.100M Ce^4+.Given the formal potential for the reduction E^0 of Fe^3+ to Fe^2+ is +0.77V, and the formal potential for the reduction of E^0 Ce^4+ to Ce^3+ is 1.45V.

A solution prepared by mixing 60.0 mL of 0.360 M AGNO, and 60.0 mL of 0.360 M TINO, was titrated with 0.720 M NaBr in a cell containing a silver indicator electrode and a reference electrode with a constant potential of 0.175 V. The reference electrode is attached to the positive terminal of the potentiometer, and the silver electrode is attached to the negative terminal. The solubility constant of TIBr is Kp = 3.6 × 10-6 and the solubility constant of AgBr is = 5.0 x 10-13. Ksp Which precipitate forms first? O TIBI O AgBr Which of the expressions shows how the cell potential, E, depends on [Ag*]? E = 0.175 V – [0.799 V – 0.05916log ([Ag+])] 1 E = |0.799 V – 0.05916log - 0.175 V [Ag*) E = [0.799 V – 0.05916log ([Ag*]D] – 0.175 V E = 0.175 V - |0.799 V – 0.05916log [Ag* * (ল) Calculate the first and second equivalence points of the titration. first equivalence point: mL second equivalence point: mL What is the cell potential after each of the given volumes of 0.720 M NaBr have been…