19.1 Redox Processes (AHL)

What we call a battery is actually a cell.

EMF stands for Electromotive Force - the force used to make electrons move - but it isn't a force.

It is the energy supplied by the cell divided by the charge carried through the circuit.

In simpler terms - it's the measured voltage across the terminals of the cell when _______ current flows.

We can look up standard electrode potentials in the data book.

If we produced a cell with Li⁺/Li as one electrode and Ni²⁺/Ni as the other, we can work out the cell potential as below

From the list:

Li⁺ + e^- --> Li -3.04v

Ni²⁺ + 2e^- --> Ni -0.26v

The Standard Cell potential is simply POSITIVE(cathode) - NEGATIVE(anode)

In this case, we don't have a positive value so the least negative is used as the positive value.

E^ocell = E^ocathode(+) -E^oanode(-) = -0.26 - - 3.04 = 2.78 v

QUESTION: Why is it impossible for E^ocell to be negative?

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But what's actually happening in this Nickel/Lithium cell?

In the simplest possible terms, the cell with most positive E^o goes forwards (naturally).

In this case,

Ni²⁺ + 2e^- --> Ni -0.26v

And the cell with most negative E^o goes backwardswards.

In this case,

Li --> Li⁺ + e^- -3.04v

But as written the 2 half-equations have unequal numbers of electrons so we simply double the Lithium half equation

Ni²⁺ + 2e^- --> Ni -0.26v

2Li --> 2Li⁺ + 2e^- -3.04v

NOTE: THIS DOES NOT CHANGE THE E^o DATA

The overall reaction is calculated by adding the two half equations

2Li + Ni²⁺ + 2e^- --> Ni + 2Li⁺ + 2e^-

2Li + Ni²⁺ --> Ni + 2Li⁺

So, the Lithium is oxidised and the Nickel is reduced.

Oxidation always happens at the anode; reduction at the cathode.

1. Work out the overall reaction and the EMF for the following combinations. State which process happens at the anode, which at the cathode.

a) Sn²⁺/Sn and Cu²⁺/Cu

b) K⁺/K and Mn²⁺/Mn

c) Ag⁺/Ag and Cu²⁺/Cu

d) Zn²⁺/Zn and Cu²⁺/Cu

e) K⁺/K and Fe²⁺/Fe

2. Which half-cell(s) could be used to reduce Ca²⁺ to Ca?

3. Which half-cell(s) could be used to oxidise Cl^- to Cl₂?

4. Which half-cell(s) could be used to oxidise Mn²⁺ to MnO₄^-?

5. Which half cell would oxidise Cu to Cu²⁺ but not Cu to Cu⁺?

E^o for every half-cell has to be measured against another cell or no voltage will be generated.

So, the standard cell is for Hydrogen, which is conveniently in the middle of the list.

And measurements have to betaken under standard conditions;

The Standard Temperature is ________^oC or _______K

The Standard Pressure is ________atm or _______kPa

The Standard Concentration is ________ moldm^-3

The Hydrogen cell is inconvenient because we can't have a stick of solid Hydrogen to act as an electrode.

Platinum is used because it can _______________________________________ but won't ________________ (is inert)

The Hydrogen gas is kept at 1 atm so that the electrode is always surrounded by H⁺ ions and H₂ gas, whichever way the 2H⁺ + 2e^- -> H₂ half-cell goes.

Measuring E^o for Zn/Zn²⁺

You should now know that if E^o is +ve the cell reaction will happen.

And you should remember that if ΔG is _________ the reaction is spontaneous.

So, it won't be a surprise to know that there's an equation to relate the two terms.

ΔG^o = -nFE^ocell

where n = No moles of electrons in the balanced equation

and F = Faraday Constant (96500 Cmol^-1)

EXAMPLE.

For a cell containing Zn²⁺ + 2e^- -> Zn -0.76v and Sn⁴⁺ + 2e^- -> Sn²⁺ +0.15v.

Balanced Equation:__________________________________________________________________

No. moles of electrons in balanced equation:_________________________________________

E^o:_________________

ΔG:_________________

QUESTIONS:

Write the balanced equation for the spontaneous reaction and calculate ΔG for the following pairs of half cells. Identify the anode and cathode reaction in each case.

Electrolytic cells are, in some ways, the opposite of Galvanic cells.

in stead of turning chemical energy into electrical energy, they use electrical energy to break up Ionic Compounds.

Molten electrolysis is easier than aqueous electrolysis because the only ions available are from the compound being electrolysed and not from the water.

Inevitably, the IB only cares about Aqueous Electrolysis!!!!

QUESTION: Which two ions are available from the water in aqueous electrolysis?

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There are two possible reactions at both the anode and the cathode and you have to learn which will happen.

Cathode

Na⁺ + e^- --> Na -2.71v

H₂O + e^- --> ¹/₂ H₂ + OH^- -0.83v

This would seem to favour the Sodium half equation.

But Sodium would immediately react with water to reform Sodium ions.

This would form Hydrogen gas and Hydroxide ions anyway - which is what the water half-cell would make.

So, we might as well assume that the bottom half-reaction is occuring.

QUESTION: Write a chemical equation for the reaction of Sodium with water.

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Anode

2 Cl^- --> Cl₂ + 2e^- -1.36v

H₂O --> ¹/₂ O₂ + 2H⁺ + 2e^- -1.23v

This favours the Chlorine half equation

QUESTION: Write a chemical equation for the electrolysis of aqueous Sodium Chloride.

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Concentrated or Dilute Aqueous Sodium Chloride?

For Dilute Sodium Chloride solution we work out the Cell Potential the conventional way

2 Cl^- --> Cl₂ + 2e^- -1.36v

H₂O + e^- --> ¹/₂ H₂ + OH^- -0.83v

QUESTION: What is the Standard Cell Potential for the combination above.

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For Concentrated Sodium Chloride solution we can't work out the Cell Potential this way

2 Cl^- --> Cl₂ + 2e^- -1.36v

The -1.36 v in this is measured under standard conditions of 1 moldm^-3 Cl^-

But, concentrated Sodium Chloride would have a far higher concentration of Chloride ions.

QUESTION: According to Le Chatalier's Principle, what would a higher concentration of Chloride ions do to the Cl^-/Cl₂ equilibrium?

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QUESTION: What would this do to the 1.36 v value of E^o? What effect would it have on the overall cell potential?

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This effect is called OVERVOLTAGE - and it depends mostly on the concentration.

Observing the electrolysis

At the anode you would expect to see _____________________________________________

At the cathode you would expect to see _____________________________________________

You could prove that one gas was Chlorine by _______________________________________

You could prove that the other gas was Hydrogen by _______________________________________

The pH of the solution would slowly ____________________ during the reaction due to the production of ___________ ions.

Industry could use the Chlorine gas made to produce ________________________________________________ and to _____________________________ water.

Industry could use the Hydrogen gas made to make _____________________ in the Haber Process and as a fuel.

The concentration of Sodium ions remains unchanged during the process, so at the end we would have concentrated _______________________________ solution that could be used to make _________________.

This electrolysis is best carried out with Platinum electrodes because _______________________

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Cathode

The choice is:

Cu²⁺ + 2e^- --> Cu +0.34v

H₂O + e^- --> ¹/₂H₂ + OH^- -0.83v

So _______________ is favoured and we will see _____________________

Anode

There really isn't a choice because it's hard to oxidise Sulphur from the +6 Oxidation State

H₂O --> ¹/₂O₂ + 2H⁺ + e^2- -1.23v

QUESTION: What would you see at the anode?

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QUESTION: Write the overall chemical equation and calculate the EMF

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QUESTION: What happens to the pH during the electrolysis?

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QUESTION: What happens to the colour of the solution during the electrolysis? Why?

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With Copper electrodes

Cathode

Again, the choice is:

Cu²⁺ + 2e^- --> Cu +0.34v

H₂O + e^- --> ¹/₂H₂ + OH^- -0.83v

And so _______________ is favoured and we will see _____________________

Anode

This time the Copper anode can replace the copper ions being deposited at the Cathode

Cu --> Cu²⁺ + 2e^-

QUESTION: What would you see at the anode?

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QUESTION: What happens to the colour of the solution during the electrolysis? Why?

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QUESTION: What happens to the pH during the electrolysis?

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QUESTION: How could we use this cell to make impure Copper purer?

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QUESTION: What is the advantage of using purer Copper in wires?

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QUESTION: We could use this cell to electroplate (cover) metals with Copper, but which metals to we usually electroplate onto other metals? What for?

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To allow us to electrolyse water we first add a very small quantity of dilute Sulphuric Acid (or Sodium Chloride)

Cathode

There are H⁺ ions from both the acid and very few from the water

H⁺ + e^- --> ¹/₂H₂ 0.00v

So, we will see _____________________

Anode

AS before, there isn't a real choice because it's hard to oxidise Sulphur from the +6 Oxidation State

H₂O --> ¹/₂O₂ + 2H⁺ + e^2- -1.23v

QUESTION: What would you see at the anode?

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QUESTION: Write the overall chemical equation and calculate the EMF

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QUESTION: What happens to the pH during the electrolysis?

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QUESTION: Why would we get twice as much Hydrogen than water?

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QUESTION: What might we use the Hydrogen gas for?

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Calculations involving Electrolysis

How much product you can make by electrolysis depends on:

i) Time - the _______________ the current is left on the more product should be made.

ii) Current - the _______________ the current the more product we should expect. Why?

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iii) Ionic Charge - The __________________ the charge of the ion being discharged the more product we should expect. Why?

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Example.

If a 1.0 amp current passes through Copper (II) Sulphate for 1 hour, how many grams of copper should be deposited?

Step 1 - calculate total charge passed.

Q = It (Charge = Current (amps) x time (seconds))

Q = 1.0 x (1 x 60 x 60) = 3600 C

Step 2 - calculate the number of moles of electrons in this charge.

Moles of Electrons = Total Charge passed / 96500 (the Faraday constant)

3600/96500 = 0.037 moles of electrons

Step 3 - Write the equation for making the product to calculate moles of product

Cu²⁺ + 2e^- --> Cu

It takes 2 moles of electrons to make 1 mole of product.

So, we would make 0.037/2 = 0.0185 moles of Copper

Step 4 - Calculate the mass of the product

Mass = Moles x Molar Mass

= 0.0185 x 63.5 = 2.37 g

QUESTIONS

1. Calculate the mass of Silver deposited if a 1.25 amp current passes through Silver Nitrate solution for 3 hours.

2. Calculate the mass of Aluminium deposited if a 1.5 amp current passes through Aluminium Nitrate solution for 4 hours.

3. A current passes through Copper (II) Sulphate solution for 3.25 hours and deposits 5.8g of Copper. If the same current passes through Silver Nitrate solution for the same time what mass of Silver would be deposited.

i) How many moles of copper were deposited?

ii) How many moles of electrons would be required?

iii) How many moles of Silver would this cause to be deposited?

iv) What mass of Silver would be deposited?