Question

Use the following half-equations to write three spontaneous reactions. Justify your answers by calculating \(E^{\circ}\) for the cells. \(\begin{array}{ll}\text { 1. } \mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 e^{-} \longrightarrow & \\ \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O} & E^{\circ}=+1.512 \mathrm{~V} \\ \text { 2. } \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 e^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} & \\\ \text { 3. } \mathrm{Co}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Co}(s) & E^{\circ}=+1.229 \mathrm{~V} \\ & E^{\circ}=-0.282 \mathrm{~V}\end{array}\)

Short answer

Answer: All three reactions are spontaneous when the half-equations are combined, as their calculated cell potentials are positive.

Step-by-step solution

Combine half-reactions with a proper electron balance

First, let's find pairs of half-reactions that we can combine while keeping the number of electrons lost and gained equal. 1. Combining half-reaction 1 and half-reaction 2: \\ Here, half-reaction 1 loses 5 electrons, and half-reaction 2 gains 4 electrons. To balance the electrons, multiply half-reaction 1 by 4 and half-reaction 2 by 5. So, both half-reactions will involve 20 electrons. 2. Combining half-reaction 1 and half-reaction 3: \\ Half-reaction 1 loses 5 electrons, and half-reaction 3 gains 2 electrons. To balance the electrons, multiply half-reaction 1 by 2 and half-reaction 3 by 5. So, both half-reactions will involve 10 electrons. 3. Combining half-reaction 2 and half-reaction 3: \\ Both half-reactions 2 and 3 have an even number of electrons; thus, they can be directly combined without any balancing factor. Now let's calculate the cell potentials for each combination:

Calculate cell potentials

To calculate the cell potentials, add the standard potentials of the relevant half-reactions. 1. For the reaction between half-reaction 1 and half-reaction 2: \\ \(E^{\circ}_{cell} = E^{\circ}_{1} + E^{\circ}_{2} = +1.512 \mathrm{V} - 0.282 \, \mathrm{V} = +1.230 \, \mathrm{V}\) 2. For the reaction between half-reaction 1 and half-reaction 3: \\ \(E^{\circ}_{cell} = E^{\circ}_{1} + E^{\circ}_{3} = +1.512 \mathrm{V} + 1.229 \, \mathrm{V} = +2.741 \, \mathrm{V}\) 3. For the reaction between half-reaction 2 and half-reaction 3: \\ \(E^{\circ}_{cell} = E^{\circ}_{2} + E^{\circ}_{3} = -0.282 \mathrm{V} + 1.229 \, \mathrm{V} = +0.947 \, \mathrm{V}\) All three calculated cell potentials are positive, which means all three reactions are spontaneous.