${\rm{MnO}}_4^ - + 8{{\rm{H}}^ + } + 5{e^ - } \to {\rm{M}}{{\rm{n}}^{2 + }} + 4{{\rm{H}}_2}{\rm{O}};{E^ \circ } = 1.51{\rm{V}}$ ${\rm{Mn}}{{\rm{O}}_4} + 4{{\rm{H}}^ + } + 2{e^ - } \to {\rm{M}}{{\rm{n}}^{2 + }} + 2{{\rm{H}}_2}{\rm{O}};{E^ \circ } = 1.23{\rm{V}}$ $E_{{\rm{MnO}}_4^ - |{\rm{Mn}}{{\rm{O}}_2}}^ \circ $ is

NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

1 Hard

1 of 1

Analyzing +4 | -1

\mathrm{MnO}_4^ - + 8{\mathrm{H}^ + } + 5{e^ - } \to \mathrm{M}{\mathrm{n}^{2 + }} + 4{\mathrm{H}_2}\mathrm{O};{E^ \circ } = 1.51\mathrm{V}

\mathrm{Mn}{\mathrm{O}_4} + 4{\mathrm{H}^ + } + 2{e^ - } \to \mathrm{M}{\mathrm{n}^{2 + }} + 2{\mathrm{H}_2}\mathrm{O};{E^ \circ } = 1.23\mathrm{V}

E_{\mathrm{MnO}_4^ - |\mathrm{Mn}{\mathrm{O}_2}}^ \circ

View answer and explanation

Correct option: A

1.70 V

Option A (1.70 V) is correct because it is calculated by combining the given half-reactions to obtain the desired half-reaction and using the relationship between Gibbs free energy and standard potential.

NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

More Questions from ELECTRO CHEMISTRY