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The value of 'n' in, MnO4^(-) + 8H^(+) +...

The value of 'n' in, `MnO_4^(-) + 8H^(+) + n e to Mn^(2+) + 4H_2O` is:

A

5

B

4

C

3

D

2

Text Solution

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The correct Answer is:
A
5
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Using the following E^(@) values for electrode potentials, calculate triangle G^(@) in kJ for the indicated reaction: 5Ce^(4+) (aq) + Mn^(2+) (aq) + 4H_(2)O(l) rarr 5Ce^(3+) (aq) + MnO_(4)^(-) (aq) +8H^(+) (aq) MnO_(4)^(-) (aq) +8H^(+) (aq) +5e^(-) rarr Mn^(2+) (aq) + 4H_(2) O(l), E^(@) = +1.51V Ce^(4+) (aq) + e^(-)rarr Ce^(3+)(aq), E^(@) = +1.61V

Write ionic equation for the reduction of MnO_(4)^(-) to Mn^(2+) by H_(2)O_(2) , in acidic medium.

Oxidation nümber of 'Mn' in 'MnO4- is: '

Consider the half reactions of a galvanic cell given below: MnO_(2) + H^(+) hArr Mn^(2+) + 2H_(2)O, E_(MnO_(2))^(@)= 1.23V, PbCl_(2) hArr Pb+ 2Cl^(-), E_(PbCl_(2)^(@)= -0.27V . The correct statement about the cell is

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