Potentiometric and redox titration potentials

9 questions on Potentiometric and redox titration potentials from the Electrochemistry chapter: 2 NEET level, 3 JEE Main, 4 JEE Advanced. Each one opens with its answer and solution.

• Fe^2+ in 1 M H2SO4 is titrated with Ce^4+. Formal potentials: E deg(Fe^3+/Fe^2+) = 0.68 V, E deg(Ce^4+/Ce^3+) = 1.44 V. The potential of the platinum indicator ... NEET level
• 25 mL of 0.1 M Fe^2+ is titrated potentiometrically with 0.1 M Ce^4+ (E deg(Fe^3+/Fe^2+) = 0.68 V, E deg(Ce^4+/Ce^3+) = 1.44 V). After 12.5 mL of titrant has be... JEE Main level
• In the potentiometric titration of Fe^2+ with Ce^4+, the indicator electrode normally used is NEET level
• 25 mL of 0.1 M Fe^2+ is titrated with 0.1 M Ce^4+ at 298 K. E deg(Fe^3+/Fe^2+) = 0.68 V, E deg(Ce^4+/Ce^3+) = 1.44 V, 2.303RT/F = 0.0591 V, log 2 = 0.301. After... JEE Main level
• 25 mL of 0.1 M Fe^2+ is titrated with 0.1 M Ce^4+ at 298 K. E deg(Fe^3+/Fe^2+) = 0.68 V, E deg(Ce^4+/Ce^3+) = 1.44 V, 2.303RT/F = 0.0591 V, log 2 = 0.301. After... JEE Main level
• Fe^2+ is titrated potentiometrically with Ce^4+ using a platinum indicator electrode. Both couples are one-electron couples. Which statements are correct? JEE Advanced level
• Sn^2+ is titrated with Fe^3+ according to Sn^2+ + 2Fe^3+ -> Sn^4+ + 2Fe^2+ at 298 K. E deg(Sn^4+/Sn^2+) = 0.15 V and E deg(Fe^3+/Fe^2+) = 0.77 V. The potential ... JEE Advanced level
• Fe^2+ is titrated with KMnO4 in a solution maintained at pH 1 ([H+] = 0.1 M) at 298 K: MnO4- + 5Fe^2+ + 8H+ -> Mn^2+ + 5Fe^3+ + 4H2O. E deg(MnO4-, H+/Mn^2+) = 1... JEE Advanced level
• Fe^2+ is titrated with Ce^4+ at 298 K; E deg(Fe^3+/Fe^2+) = 0.68 V, E deg(Ce^4+/Ce^3+) = 1.44 V, 2.303RT/F = 0.0591 V. The change in indicator electrode potenti... JEE Advanced level