Post Closed as "duplicate" by orthocresol of
4 Corrected stoichiometry, added symbology, spelling, clarification, and grammar.

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.

# 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.

2 the use of the \ce{...} environment saves a lot of typing

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