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Calculate the cell potential of the electrochemical cell in which the following reaction takes place, and represent the cell using standard notation.
(E V = -2.37)
(E V = 0.00)
From the overall equation, it is clear that magnesium is oxidised and hydrogen ions are reduced in this reaction. Magnesium is therefore the anode reaction and hydrogen is the cathode reaction.
E = E (cathode) - E (anode)
= 0.00 - (-2.37)
= +2.37 V
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,
the two half reactions are as follows:
(-0.13 V)
(+1.06 V)
You will see that the half reactions are written as they appear in the table of standard electrode potentials. It may be useful to highlight the reacting substance in each half reaction. In this case, the reactants are Pb and Br ions.
Look at the electrode potential for the first half reaction. The negative value shows that lead loses electrons easily, in other words it is easily oxidised. The reaction would normally proceed from right to left (i.e. the equilibrium lies to the left), but in the original equation, the opposite is happening. It is the Pb ions that are being reduced to lead. This part of the reaction is therefore not spontaneous. The positive electrode potential value for the bromine half-reaction shows that bromine is more easily reduced, in other words the equilibrium lies to the right. The spontaneous reaction proceeds from left to right. This is not what is happening in the original equation and therefore this is also not spontaneous. Overall it is clear then that the reaction will not proceed spontaneously.
Will copper react with dilute sulfuric acid (H SO )? You are given the following half reactions:
(E = +0.34 V)
(E = 0 V)
In the first half reaction, the positive electrode potential means that copper does not lose electrons easily, in other words it is more easily reduced and the equilibrium position lies to the right. Another way of saying this is that the spontaneous reaction is the one that proceeds from left to right, when copper ions are reduced to copper metal.
In the second half reaction, the spontaneous reaction is from right to left.
What you should notice is that in the original reaction, the reactants are copper (Cu) and sulfuric acid (2H ). During the reaction, the copper is oxidised and the hydrogen ions are reduced. But from an earlier step, we know that neither of these half reactions will proceed spontaneously in the direction indicated by the original reaction. The reaction is therefore not spontaneous.
A second method for predicting whether a reaction is spontaneous
Another way of predicting whether a reaction occurs spontaneously, is to 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 .
We will look at this in more detail in the next section.
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