Chapter 19 Worksheet 2Electrochemical cells, EMF, G, K, Nernst equation

VocabularyElectrochemical Cell (voltaic cell, galvanic cell, battery)Redox reactionRedox coupleHalf-reactionHalf-cellElectrodeAnodeCathodeCell potential [synonyms: voltage, electromotive force (emf, Ecell)]Standard cell potential (Eo

cell)Coulomb (C)Volt (1 J/C)Standard reduction potential (Eo

red)

Introduction

Redox reactions involve electron transfer between species, causing changes in their oxidation states. We usually talk of redox couples such as Zn / Zn2+ and Fe2+ / Fe3+ (reduced form / oxidized form) when characterizing redox reactions. (Recall the half-reaction method of balancing redox equations). In this exercise we deal with electrochemical cells, in which two redox couples are placed in separate compartments as opposed to their direct contact.

Cell potential

The cell voltage is a measure of the potential energy difference between the two electrodes so it is also called cell potential. A potential difference of 1 volt (V) is sufficient to impart 1 J of energy to a charge of 1 coulomb (C) (In other words: 1 V = 1 J/C). Cell potential is the driving force for a redox reaction so it is also called electromotive force (emf, E). It is a measure of how far a redox reaction is from equilibrium. You will see that cell potential is directly proportional to G for the redox reaction that is occurring!

Standard cell potential (standard emf, Eocell) is the potential obtained when all reactants and

products are at standard concentrations or pressures (1 M, or 1 atm). That is, when Q = 1. Cells are not usually run with standard concentrations but we will be able to calculate any non-standard Ecell if we know Eo

cell (guess how?).

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Gravitational Potential Energy Cell Potential

Gravitational potential energyEgrav = mgh

It takes 1 J of work to lift a 1 kg mass, 1 m

Cell potentialIt takes 1 J of work to move 1 coulomb of charge “up” a 1 V potential difference.

1 J = 1 C x 1 V

1 V = 1 J/C (or 1 C = 1 J/V)

Charge on an electron = 1.602 x 10-19 C

Charge on one mole of electrons = Faraday’s constant= (1.602 x 10-19 C/e-)(6.022 x 1023 e-/mol) = 9.65 x 104 C/mol

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Although we can only measure differences in potential, a number called the standard reduction potential (Eo

red or simply Eo ) can be assigned to every half-reaction by assigning a value of zero to a reference half-reaction. The reference half-reaction is:

2H+(aq) + 2e- → H2 (g) Eo = 0 V

Reductions that occur more easily than this one are given a positive Eored. Reductions that occur less

easily have a negative Eored.

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The standard cell potential for any cell can be calculated as the difference between the standard reduction potentials for the reactions occurring at the two electrodes.

Eocell = Eo

cathode - Eoanode

(Some textbooks use the symbol Eo instead of Eo

cell.)

Zinc-Copper Cell: Zn/Zn2+ ׀׀ Cu/Cu2+

Overall cell reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

Memory tricksOxidation takes place at the Anode (vowels!)

Reduction takes place at the Cathode (consonants! Red wire!)Cathode and Cations are +; Anode and Anions are -

Cations are attracted to the Cathode Anions are attractded to the Anode

Standard Hydrogen Electrode (SHE)

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red wire

The relationship between G and E cell

Like G, Ecell is a measure of how far the reaction is from equilibrium. It is the “driving force”.

Like G, Ecell depends on:The identity of the reactionThe initial concentration of reactants and products (Q)Temperature

There is a simple mathematical relationship between G and Ecell:

G = -nFEcell

n = number of electrons transferred in the balanced equation (unitless)F = Faraday’s constant = charge on a mole of electrons = 96,500 C/mol e-

What is the unit on G? (remember: 1V = 1 J/C)

For a spontaneous redox reaction: G < 0, Ecell > 0

At equilibrium: G = 0, and Ecell = 0 (For a cell, equilibrium is death!)

EMF of a non-standard cell (Nernst equation)

G = Go + RTln Q

-nFEcell = -nFEocell + RTln Q

Ecell = Eocell – (RT/nF)ln Q (Nernst equation)

By convention, the Nernst equation is written in terms of log Q rather than ln Q:

ln Q = (ln 10)(log Q) = 2.303log Q, so

Ecell = Eocell – (2.303RT/nF)log Q

Assuming T = 298 K and combining constants:

2.303RT/F = [(2.303)(8.314 J/mol-K)(298 K)] / 96,500 C/mol = 0.0592 J/C = 0.0592 V

Ecell = Eocell – (0.0592/n)log Q (Nernst equation at T = 298 K)

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Calculating cell potential ( E cell)

1. Consider an electrochemical cell (a.k.a. galvanic cell, voltaic cell, battery) with Zn(s) and 0.25 M Zn(NO3)2(aq) in one compartment and Cu(s) and 0.25 M Cu(NO3)2(aq) in the other compartment.

a. Calculate the cell potential (EMF, voltage) at 298 K.

Reduction (cathode): Eocathode =

Oxidation (anode): Eoanode =

Net (overall cell reaction): Eocell =

Ecell = Eocell – (0.0592/n)log Q =

b. What would be the cell potential if the concentration of Zn(NO3)2 was increased to 2.5 M?

c. Calculate the equilibrium constant (at 298 K) for the overall cell reaction.

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2. Consider a voltaic cell with Cr(s) and Cr3+(aq) in one compartment and Zn(s) and Zn2+(aq) in the other compartment.

a. Draw and label a diagram of this cell.

b. Calculate the standard cell potential and the equilibrium constant for the overall reaction.

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c. Calculate the cell potential if [Zn2+] = 5.00 M and [Cr3+] = 0.50 M? What does this result tell you? How would your diagram change? How does Q compare to K?

d. If zinc is the anode and [Cr3+] = 0.010 M, what must the [Zn2+] be in order to achieve 0.050V?

e. What must the value of Q be in order for the voltage to be zero?

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3. A standard zinc half-cell was coupled to a standard hydrogen electrode. When the red wire was attached to the hydrogen electrode, the cell potential was 0.763 V. What is the standard reduction potential for the zinc half-cell?

Zn/Zn2+ ׀׀ H2/H+ [Zn2+] = 1.0 M, [H+] = 1.0 M, PH2 = 1 atm (standard hydrogen electrode)

Reduction (cathode): Eocathode =

Oxidation (anode): Eoanode =

Net (overall cell reaction): Eocell =

4. A Zn/Zn2+ half-cell was coupled to a hydrogen electrode in which PH2 = 1 atm. The [Zn2+] in the anode compartment was 0.10 M and the cell emf was 0.542 V. Calculate the pH in the cathode compartment.

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