1. Tardigrade
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  3. Chemistry
  4. MnO4-+8H++5e- xrightarrow[]Mn2++4H2O; E° =1.51 V MnO2+4H++2e- xrightarrow[]Mn2++2H2O; E° =1.23 V

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Q. $ MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\xrightarrow[{}]{{}}M{{n}^{2+}}+4{{H}_{2}}O; $ $ E{}^\circ =1.51\,\,V $ $ Mn{{O}_{2}}+4{{H}^{+}}+2{{e}^{-}}\xrightarrow[{}]{{}}M{{n}^{2+}}+2{{H}_{2}}O; $ $ E{}^\circ =1.23\,\,V $

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Solution:

For the reaction, $ MnO_{4}^{-}+4{{H}^{+}}+3{{e}^{-}}\xrightarrow{{}}Mn{{O}_{2}}+2{{H}_{2}}O $ $ -{{E}_{3}}=\frac{-1.51\times 5+2\times 1.23}{3} $ $ \therefore $ $ {{E}_{3}}=1.70\,V $