Post Closed as "duplicate" by orthocresol of
4 Corrected stoichiometry, added symbology, spelling, clarification, and grammar.
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I mean by "combining" is to make a new half-reaction equation and not an overall equation for a reaction in whilewhole.

For instance, I was trying to arrive at the following half-reaction:

$\ce{MnO2(s) + 4H+(aq) + 2e- -> Mn^{2+}(aq) + 2H2O(l)}\quad E^\circ= 1.23\,\mathrm{V}$

by combining

$\ce{MnO4^{-}(aq) + 8H^+(aq) + 5e- -> Mn^{2+}(aq) + 4H_2O(l)}\quad\quad E^\circ= 1.51 \,\mathrm{V}$

and

$\ce{MnO4^{-}(aq) + 4H+(aq) + 3e- -> MnO2(s) + 2H_2O(l)}\quad\quad E^\circ= 1.7\,\mathrm{V}$

using Hess's Law. My calculation yielded +0.19 V, and not +1.23 V as I would expect.

Why caused this? I know that there are half-reaction potentialsreactions that can be constructed in this way (aboveas above, by combining two or more other half-reactions) that result in the correct potential being calculated.

I mean by "combining" is to make a new half-reaction equation and not an overall equation for a reaction in while.

For instance, I was trying to arrive at the following half-reaction:

$\ce{MnO2(s) + 4H+(aq) + 2e- -> Mn^{2+}(aq) + 2H2O(l)}\quad E^\circ= 1.23\,\mathrm{V}$

by combining

$\ce{MnO4^{-}(aq) + 8H^+(aq) + 5e- -> Mn^{2+}(aq) + 4H_2O(l)}\quad\quad E^\circ= 1.51 \,\mathrm{V}$

and

$\ce{MnO4^{-}(aq) + 4H+(aq) + 3e- -> MnO2(s) + 2H_2O(l)}\quad\quad E^\circ= 1.7\,\mathrm{V}$

using Hess's Law. My calculation yielded +0.19 V, and not +1.23 V as I would expect.

Why caused this? I know that there are half-reaction potentials that can be constructed in this way (above, by combining two or more other half-reactions) that result in the correct potential being calculated.

I mean by "combining" is to make a new half-reaction equation and not an overall equation for a reaction in whole.

For instance, I was trying to arrive at the following half-reaction:

$\ce{MnO2(s) + 4H+(aq) + 2e- -> Mn^{2+}(aq) + 2H2O(l)}\quad E^\circ= 1.23\,\mathrm{V}$

by combining

$\ce{MnO4^{-}(aq) + 8H^+(aq) + 5e- -> Mn^{2+}(aq) + 4H_2O(l)}\quad\quad E^\circ= 1.51 \,\mathrm{V}$

and

$\ce{MnO4^{-}(aq) + 4H+(aq) + 3e- -> MnO2(s) + 2H_2O(l)}\quad\quad E^\circ= 1.7\,\mathrm{V}$

using Hess's Law. My calculation yielded +0.19 V, and not +1.23 V as I would expect.

Why caused this? I know that there are half-reactions that can be constructed in this way (as above, by combining two or more other half-reactions) that result in the correct potential being calculated.

3 Corrected stoichiometry, added symbology, spelling, clarification, and grammar.
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Why can some redox half-equations canreactions be combined and some notcannot?

I mean by "combining" makingis to make a new half-reaction equation. Not a and not an overall equation for a reaction in while.

One that cannot be made by combiningFor instance, I was trying to arrive at the following half-reaction:

$\ce{MnO2(s) + 4H+(aq) + 2e- -> Mn^{2+}(aq) + 2H2O(l)}\quad E^\circ= 1.23\,\mathrm{V}$

I was trying to combine these two but couldn't get the same same potential by adding:combining

$\ce{MnO4^{-}(aq) + 4H^+(aq) + 5e- -> Mn^{2+}(aq) + 2H_2O(l)}\quad\quad 1.51 \,\mathrm{V}$$\ce{MnO4^{-}(aq) + 8H^+(aq) + 5e- -> Mn^{2+}(aq) + 4H_2O(l)}\quad\quad E^\circ= 1.51 \,\mathrm{V}$

Andand

$\ce{MnO4^{-}(aq) + 4H+(aq) +3e- -> MnO2(s) + 2H_2O(l)}\quad\quad 1.7\,\mathrm{V}$$\ce{MnO4^{-}(aq) + 4H+(aq) + 3e- -> MnO2(s) + 2H_2O(l)}\quad\quad E^\circ= 1.7\,\mathrm{V}$

By combining I got 0using Hess's Law.19V My calculation yielded +0.  19 V, and not +1.23 V as I would expect.

Why caused this?(there I know that there are half reaction equation-reaction potentials that can be obtained with the right potentialconstructed in this way (above, by combining two or more other half reaction equations-reactions) that result in the correct potential being calculated.

Why some redox half-equations can be combined and some not?

I mean by "combining" making a new half equation. Not a overall equation.

One that cannot be made by combining:

$\ce{MnO2(s) + 4H+(aq) + 2e- -> Mn^{2+}(aq) + 2H2O(l)}\quad E^\circ= 1.23\,\mathrm{V}$

I was trying to combine these two but couldn't get the same same potential by adding:

$\ce{MnO4^{-}(aq) + 4H^+(aq) + 5e- -> Mn^{2+}(aq) + 2H_2O(l)}\quad\quad 1.51 \,\mathrm{V}$

And

$\ce{MnO4^{-}(aq) + 4H+(aq) +3e- -> MnO2(s) + 2H_2O(l)}\quad\quad 1.7\,\mathrm{V}$

By combining I got 0.19V.  Why caused this?(there are half reaction equation that can be obtained with the right potential by combining two or more other half reaction equations)

Why can some redox half-reactions be combined and some cannot?

I mean by "combining" is to make a new half-reaction equation and not an overall equation for a reaction in while.

For instance, I was trying to arrive at the following half-reaction:

$\ce{MnO2(s) + 4H+(aq) + 2e- -> Mn^{2+}(aq) + 2H2O(l)}\quad E^\circ= 1.23\,\mathrm{V}$

by combining

$\ce{MnO4^{-}(aq) + 8H^+(aq) + 5e- -> Mn^{2+}(aq) + 4H_2O(l)}\quad\quad E^\circ= 1.51 \,\mathrm{V}$

and

$\ce{MnO4^{-}(aq) + 4H+(aq) + 3e- -> MnO2(s) + 2H_2O(l)}\quad\quad E^\circ= 1.7\,\mathrm{V}$

using Hess's Law. My calculation yielded +0.19 V, and not +1.23 V as I would expect.

Why caused this? I know that there are half-reaction potentials that can be constructed in this way (above, by combining two or more other half-reactions) that result in the correct potential being calculated.

    Tweeted twitter.com/#!/StackChemistry/status/597574969892143104
2 the use of the \ce{...} environment saves a lot of typing
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I mean by "combining" making a new half equation. Not a overall equation.

One that cannot be made by combining:

$MnO_2(s) + 4H^+(aq) + 2e^- \rightarrow Mn^{2+}(aq) + 2H_2O(l)$ $E^。= 1.23V$$\ce{MnO2(s) + 4H+(aq) + 2e- -> Mn^{2+}(aq) + 2H2O(l)}\quad E^\circ= 1.23\,\mathrm{V}$

I was trying to combine these two but couldn't get the same same potential by adding:

$MnO_4^-(aq) + 4H^+(aq) + 5e^- \rightarrow Mn^{2+}(aq) + 2H_2O(l)$ 1.51V$\ce{MnO4^{-}(aq) + 4H^+(aq) + 5e- -> Mn^{2+}(aq) + 2H_2O(l)}\quad\quad 1.51 \,\mathrm{V}$

And

$MnO_4^-(aq) + 4H^+(aq) +3e^- \rightarrow MnO_2(s) + 2H_2O(l)$ 1.7V$\ce{MnO4^{-}(aq) + 4H+(aq) +3e- -> MnO2(s) + 2H_2O(l)}\quad\quad 1.7\,\mathrm{V}$

By combining I got 0.19V. Why caused this?(there are half reaction equation that can be obtained with the right potential by combining two or more other half reaction equations)

I mean by "combining" making a new half equation. Not a overall equation.

One that cannot be made by combining:

$MnO_2(s) + 4H^+(aq) + 2e^- \rightarrow Mn^{2+}(aq) + 2H_2O(l)$ $E^。= 1.23V$

I was trying to combine these two but couldn't get the same same potential by adding:

$MnO_4^-(aq) + 4H^+(aq) + 5e^- \rightarrow Mn^{2+}(aq) + 2H_2O(l)$ 1.51V

And

$MnO_4^-(aq) + 4H^+(aq) +3e^- \rightarrow MnO_2(s) + 2H_2O(l)$ 1.7V

By combining I got 0.19V. Why caused this?(there are half reaction equation that can be obtained with the right potential by combining two or more other half reaction equations)

I mean by "combining" making a new half equation. Not a overall equation.

One that cannot be made by combining:

$\ce{MnO2(s) + 4H+(aq) + 2e- -> Mn^{2+}(aq) + 2H2O(l)}\quad E^\circ= 1.23\,\mathrm{V}$

I was trying to combine these two but couldn't get the same same potential by adding:

$\ce{MnO4^{-}(aq) + 4H^+(aq) + 5e- -> Mn^{2+}(aq) + 2H_2O(l)}\quad\quad 1.51 \,\mathrm{V}$

And

$\ce{MnO4^{-}(aq) + 4H+(aq) +3e- -> MnO2(s) + 2H_2O(l)}\quad\quad 1.7\,\mathrm{V}$

By combining I got 0.19V. Why caused this?(there are half reaction equation that can be obtained with the right potential by combining two or more other half reaction equations)

1
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