To know the relationship in between cell potential and the equilibrium constant. To use cell potentials to calculate systems concentrations.

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Changes in reaction conditions can have a tremendous effect on the course of a redox reaction. Because that example, under standard conditions, the reaction that (ceCo(s)) v (ceNi^2+(aq)) to type (ceNi(s)) and(ceCo^2+(aq)) wake up spontaneously, however if we minimize the concentration of (ceNi^2+) by a variable of 100, so that(ce) is 0.01 M, climate the turning back reaction occurs spontaneously instead. The relationship in between voltage and concentration is just one of the determinants that have to be construed to predict even if it is a reaction will certainly be spontaneous.

## The Relationship in between Cell Potential & Gibbs Energy

Electrosnucongo.orgical cells convert snucongo.orgical power to electrical energy and also vice versa. The complete amount the energy developed by an electrosnucongo.orgical cell, and also thus the amount of energy available to do electric work, depends on both the cabinet potential and the total variety of electrons that are transferred native the reductant come the oxidant during the course of a reaction. The resulting electric current is measure in coulombs (C), an SI unit that procedures the number of electrons passing a given suggest in 1 s. A coulomb relates power (in joules) to electrical potential (in volts). Electric present is measure in amperes (A); 1 A is identified as the flow of 1 C/s previous a given point (1 C = 1 A·s):

In snucongo.orgistry reactions, however, we have to relate the coulomb come the fee on a mole of electrons. Multiplying the fee on the electron by Avogadro’s number provides us the charge on 1 mol the electrons, i m sorry is dubbed the faraday (F), called after the English physicist and also snucongo.orgist Michael Faraday (1791–1867):

<eginalignF &=(1.60218 imes10^-19 extrm C)left(dfrac6.02214 imes 10^23, J extrm1 mol e^- ight)\<4pt> &=9.64833212 imes10^4 extrm C/mol e^- \<4pt> &simeq 96,485, J/(mathrmVcdot mol;e^-)endalign label20.5.2>

The total charge moved from the reductant to the oxidant is thus (nF), whereby (n) is the variety of moles of electrons.

Faraday to be a british physicist and also snucongo.orgist that was arguably one of the greatest experimental scientists in history. The boy of a blacksmith, Faraday was self-educated and also became an apprentice bookbinder at period 14 before turning to science. His experiments in electricity and magnetism made power a routine tool in science and led to both the electrical motor and the electrical generator. He discovered the phenomenon the electrolysis and also laid the foundations of electrosnucongo.orgistry. In fact, many of the devoted terms introduced in this thing (electrode, anode, cathode, and so forth) are because of Faraday. In addition, he uncovered benzene and invented the system of oxidation state numbers the we use today. Faraday is probably best known because that “The snucongo.orgical history of a Candle,” a collection of windy lectures ~ above the snucongo.orgistry and physics the flames.

The maximum lot of work that deserve to be produced by one electrosnucongo.orgical cell ((w_max)) is equal to the product that the cell potential ((E^°_cell)) and the full charge transferred throughout the reaction ((nF)):

< w_max = nFE_cell label20.5.3>

Work is expressed together a an adverse number due to the fact that work is being excellent by a mechanism (an electrosnucongo.orgical cell with a positive potential) top top its surroundings.

The readjust in complimentary energy ((DeltaG)) is additionally a measure up of the maximum quantity of work-related that deserve to be performed throughout a snucongo.orgical procedure ((ΔG = w_max)). Consequently, there should be a relationship between the potential of one electrosnucongo.orgical cell and (DeltaG); this relationship is as follows:

A spontaneous redox reaction is thus characterized by a an adverse value of (DeltaG) and also a optimistic value of (E^°_cell), constant with our earlier discussions. When both reactants and products are in their conventional states, the relationship in between ΔG° and also (E^°_cell) is together follows:

A spontaneous oxidization reaction is characterized by a an unfavorable value the ΔG°, which corresponds to a hopeful value of E°cell.

Example (PageIndex1)

Suppose you want to prepare elemental bromine native bromide utilizing the dichromate ion as an oxidant. Making use of the data in Table P2, calculate the free-energy adjust (ΔG°) for this oxidization reaction under conventional conditions. Is the reaction spontaneous?

Given: redox reaction

Asked for: (ΔG^o) for the reaction and also spontaneity

Strategy:

native the appropriate half-reactions and the equivalent values of (E^o), write the in its entirety reaction and calculate (E^°_cell). Recognize the number of electrons transferred in the overall reaction. Then use Equation ef20.5.5to calculate (ΔG^o). If (ΔG^o) is negative, climate the reaction is spontaneous.

Solution

A

As always, the first step is to compose the appropriate half-reactions and also use castle to achieve the overall reaction and also the size of (E^o). Indigenous Table P2, us can uncover the reduction and also oxidation half-reactions and also corresponding (E^o)values:

<eginalign*& extrmcathode: &quad & mathrmCr_2O_7^2-(aq) + mathrm14H^+(aq)+mathrm6e^- ightarrow mathrm2Cr^3+(aq) +mathrm7H_2O(l)&quad & E^circ_ extrmcathode = extrm1.36V \& extrmanode: &quad & mathrm2Br^-(aq) ightarrow mathrmBr_2(aq) +mathrm2e^-&quad & E^circ_ extrmanode = extrm1.09 Vendalign*>

To obtain the all at once balanced snucongo.orgical equation, we should multiply both political parties of the oxidation half-reaction through 3 to acquire the same variety of electrons as in the reduction half-reaction, remembering the the magnitude of (E^o)is not affected:

<eginalign*& extrmcathode: &quad & mathrmCr_2O_7^2-(aq) + mathrm14H^+(aq)+mathrm6e^- ightarrow mathrm2Cr^3+(aq) +mathrm7H_2O(l)&quad & E^circ_ extrmcathode = extrm1.36V \<4pt>& extrmanode: &quad & mathrm6Br^-(aq) ightarrow mathrm3Br_2(aq) +mathrm6e^-&quad & E^circ_ extrmanode = extrm1.09 V \<4pt> hline& extrmoverall: &quad & mathrmCr_2O_7^2-(aq) + mathrm6Br^-(aq) + mathrm14H^+(aq) ightarrow mathrm2Cr^3+(aq) + mathrm3Br_2(aq) +mathrm7H_2O(l)&quad & E^circ_ extrmcell = extrm0.27Vendalign*>

B

We deserve to now calculation ΔG° using Equation ( ef20.5.5). Since six electrons room transferred in the in its entirety reaction, the worth of (n) is 6:

<eginalign*Delta G^circ &=-(n)(F)(E^circ_ extrmcell) \<4pt> & =-( extrm6 mole)<96,485;mathrmJ/(Vcdot mol)( extrm0.27V)> \& =-15.6  imes 10^4 extrm J \ & =-156;mathrmkJ/mol;Cr_2O_7^2- endalign*>

Thus (ΔG^o) is −168kJ/molfor the reaction as written, and the reaction is spontaneous.

Exercise (PageIndex1)

Use the data in Table P2 to calculation (ΔG^o) because that the palliation of ferric ion by iodide:

Is the reaction spontaneous?

−44 kJ/mol I2; yes

## Potentials for the Sums of Half-Reactions

Although Table P2 list number of half-reactions, many much more are known. As soon as the standard potential because that a half-reaction is no available, we have the right to use relationships in between standard potentials and complimentary energy to attain the potential of any kind of other half-reaction that deserve to be created as the sum of 2 or much more half-reactions whose standard potentials room available. For example, the potential for the palliation of (ceFe^3+(aq)) come (ceFe(s)) is not listed in the table, however two related reductions are given:

Fe^2+(aq) ;;;E^° = +0.77 V label20.5.6 onumber >

Fe(s) ;;;E^° = −0.45 V label20.5.7 onumber >

Although the sum of these 2 half-reactions offers the desired half-reaction, we cannot simply include the potentials of 2 reductive half-reactions to achieve the potential that a third reductive half-reaction because (E^o)is not a state function. However, due to the fact that (ΔG^o)is a state function, the sum of the (ΔG^o)values for the individual reactions gives us (ΔG^o)for the overall reaction, i beg your pardon is proportional to both the potential and also the variety of electrons ((n)) transferred. To acquire the value of (E^o)for the all at once half-reaction, we very first must add the values of (ΔG^o(= −nFE^o)) for each separation, personal, instance half-reaction to acquire (ΔG^o) for the overall half-reaction:

<eginalign*ceFe^3+(aq) + cee^- & ightarrow cemathrmFe^2+(aq) &quad Delta G^circ&=-(1)(F)( extrm0.77 V)\<4pt>ceFe^2+(aq)+ce2e^- & ightarrowceFe(s) &quadDelta G^circ &=-(2)(F)(- extrm0.45 V)\<4pt>ceFe^3+(aq)+ce3e^- & ightarrow ceFe(s) &quadDelta G^circ & =<-(1)(F)( extrm0.77 V)>+<-(2)(F)(- extrm0.45 V)> endalign*>

Solving the last expression because that ΔG° because that the in its entirety half-reaction,

= F(0.13 V) label20.5.9 onumber >

Three electron ((n = 3)) space transferred in the all at once reaction, for this reason substituting into Equation ( ef20.5.5) and solving for (E^o)gives the following:

<eginalign*Delta G^circ & =-nFE^circ_ extrmcell \<4pt>F( extrm0.13 V) & =-(3)(F)(E^circ_ extrmcell) \<4pt>E^circ & =-dfrac0.13 extrm V3=-0.043 extrm Vendalign*>

This value of (E^o)is really different from the value that is obtained by simply adding the potentials because that the two half-reactions (0.32 V) and also even has actually the the contrary sign.

Values that (E^o)for half-reactions can not be included to provide (E^o)for the sum of the half-reactions; just values that (ΔG^o= −nFE^°_cell) because that half-reactions have the right to be added.

## The Relationship between Cell Potential & the Equilibrium Constant

We have the right to use the relationship in between (DeltaG^°) and also the equilibrium constant (K), to obtain a relationship in between (E^°_cell) and (K). Recall the for a basic reaction the the type (aA + bB ightarrow cC + dD), the traditional free-energy change and the equilibrium constant are related by the following equation:

Given the relationship between the conventional free-energy change and the traditional cell potential (Equation ( ef20.5.5)), we deserve to write

<−nFE^°_cell = −RT ln K label20.5.12>

Rearranging this equation,

For (T = 298, K), Equation ( ef20.5.12B) deserve to be streamlined as follows:

< eginalign E^circ_ extrmcell &=left(dfracRTnF ight)ln K \<4pt> &=left< dfrac<8.314;mathrmJ/(molcdot K)( extrm298 K)>n<96,485;mathrmJ/(Vcdot mol)> ight>2.303 log K \<4pt> &=left(dfrac extrm0.0592Vn ight)log K label20.5.13 endalign>

Thus (E^°_cell) is directly proportional to the logarithm that the equilibrium constant. This way that large equilibrium constants exchange mail to large positive values of (E^°_cell) and vice versa.

Example (PageIndex2)

Use the data in Table P2 to calculation the equilibrium consistent for the reaction of metallic lead with PbO2 in the visibility of sulfate ions to offer PbSO4 under standard conditions. (This reaction occurs once a auto battery is discharged.) Report your answer come two far-reaching figures.

Given: oxidation reaction

Strategy:

write the pertinent half-reactions and potentials. Native these, achieve the overall reaction and (E^o_cell). Determine the variety of electrons transferred in the overall reaction. Use Equation ( ef20.5.13) to solve for (log K) and then (K).

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Solution

A The appropriate half-reactions and potentials indigenous Table P2 room as follows:

<eginalign* & extrm cathode:& & mathrmPbO_2(s)+mathrmSO_4^2-(aq)+mathrm4H^+(aq)+mathrm2e^- ightarrowmathrmPbSO_4(s)+mathrm2H_2O(l)& & E^circ_ extrmcathode= extrm1.69 V \<4pt>& extrmanode:& & mathrmPb(s)+mathrmSO_4^2-(aq) ightarrowmathrmPbSO_4(s)+mathrm2e^-& & E^circ_ extrmanode=- extrm0.36 V \<4pt> hline& extrm overall:& & mathrmPb(s)+mathrmPbO_2(s)+mathrm2SO_4^2-(aq)+mathrm4H^+(aq) ightarrowmathrm2PbSO_4(s)+mathrm2H_2O(l)& & E^circ_ extrmcell= extrm2.05 V endalign*>

B two electrons space transferred in the as whole reaction, therefore (n = 2). Addressing Equation ( ef20.5.13) because that log K and also inserting the values of (n) and also (E^o),

<eginalign*log K & =dfracnE^circ extrm0.0591 V=dfrac2( extrm2.05 V) extrm0.0591 V=69.37 \<4pt>K & =2.3 imes10^69endalign*>

Thus the equilibrium lies far to the right, donate a discharged battery (as anyone that has ever before tried unsuccessfully to begin a car after letting it sit for a long time will know).

Exercise (PageIndex2)

Use the data in Table P2 to calculation the equilibrium continuous for the reaction the (ceSn^2+(aq)) through oxygen to produce (ceSn^4+(aq)) and water under conventional conditions. Report your answer to two far-ranging figures. The reaction is together follows: