Under standard conditions, what reaction occurs, if any, when each of the following operations i

Under standard conditions, what reaction occurs, if any,...

Key Concepts

Balanced Redox Equations

Formulating balanced redox equations involves writing separate half?reactions for oxidation and reduction processes and then combining them to ensure both mass and charge are balanced. This technique is fundamental for quantitative analyses in electrochemistry, such as computing overall cell potentials, free energy changes, and equilibrium constants.

Electrochemical Thermodynamics

This concept connects the electrochemical potentials to thermodynamic functions. The standard free energy change (?G°) of a redox reaction is related to its cell potential by the equation ?G° = –nFE°, where n is the number of electrons transferred and F is the Faraday constant. This free energy change, in turn, is linked to the equilibrium constant (K) through the relation ?G° = –RT ln K, allowing prediction of reaction completeness and direction under standard conditions.

Halogen Displacement Reactions

Halogen displacement reactions are a specific type of redox process where a more reactive halogen oxidizes the halide ion of a less reactive halogen. An understanding of the reactivity series of halogens is crucial here, as it helps to predict whether one halogen can displace another from its ionic compound in solution.

Redox Reaction Spontaneity

Redox reaction spontaneity is determined by the cell potential (E°_cell), which is calculated from the standard electrode potentials of the half?reactions. A positive E°_cell indicates a spontaneous reaction under standard conditions, while a negative value suggests that the reaction is nonspontaneous. This analysis is essential for deciding which reactions occur when two species interact.

Standard Electrode Potentials

This concept involves using tabulated standard reduction potentials to predict the direction of redox reactions under standard conditions. By comparing the potential values of the reduction half?reactions, one can determine whether a given redox transformation will proceed spontaneously, with the higher (more positive) potential favorably undergoing reduction and the lower one undergoing oxidation.

Transcript

00:01 In this problem, we are given descriptions of four different reactions, and we want to determine first whether or not the reaction will occur.

00:09 And we secondly want to determine for the reactions that will indeed occur what the values are for the cell potential and change in gives free energy at standard conditions, as well as the value for the equilibrium constant k at 25 degrees celsius.

00:24 So the way that we need to determine whether these reactions will occur is whether or not they are.

00:30 Are spontaneous.

00:32 So we need to first calculate the cell potential, and we know that for positive values of the cell potential, that that corresponds to a spontaneous reaction that will indeed occur.

00:41 So in part a, we have i2 crystals being added to a solution of nacl.

00:47 So i2 in the form of crystals must mean that it is a solid, and nacl means that we have cl minus ions and aqueous solution.

00:56 So we can look up the two half reactions of that corresponds to from our standard reduction potentials table.

01:05 And we know that we have to have i2 being reduced.

01:11 And we also have to have cl minus being oxidized to cl2.

01:18 So we have to flip the sign of the reduction potential for that half reaction.

01:23 And when we do that, we can now combine these two equations to find the overall cell potential.

01:30 And when we add those two standard reduction potentials together, we see that the cell potential comes out to negative 0 .82 volts.

01:48 And since that cell potential at standard conditions is less than zero, this is non -spontaneous.

02:00 And that means that this reaction will not proceed unless we supply some kind of energy.

02:10 So that reaction will not occur for that reason, that it is non -spontaneous.

02:16 Now, part b, we are told that we are putting cl2 gas into a solution of nai.

02:23 So this time we start with cl2 gas, and we have i -minus in solution.

02:30 So from the half -reactions in the standard reduction potentials table, we have to have cl2 being reduced this time, and we also have to have i -minus being oxidized to i -2.

02:45 So that is why we have to flip the first half reaction, and when we do that, we can add together the standard reduction potentials, and this time we see that the overall cell potential at standard conditions comes out to a positive value of 0 .82 volts.

03:10 And so now since the value of the cell potential is positive greater than zero, we know that this reaction will proceed spontaneously, and so this reaction will occur.

03:22 And now we need to calculate the value for the change in gibbs free energy.

03:29 Again, this is the overall reaction, so we can eliminate those two electrons.

03:35 So when we combine that together, we have cl2 gas plus 2 i minus aqueous, going to i2 solid plus 2cl minus aqueous.

04:00 That is the overall reaction.

04:03 And now to find the change in gibbs free energy, delta g, standard conditions, that's just negative n -f -e.

04:15 N is the number of moles of electrons, and from that equation we canceled out two electrons, and faraday's constant is 96 ,485 coolums per moles of electrons, and the cell potential we found was 0 .82 joules per quorum, which corresponds to units of volts...

Please give Ace some feedback