Home
Class 12
CHEMISTRY
Calculate the potential of an indicator ...

Calculate the potential of an indicator electrode versus the standard hydrogen electrode, which originally contained `0.1M MnO_(4)^(-)` and `0.8M H^(+)` and which was treated with `Fe^(2+)` necessary to reduce `90%` of the `MnO_(4)` to `Mn^(2+)`
`MnO_(4)^(-) +8H^(+) +5e rarr Mn^(2+) +H_(2)O, E^(@) = 1.51V`

Text Solution

Verified by Experts

The correct Answer is:
`1.39V`
`MnO_(4)^(-) +8H^(+) +2e^(-) rarr Mn^(2+) +4H_(2)O +E^(@) = 1.51V`
`{:(0.1,0.8),(0.01,0.08):}`
`bar(E=1.51 -(0.0591)/(5)log.(0.09)/(0.01xx(0.08)^(8))`
`E = 1.51 -0.15`
`= 1.39V`
Promotional Banner

Similar Questions

Explore conceptually related problems

The standard electrode potentials (reduction) of Pt//Fe^(3+),Fe^(+2) and Pt//Sn^(4+),Sn^(+2) are +0.77V and 0.15V respectively at 25^(@)C . The standard EMF of the reaction Sn^(4+) +2Fe^(2+) rarr Sn^(2+) +2Fe^(3+) is

Find the standard electrode potential of MnO_(4)^(c-)|MnO_(2). The standard electrode potential of MnO_(4)^(c-)|Mn^(2+)=1.51V and MnO_(2)|MnO_(2)|Mn^(2+)=1.23V .