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To find the standard potential of M3+/M electrode, the following cell is constituted : Pt/M/M3+(0.001 mol L−1)/Ag+(0.01 mol L−1)/AgThe emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half reaction M3++3e−→ M at 298 K will be :(Given EAg+/Ag at 298 K = 0.80 Volt)a)0.38 Voltb)0.32 Voltc)1.28 Voltd)0.66 VoltCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about To find the standard potential of M3+/M electrode, the following cell is constituted : Pt/M/M3+(0.001 mol L−1)/Ag+(0.01 mol L−1)/AgThe emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half reaction M3++3e−→ M at 298 K will be :(Given EAg+/Ag at 298 K = 0.80 Volt)a)0.38 Voltb)0.32 Voltc)1.28 Voltd)0.66 VoltCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for To find the standard potential of M3+/M electrode, the following cell is constituted : Pt/M/M3+(0.001 mol L−1)/Ag+(0.01 mol L−1)/AgThe emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half reaction M3++3e−→ M at 298 K will be :(Given EAg+/Ag at 298 K = 0.80 Volt)a)0.38 Voltb)0.32 Voltc)1.28 Voltd)0.66 VoltCorrect answer is option 'C'. Can you explain this answer?.

Solutions for To find the standard potential of M3+/M electrode, the following cell is constituted : Pt/M/M3+(0.001 mol L−1)/Ag+(0.01 mol L−1)/AgThe emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half reaction M3++3e−→ M at 298 K will be :(Given EAg+/Ag at 298 K = 0.80 Volt)a)0.38 Voltb)0.32 Voltc)1.28 Voltd)0.66 VoltCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Here you can find the meaning of To find the standard potential of M3+/M electrode, the following cell is constituted : Pt/M/M3+(0.001 mol L−1)/Ag+(0.01 mol L−1)/AgThe emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half reaction M3++3e−→ M at 298 K will be :(Given EAg+/Ag at 298 K = 0.80 Volt)a)0.38 Voltb)0.32 Voltc)1.28 Voltd)0.66 VoltCorrect answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of To find the standard potential of M3+/M electrode, the following cell is constituted : Pt/M/M3+(0.001 mol L−1)/Ag+(0.01 mol L−1)/AgThe emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half reaction M3++3e−→ M at 298 K will be :(Given EAg+/Ag at 298 K = 0.80 Volt)a)0.38 Voltb)0.32 Voltc)1.28 Voltd)0.66 VoltCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for To find the standard potential of M3+/M electrode, the following cell is constituted : Pt/M/M3+(0.001 mol L−1)/Ag+(0.01 mol L−1)/AgThe emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half reaction M3++3e−→ M at 298 K will be :(Given EAg+/Ag at 298 K = 0.80 Volt)a)0.38 Voltb)0.32 Voltc)1.28 Voltd)0.66 VoltCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of To find the standard potential of M3+/M electrode, the following cell is constituted : Pt/M/M3+(0.001 mol L−1)/Ag+(0.01 mol L−1)/AgThe emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half reaction M3++3e−→ M at 298 K will be :(Given EAg+/Ag at 298 K = 0.80 Volt)a)0.38 Voltb)0.32 Voltc)1.28 Voltd)0.66 VoltCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice To find the standard potential of M3+/M electrode, the following cell is constituted : Pt/M/M3+(0.001 mol L−1)/Ag+(0.01 mol L−1)/AgThe emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half reaction M3++3e−→ M at 298 K will be :(Given EAg+/Ag at 298 K = 0.80 Volt)a)0.38 Voltb)0.32 Voltc)1.28 Voltd)0.66 VoltCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.