Electrolytic Cells (a) When the power is turned on, what happens at the cathode? (b) When the power is turned on, what happens at the anode? (c) Write oxidation and reduction half-reactions. (d) Write an overall equation for the reaction.

Question

Electrolytic Cells

The electrolysis of molten potassium iodide produces liquid potassium and gaseous iodine. Consider the following diagram:

(a) When the power is turned on, what happens at the cathode?
(b) When the power is turned on, what happens at the anode?
(c) Write oxidation and reduction half-reactions.
(d) Write an overall equation for the reaction.

Accepted Answer

(a) Since an oxidation-reduction reaction occurs, one element must be oxidized and the other reduced. Potassium iodide contains K⁺ ions and I^– ions. Based on its position in the periodic table, we wouldn’t expect potassium to become more oxidized than K⁺, so it is reduced. Since reduction always occurs at the cathode, potassium metal is produced there.

(b) Iodide ions form iodine atoms at the anode, because it’s the site of oxidation in an electrochemical cell. Two of these atoms combine to form an iodine molecule.

(c) To begin writing half-reactions, first consider the skeletal equation for the overall reaction. The high temperature required to keep the KI molten will make the potassium liquid and the iodine gaseous. Recall that iodine is one of the elements that exists as diatomic molecules:

KI(l) → K(l) + I₂(g)

Iodine is oxidized in this reaction, so the half-reaction for the oxidation process is

2I^–(l) → I₂(g) + 2e^–

The half-reaction for the reduction of potassium is

K⁺(l) + e^– → K(l)

d) To write an overall equation, combine the oxidation and reduction half-reactions:

Oxidation 2I^–(l) → I2(g) + 2e^–

Reduction K⁺(l) + e^– → K(l)

To equalize electrons lost and gained, multiply the reduction half-reaction by 2:

2[K⁺(l) + e^–→ K(l)]

2K⁺(l) + 2e^–→ 2K(l)

Now add the two half-reactions:

[latex]\begin{matrix}2I^{-}(l)\longrightarrow I_{2}(g)+2e^{-} \ 2K^{+}(l)+2e^{-}\longrightarrow 2K(l) \ \hline 2K^{+}(l)+2I^{-}(l)+2e^{-}\longrightarrow 2K(l)+I_{2}(g)+2e^{-} \end{matrix}[/latex]

The first two terms, K⁺(l) and I^–(l), can be represented as KI(l). We can also cancel 2e^– from both sides of the equation:

2KI(l) 2K(l) + I₂(g)

Question 14.9: Electrolytic Cells (a) When the power is turned on, what hap...

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