The Foundations of Chemistry

EXAMPLE 21-4 Predicting the Direction of Reactions

In an acidic solution at standard conditions, will tin(IV) ions, Sn^4 , oxidize gaseous nitrogen

oxide, NO, to nitrate ions, NO 3 , or will NO 3 oxidize Sn^2 to Sn^4 ions? Write the cell

reaction and calculate E^0 cellfor the spontaneous reaction.

Plan

We refer to the table of standard reduction potentials (see Table 21-3) and choose the appro-

priate half-reactions.

Solution

The NO 3 /NO reduction half-reaction has the more positive E^0 value, so we write it first and

write the Sn^4 /Sn^2 half-reaction as an oxidation. We balance the electron transfer and add

the two half-reactions to obtain the equation for the spontaneousreaction. Then we add the

half-reaction potentials to obtain the overall cell potential.

E^0

2(NO 3 4H 3 e88nNO2H 2 O) (reduction) 0.96 Vo

3(Sn^2 88nSn^4  2 e) (oxidation) 0.15 Vo

o2NO 3 8H3Sn^2 88n2NO4H 2 O3Sn^4  E^0 cell0.81 Vo

Because E^0 cellis positive for this reaction,

nitrate ions spontaneously oxidize tin(II) ions to tin(IV) ions and are reduced to nitrogen

oxide in acidic solution.

You should now work Exercises 55 and 56.

872 CHAPTER 21: Electrochemistry

Problem-Solving Tip:A Common Error in E^0 cellCalculations

Remember the italicized warning in step 4 of the procedure set out in Section 21-15.

We do not multiply the potentials by the numbers used to balance the electron transfer!This is

a very common error.

Problem-Solving Tip:Remember What We Mean by Standard

Conditions

When we say that a reaction takes place at standard conditions,we mean the following:

1.The temperature is the standard thermodynamic temperature, 25°C, unless stated

otherwise.

2.All reactants and products are at unit activity.This means that:

a.Any solution species that takes part in the reaction is at a concentration of exactly

1 M;

b.Any gas that takes part in the reaction is at a pressure of exactly 1 atm;

c.Any other substance that takes part in the reaction is pure.

(When we say “takes part in the reaction” we mean either as a reactant or a product.)

These are the same conditions that were described as standard conditionsfor thermody-

namic purposes (Section 15-6). When one or more of these conditions is not satisfied,

we must adjust our calculations for nonstandard conditions. We shall learn how to do

this in Section 21-19.