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How would one convert on the right side of the cell diagram:

$$\ce{Zn(s) | Zn^2+(aq) || MnO4-(aq), Mn^2+(aq), H+(aq) | Pt(s)}$$

$$ \begin{align} \ce{Zn(s) &-> Zn^2+(aq)} \tag{ox}\\ \ce{MnO4-(aq) &-> Mn^2+(aq)}\tag{red} \end{align} $$

Obviously, one would then balance and combine the half-equations. I just need to know if the unbalanced half-equations are correct.

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Assuming you meant $\ce{MnO^-4}$ and not $\ce{MnO^{4-}}$, your reduction half reaction is incorrect. Because according to your equation $\ce{Mn^{2+}}$ is being oxidised to $\ce{MnO^-4}$ rather than being reduced.

The correct half cell reactions would be:

$$ \ce{Zn -> Zn^{2+} + 2e-}\\ \ce{MnO^-4 + 8H+ + 5e- -> Mn^{2+} + 4H2O}\\ $$

And so the overall balanced reaction would come out to be:

$$\ce{5Zn + 2MnO^-4 + 16H+ -> 5Zn^{2+} + 2Mn^{2+} + 8H2O}$$

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    $\begingroup$ It was typed wrong on the tutorial sheet and I did not spot the error before submitting the question. The corrected cell diagram makes far more sense; before it was conflicting against cell diagram convention. $\endgroup$ – user108552 Apr 16 at 6:30
  • $\begingroup$ To clarify: The H+ on the RHS of the cell diagram simply means the reaction is being performed under acidic conditions? $\endgroup$ – user108552 Apr 16 at 6:37
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    $\begingroup$ @jseissmann Yes, it shows that the medium is acidic. $\endgroup$ – Nisarg Bhavsar Apr 16 at 6:46
  • $\begingroup$ @jseissmann For a reaction equation to be balanced, the number of atoms on the right, and on the left hand side must be ... equal. Simultaneously, the sum of the charges on the left must equate the sum of the charges on the right hand side of the arrow, too. The reduction, as written by you: $\ce{ MnO^-4 -> Mn^{2+} }$. Left hand side: four atoms of oxygen, right hand side: none; thus not balanced. $\endgroup$ – Buttonwood Apr 16 at 6:57
  • $\begingroup$ @NisargBhavsar This is true, I forgot to set the @ sign; the comment was to the OP. $\endgroup$ – Buttonwood Apr 16 at 6:58

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