Question #f0862

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Question #f0862

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Answer 1 · 2018-02-02T00:35:37

All the info you request is in the analysis below...

Explanation:

The task here is to create the Mn half-reaction (as it does not seem to appear on any of the standard tables I could find).

The zinc (anode) is quite simple:

$Z n (s) \to Z n^{2 +} + 2 e^{-}$ $Zn(s) rarr Zn^(2+) + 2e^-$ $E^{o} = - 0.76 V$ $E^o=-0.76 V$

To build the reduction half reaction, we start with the Mn species:

$M n O_{2} \to M n {(O H)}_{3}$ $MnO_2 rarr Mn(OH)_3$

Next, we balance the hydrogen and oxygen atoms by adding $H_{2} O$ $H_2O$ and $O H^{-}$ $OH^-$ as needed:

$M n O_{2} + 2 H_{2} O \to M n {(O H)}_{3} + O H^{-}$ $MnO_2 + 2H_2O rarr Mn(OH)_3 + OH^-$

Finally, add electrons to balance the charge:

$M n O_{2} + 2 H_{2} O + e^{-} \to M n {(O H)}_{3} + O H^{-}$ $MnO_2 + 2H_2O + e^(-) rarr Mn(OH)_3 + OH^-$

which is a reduction, as expected.

Finally, since we know the net reaction potential (the cell voltage) is 1.50 V, the potential of the reduction must be 0.74 V, because the difference in potential

$E^{o} (c a t h o d e) - E^{o} (a n o d e) = 1.5 V$ $E^o (cathode) - E^o (anode) = 1.5 V$

$0.74 - (- 0.76) = 1.5$ $0.74 - (-0.76) = 1.5$

Finally, to balance the equation, we must balance the # of electrons. This means multiplying the reduction half-reaction by 2, then adding them together:

$Z n + 2 M n O_{2} + 4 H_{2} O \to Z n^{2 +} + 2 M n {(O H)}_{3} + 2 O H^{-}$ $Zn + 2MnO_2 + 4H_2O rarr Zn^(2+) + 2Mn(OH)_3 + 2OH^-$

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Question #f0862

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