The amount of manganese in steel is determined by changing...

The amount of manganese in steel is determined by changing it to permanganate ion. The steel is first dissolved in nitric acid, producing $\mathrm{Mn}^{2+}$ ions. These ions are then oxidized to the deeply colored $\mathrm{MnO}_{4}^{-}$ ions by periodate ion $\left(\mathrm{IO}_{4}^{-}\right)$ in acid solution. a. Complete and balance an equation describing each of the above reactions. b. Calculate $\mathscr{E}^{\circ}$ and $\Delta G^{\circ}$ at $25^{\circ} \mathrm{C}$ for each reaction.

Transcript

00:01 In this question, we're going to be applying some concepts that we learned in the topic of electrochemistry.

00:06 So for the first part, it is asking for the balanced chemical equations, representing the two equations.

00:12 So we need three moles of mn reacting with eight moles of h plus and two moles of n0 to produce three moles of mn2 plus and two moles of n0 and four modes of h2o.

00:29 This is equation 1 and then the second equation, we need 2 moles of mn2 plus that are going to rate with 5 moles of 04 minus and 3 moles of h2o to produce 2 moles of mno 4 plus plus 5 moles of io3 minus and 6 moles of h plus.

00:52 And then looking at the next equation, what would want to do is to determine the gibbs energy change and the equilibrium constant at a temperature.

00:59 Of 25 degrees celsius.

01:02 So looking at the information that we've been given, looking at, for example, equation one, this is the overall equation and for us to look into the gibbs energy change, we need to first of all determine the cell potential.

01:17 So the cell potential for the first equation would need the oxidation reaction and the reduction reaction.

01:24 So what we have here is the oxidation reaction.

01:27 This is 3 moles of mn being oxidized.

01:30 Into 3 moles of mn2 plus plus 6 electrons.

01:35 And then we've got the reduction equation where we've got 2 moles of n0 3, reacting with 8 moles of h plus the 6 electrons that are being lost from the anode to produce 2n0 plus 4 of h2o.

01:54 Now these two equations are what gave us equation 1.

01:57 So the cell potential for this process, this equation 1, this equation 1, this equation 1, this is going to be equal to the carpet and the annod potential, which is 0 .96 minus negative 1 .18, giving us a standard cell potential for the first reaction that is equal to 2 .14 volts...

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