Chemistry » Chemistry 111: Electrochemical Reactions » Standard Electrode Potentials

Spontaneity

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Spontaneity

Tip:

Spontaneous

positive EMF

Non-spontaneous

negative EMF

Table: Using EMF to determine cell spontaneity.

You can see from the table of reduction potentials (see table from previous lesson) that different metals have different reactivities. Some are reduced more easily than others. You can also say that some are oxidised more easily than others.

For example, copper (E° = \(\text{+0.34}\) \(\text{V}\)) is more easily reduced than zinc (E° = \(-\text{0.76}\) \(\text{V}\)). Therefore, if a reaction involves the reduction of copper and the oxidation of zinc, it will occur spontaneously. If, however, it requires the oxidation of copper and the reduction of zinc, it will not occur spontaneously.

To predict whether a reaction occurs spontaneously you can look at the sign of the EMF value for the cell. If the EMF is positive then the reaction is spontaneous. If the EMF is negative then the reaction is not spontaneous.

Fact:

One can perform experiments to predict whether a reaction will be spontaneous or not. It turns out that the sign of the EMF is equivalent to whether a cell reaction is spontaneous or not. Those reactions that are spontaneous have a positive EMF and those reactions that are non-spontaneous have a negative EMF.

Look at the following example to help you to understand how to predict whether a reaction will take place spontaneously or not.

In the reaction: \(\color{red}{\text{Pb}^{2+}{\text{(aq)}}} + \color{blue}{\text{2Br}^{-}{\text{(aq)}}} \rightleftharpoons \color{blue}{\text{Br}_{2}{\text{(l)}}} + \color{red}{\text{Pb(s)}}\)

the two half-reactions are as follows:

\(\color{red}{\text{Pb}^{2+}{\text{(aq)}} + {\text{2e}}^{-} \rightleftharpoons \text{Pb(s)}}\) (E° = \(-\text{0.13}\) \(\text{V}\))

\(\color{blue}{\text{Br}_{2}{\text{(l)}} + {\text{2e}}^{-} \rightleftharpoons \text{2Br}^{-}{\text{(aq)}}}\) (E° = \(\text{+1.06}\) \(\text{V}\))

EMF = E°(reduction half-reaction) – E°(oxidation half-reaction)

EMF = E°(lead) – E°(bromide)

EMF = \(-\text{0.13}\) \(\text{V}\) – \(\text{1.06}\) = \(-\text{1.19}\) \(\text{V}\)

The sign of the EMF is negative, therefore this reaction will not take place spontaneously. Let’s look at the reasoning behind this in more detail.

Look at the electrode potential for the first half-reaction. The negative value shows that lead loses electrons more easily than bromine, in other words it is easily oxidised. However, in the original equation, lead ions (\(\text{Pb}^{2+}\)) are being reduced. This part of the reaction is not spontaneous.

For the second half-reaction the positive electrode potential value shows that bromine is more easily reduced than lead: bromine will more easily gain electrons to become \(\text{Br}^{-}\). This is not what is happening in the original equation and therefore this is also not spontaneous.

It therefore makes sense that the reaction will not proceed spontaneously.

Example: Determining Spontaneity

Question

Will copper react with dilute sulfuric acid (\(\text{H}_{2}\text{SO}_{4}\))? You are given the following half-reactions:

\(\text{Cu}^{2+}(\text{aq}) + 2\text{e}^{-}\) \(\rightleftharpoons\) \(\text{Cu}(\text{s})\) (E° = \(\text{+0.34}\) \(\text{V}\))

\(2\text{H}^{+}(\text{aq}) + 2\text{e}^{-}\) \(\rightleftharpoons\) \(\text{H}_{2}(\text{g})\) (E° = \(\text{0.00}\) \(\text{V}\))

Step 1: Determine the overall equation for this reaction

The question asked if copper metal will react with dilute sulfuric acid, therefore the reactants are \(\text{Cu}(\text{s})\) and \(\text{H}^{+}(\text{aq})\):

\(\text{Cu}(\text{s}) + 2\text{H}^{+}(\text{aq})\) \(\to\) \(\text{Cu}^{2+}(\text{aq}) + \text{H}_{2}(\text{g})\)

Step 2: Which reactant should be oxidised, and which should be reduced?

\(\text{Cu}(\text{s})\) \(\to\) \(\text{Cu}^{2+}(\text{aq}) + 2\text{e}^{-}\), therefore copper needs to be oxidised.

\(2\text{H}^{+}(\text{aq}) + 2\text{e}^{-}\) \(\to\) \(\text{H}_{2}(\text{g})\), therefore hydrogen ions needs to be reduced.

Step 3: Predict whether the reaction will be spontaneous or non-spontaneous

Copper has a larger, positive E° than hydrogen. Therefore copper is more easily reduced than hydrogen, and hydrogen is more easily oxidised than copper.

The reaction is non-spontaneous.

Step 4: Calculate the EMF of the cell

E\(^{\circ}_{\text{(cell)}}\) = E\(_{\text{(reduction)}}\) – E\(_{\text{(oxidation)}}\)

E\(^{\circ}_{\text{(cell)}}\) = E\(_{\text{(hydrogen ions)}}\) – E\(_{\text{(copper)}}\)

E\(^{\circ}_{\text{(cell)}}\) = \(\text{0.00}\) \(\text{V}\) – (\(\text{+0.34}\) \(\text{V}\)) = \(-\text{0.34}\) \(\text{V}\)

Step 5: Is the reaction spontaneous?

The EMF is negative, therefore the reaction is non-spontaneous.

Example: Determining Spontaneity

Question

Will zinc react with dilute hydrochloric acid (\(\text{HCl}\))? You are given the following half-reactions:

\(\text{Zn}^{2+}(\text{aq}) + 2\text{e}^{-}\) \(\rightleftharpoons\) \(\text{Zn}(\text{s})\) (E° = \(-\text{0.76}\) \(\text{V}\))

\(2\text{H}^{+}(\text{aq}) + 2\text{e}^{-}\) \(\rightleftharpoons\) \(\text{H}_{2}(\text{g})\) (E° = \(\text{0.00}\) \(\text{V}\))

Step 1: Determine the overall equation for this reaction

The question asked if zinc metal will react with dilute hydrochloric acid, therefore the reactants are \(\text{Zn}(\text{s})\) and \(\text{H}^{+}\):

\(\text{Zn}(\text{s}) + 2\text{H}^{+}(\text{aq})\) \(\to\) \(\text{Zn}^{2+}(\text{aq}) + \text{H}_{2}(\text{g})\)

Step 2: Which reactant should be oxidised, and which should be reduced?

\(\text{Zn}(\text{s})\) \(\to\) \(\text{Zn}^{2+}(\text{aq}) + 2\text{e}^{-}\), therefore zinc needs to be oxidised.

\(2\text{H}^{+}(\text{aq}) + 2\text{e}^{-}\) \(\to\) \(\text{H}_{2}(\text{g})\), therefore hydrogen ions need to be reduced.

Step 3: Predict whether the reaction will be spontaneous or non-spontaneous

Zinc has a larger, negative E° than hydrogen. Therefore zinc is more easily oxidised than hydrogen, and hydrogen is more easily reduced than copper.

The reaction is spontaneous.

Step 4: Calculate the EMF of the cell

E\(^{\circ}_{\text{(cell)}}\) = E\(_{\text{(reduction)}}\) – E\(_{\text{(oxidation)}}\)

E\(^{\circ}_{\text{(cell)}}\) = E\(_{\text{(hydrogen ions)}}\) – E\(_{\text{(zinc)}}\)

E\(^{\circ}_{\text{(cell)}}\) = \(\text{0.00}\) \(\text{V}\) – (\(-\text{0.76}\) \(\text{V}\)) = \(\text{+0.76}\) \(\text{V}\)

Step 5: Is the reaction spontaneous?

The EMF is positive, therefore the reaction is spontaneous.

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