Electrochemistry Chapter 18

Electrochemistry Chapter 18
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Oxidation half-reaction (lose e-)
Electrochemical processes are oxidation-reduction reactions in which: the energy released by a spontaneous reaction is converted to electricity or electrical energy is used to cause a nonspontaneous reaction to occur 2+ 2- 2Mg (s) + O2 (g) MgO (s) 2Mg Mg2+ + 4e- Oxidation half-reaction (lose e-) O2 + 4e O2- Reduction half-reaction (gain e-) 19.1

Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
Oxidation number The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, Pb, H2, O2, P4 = 0 In monatomic ions, the oxidation number is equal to the charge on the ion. Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2 The oxidation number of oxygen is usually –2. In H2O2 and O22- it is –1. 4.4

HCO3- O = -2 H = +1 3x(-2) + 1 + ? = -1 C = +4
The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1. Group IA metals are +1, IIA metals are +2 and fluorine is always –1. 6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. HCO3- Oxidation numbers of all the atoms in HCO3- ? O = -2 H = +1 3x(-2) ? = -1 C = +4 4.4

Balancing Redox Equations
The oxidation of Fe2+ to Fe3+ by Cr2O72- in acid solution? Write the unbalanced equation for the reaction ion ionic form. Fe2+ + Cr2O Fe3+ + Cr3+ Separate the equation into two half-reactions. Fe Fe3+ +2 +3 Oxidation: Cr2O Cr3+ +6 +3 Reduction: Balance the atoms other than O and H in each half-reaction. Cr2O Cr3+ 19.1

For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms. Cr2O Cr3+ + 7H2O 14H+ + Cr2O Cr3+ + 7H2O Add electrons to one side of each half-reaction to balance the charges on the half-reaction. Fe Fe3+ + 1e- 6e- + 14H+ + Cr2O Cr3+ + 7H2O If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients. 6Fe Fe3+ + 6e- 6e- + 14H+ + Cr2O Cr3+ + 7H2O 19.1

Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel. Oxidation: 6Fe Fe3+ + 6e- Reduction: 6e- + 14H+ + Cr2O Cr3+ + 7H2O 14H+ + Cr2O Fe Fe3+ + 2Cr3+ + 7H2O Verify that the number of atoms and the charges are balanced. 14x1 – 2 + 6x2 = 24 = 6x3 + 2x3 For reactions in basic solutions, add OH- to both sides of the equation for every H+ that appears in the final equation. 19.1

Dichromate ion in acidic solution is an oxidizing agent
Dichromate ion in acidic solution is an oxidizing agent. When it reacts with zinc, the metal is oxidized to Zn2+, and nitrate is reduced. Assume that dichromate ion is reduced to Cr3+. Write the balanced ionic equation for this reaction using the half-reaction method.

First we determine the oxidation numbers of N and Zn: +6 +3 0 +2
Cr2O72-(aq)  Cr3+(aq) and Zn(s)  Zn2+(aq) Cr was reduced from +6 to +3. Zn was oxidized from 0 to +2. Now we balance the half-reactions. 2

Oxidation half-reaction Zn(s)  Zn2+(aq) + 2e- Reduction half-reaction
First we balance Cr and O: Cr2O72-(aq)  2Cr3+(aq) + 7H2O(l) Next we balance H: 14H+(aq) + Cr2O72-(aq)  2Cr3+(aq) + 7H2O(l) Then we balance e-: 6e- + 14H+(aq) + Cr2O72-(aq)  2Cr3+(aq) + 7H2O(l) 2

6e- + 14H+(aq) + Cr2O72-(aq)  2Cr3+(aq) + 7H2O(l)
Now we combine the two half-reactions by multiplying the oxidation half reaction by 3. 3Zn(s)  3Zn2+(aq) + 6e- 6e- + 14H+(aq) + Cr2O72-(aq)  2Cr3+(aq) + 7H2O(l) 3Zn(s) + 14H+(aq) + Cr2O72-(aq)  3Zn2+(aq) + 2Cr3+(aq) + 7H2O(l) Check that atoms and charge are balanced. 3Zn; 14H; 2Cr; 7O; charge +12 2

To balance a reaction in basic conditions, first follow the same procedure as for acidic solution.
Then Add one OH- to both sides for each H+. When H+ and OH- occur on the same side, combine them to form H2O. Cancel water molecules that occur on both sides. 2

Pb(OH)3- + ClO-  PbO2 + Cl-
Lead(II) ion, Pb2+, yields the plumbite ion, Pb(OH)3-, in basic solution. In turn, this ion is oxidized in basic hypochlorite solution, ClO-, to lead(IV) oxide, PbO2. Balance the equation for this reaction using the half-reaction method. The skeleton equation is Pb(OH)3- + ClO-  PbO2 + Cl- try to balance the above as practice

H+(aq) + Pb(OH)3-(aq) + ClO-(aq)  PbO2(s) + Cl-(aq) + 2H2O(l)
To convert to basic solution, we add OH- to each side, converting H+ to H2O. H2O(l) + Pb(OH)3-(aq) + ClO-(aq)  PbO2(s) + Cl-(aq) + 2H2O(l) + OH-(aq) Finally, we cancel the H2O that is on both sides. Pb(OH)3-(aq) + ClO-(aq)  PbO2(s) + Cl-(aq) + H2O(l) + OH-(aq) 2

The next several topics describe battery cells or voltaic cells (galvanic cells).
An electrochemical cell is a system consisting of electrodes that dip into an electrolyte and in which a chemical reaction either uses or generates an electric current. 2

A voltaic or galvanic cell is an electrochemical cell in which a spontaneous reaction generates an electric current. An electrolytic cell is an electrochemical cell in which an electric current drives an otherwise nonspontaneous reaction. 2

Physically, a voltaic cell consists of two half-cells that are electrically connected.
Each half-cell is the portion of the electrochemical cell in which a half-reaction takes place. The electrical connections allow the flow of electrons from one electrode to the other. The cells must also have an internal cell connection, such as a salt bridge, to allow the flow of ions. 2

Galvanic Cells anode oxidation cathode reduction spontaneous
redox reaction 19.2

A salt bridge is a tube of electrolyte in a gel that is connected to the two half-cells of the voltaic cell. It allows the flow of ions but prevents the mixing of the different solutions that would allow direct reactions of the cell reactants. In the salt bridge, cations move toward the cathode and anions move toward the anode. 2

The electrode at which oxidation takes place is called the anode.
The electrode at which reduction takes place is called the cathode. Electrons flow through the external circuit from the anode to the cathode. 2

The oxidation half-cell, the anode, is written on the left.
Voltaic cell notation is a shorthand method of describing a voltaic cell. The oxidation half-cell, the anode, is written on the left. The reduction half-cell, the cathode, is written on the right. 2

Phase boundaries are shown with a single vertical bar, |.
The cell terminal is written on the outside: left for the anode and right for the cathode. Phase boundaries are shown with a single vertical bar, |. For example, at the anode, the oxidation of Cu(s) to Cu2+(aq) is shown as Cu(s) | Cu2+(aq). At the cathode, the reduction of Zn2+(aq) to Zn(s) is shown as Zn2+(aq) | Zn(s). 2

Cu(s) | Cu2+(aq) || Zn2+(aq) | Zn(s)
Between the anode and cathode the salt bridge is represented by two vertical bars, ||. The complete notation for the reaction is Cu(s) | Cu2+(aq) || Zn2+(aq) | Zn(s) 2

Because this is a reduction, the electrode appears on the far right.
When the half-reaction involves a gas, the electrode is an inert material such as platinum, Pt. It is included as a third substance in the half-cell. For example, the half-reaction of Cl2 being reduced to Cl- is written as follows: Cl2(g) | Cl-(aq) | Pt Because this is a reduction, the electrode appears on the far right. For the oxidation of H2(g) to H+(aq), the notation is Pt | H2(g) | H+(aq) In this case, the electrode appears on the far left. 2

Cu(s) | Cu2+(1.0 M) || F2(1.0 atm) | F-(1.0 M) | Pt
To fully specify a voltaic cell, it is necessary to give the concentrations of solutions or ions and the pressures of gases. In the cell notation, these values are written within parentheses for each species. For example, the oxidation of Cu(s) to Cu2+(aq) at the anode and the reduction of F2(g) to F-(aq) at the cathode is written as follows: Cu(s) | Cu2+(1.0 M) || F2(1.0 atm) | F-(1.0 M) | Pt

Al(s) | Al3+(aq) || Cu2+(aq) | Cu(s)
The cell notation for a voltaic cell is Al(s) | Al3+(aq) || Cu2+(aq) | Cu(s) Write the cell reaction. Our strategy is to write and balance each half-reaction, and then to combine the half-reactions to give the overall cell reaction. The skeleton oxidation reaction is Al(s)  Al3+(aq). The skeleton reduction reaction is Cu2+(aq)  Cu(s).

Balanced oxidation half-reaction Al(s)  Al3+(aq) + 3e-
Balanced reduction half-reaction Cu2+(aq) + 2e-  Cu(s) The common multiple of the electrons is 6. Combine the half-reactions by multiplying the oxidation half-reaction by 2 and the reduction half-reaction by 3. 2

2Al(s) + 3Cu2+(aq)  2Al3+(aq) + 3Cu(s)
2Al(s)  2Al3+(aq) + 6e- 3Cu2+(aq) + 6e-  3Cu(s) 2Al(s) + 3Cu2+(aq)  2Al3+(aq) + 3Cu(s) 2

Now we shift our focus to the movement of the electrons in the oxidation–reduction reaction.
To better understand this movement, we can compare the flow of electrons to the flow of water. 2

Just as work is required to pump water from one point to another, so work is required to move electrons. Water flows from areas of high pressure to areas of low pressure. Similarly, electrons flow from high electric potential to low electric potential. Electric potential can be thought of as electric pressure. 2

Electrical work = charge x potential difference
Potential difference is the difference in electric potential (electrical pressure) between two points. Potential difference is measured in the SI unit volt (V). Electrical work = charge x potential difference The SI units for this are J = C × V

w = –F × potential difference
The magnitude of charge on one mole of electrons is given by the Faraday constant, F. It equals × 103 C per mole of electrons. 1 F = × 103 C Substituting this into the equation for work: w = –F × potential difference The term is negative because the cell is doing work in creating the current (that is, electron flow). 2

Galvanic Cells The difference in electrical potential between the anode and cathode is called: cell voltage electromotive force (emf) cell potential Cell Diagram Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq) [Cu2+] = 1 M & [Zn2+] = 1 M Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s) anode cathode 19.2

Standard Reduction Potentials
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s) Anode (oxidation): Zn (s) Zn2+ (1 M) + 2e- Cathode (reduction): 2e- + 2H+ (1 M) H2 (1 atm) Zn (s) + 2H+ (1 M) Zn2+ + H2 (1 atm) 19.3

Standard reduction potential (E0) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm. Reduction Reaction 2e- + 2H+ (1 M) H2 (1 atm) E0 = 0 V Standard hydrogen electrode (SHE) 19.3

The maximum potential difference between the electrodes of a voltaic cell is called the cell potential or electromotive force (emf) of the cell, or Ecell. Ecell can be measured with a digital voltmeter. The anode of a voltaic cell has negative polarity; the cathode has positive polarity. 2

The maximum work is given by the following equation: Wmax = –nFEcell
2

The emf of a particular cell is 0.500 V. The cell reaction is
2Al(s) + 3Cu2+(aq)  2Al3+(aq) + 3Cu(s) Calculate the maximum electrical work of this cell obtained from 1.00 g of aluminum.

wmax = –(6 mol)(96,485 C/mol)(0.500 J/C)
To determine the value of n (that is, the number of moles of electrons involved in either half-cell reaction), we need to examine the half-reactions. 2Al(s)  2Al3+(aq) + 6e- 3Cu2+(aq) + 6e-  3Cu(s) n = 6 F = 96,485 C/mol Ecell = V = J/C wmax = –nFEcell wmax = –(6 mol)(96,485 C/mol)(0.500 J/C) wmax = 2.89 × 105 J 2

This is the energy corresponding to the balanced reaction, so we need to convert to one gram of aluminum. wmax = 5.36 kJ 2

Ecell = oxidation potential + reduction potential
A cell potential is a measure of the driving force of the cell reaction. It is composed of the oxidation potential for the oxidation half-reaction at the anode and the reduction potential, for the reduction half-reaction at the cathode. Ecell = oxidation potential + reduction potential But since we are using only the Chart of Reduction Potentials, Ecell = reduction potential of cathode – reduction potential of anode 2

Ecell = Eoxidation + Ereduction
Because the electrode potentials are for reduction: Eoxidation = –Eanode We can rewrite the equation for the cell: Ecell = - Eanode + Ecathode Ecell = Ecathode – Eanode 2

Let’s find Ecell for the following cell:
Al(s) | Al3+(aq) || Cu2+(aq) | Cu(s) At the cathode, Cu2+(aq) is reduced to Cu(aq). E = 0.34 V At the anode, Al is oxidized to Al3+. E = –[electrode potential for Al3+(aq)] E = –(–1.66 V) = 1.66 V Ecell = 0.34 V V Ecell = 2.00 V 2

A fuel cell is simply a voltaic cell that uses a continuous supply of electrode materials to provide a continuous supply of electrical energy. A fuel cell employed by NASA on spacecraft uses hydrogen and oxygen under basic conditions to produce electricity. The water produced in this way can be used for drinking. The net reaction is 2H2(g) + O2(g)  2H2O(g) Calculate the standard emf of the oxygen–hydrogen fuel cell. 2H2O(l) + 2e-  H2(g) + 2OH-(aq) E° = –0.83 V O2(g) + 2H2O(l) + 4e-  4OH-(aq) E° = 0.40 V

Ecell = Ecathode – Eanode
Anode reaction: 2H2O(l) + 2e-  H2(g) + 2OH-(aq) E°red = –0.83 V Cathode reaction: O2(g) + 2H2O(l) + 4e-  4OH-(aq) E°red = 0.40 V Overall reaction: Ecell = Ecathode – Eanode Ecell = 0.40 V – (–0.83 V) Ecell = 1.23 V 2

E0 = 0.76 V cell Standard emf (E0 ) cell E0 = Ecathode - Eanode cell Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s) E0 = EH /H - EZn /Zn cell + 2+ 2 0.76 V = 0 - EZn /Zn 2+ EZn /Zn = V 2+ Zn2+ (1 M) + 2e Zn E0 = V 19.3

E0 = 0.34 V cell E0 = Ecathode - Eanode cell Ecell = ECu /Cu – EH /H 2+ + 2 0.34 = ECu /Cu - 0 2+ ECu /Cu = 0.34 V 2+ Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s) Anode (oxidation): H2 (1 atm) H+ (1 M) + 2e- Cathode (reduction): 2e- + Cu2+ (1 M) Cu (s) H2 (1 atm) + Cu2+ (1 M) Cu (s) + 2H+ (1 M) 19.3

E0 is for the reaction as written
The more positive E0 the greater the tendency for the substance to be reduced The half-cell reactions are reversible The sign of E0 changes when the reaction is reversed Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0 19.3

Comparing Oxidizing Strengths
The oxidizing agent is itself reduced and is the species on the left of the reduction half-reaction. Consequently, the strongest oxidizing agent is the product of the half-reaction with the largest (most positive) E° value. 2

Comparing Reducing Strengths
The reducing agent is itself oxidized and is the species on the right of the reduction half-reaction. Consequently, the strongest reducing agent is the reactant in the half-reaction with the smallest (most negative) E° value. 2

Which is the stronger reducing agent under standard conditions: Sn2+ (to Sn4+) or Fe (to Fe2+)?
Which is the stronger oxidizing agent under standard conditions: Cl2 or MnO4-? The stronger reducing agent will be oxidized and has the more negative electrode potential. The standard (reduction) potentials are Sn2+ to Sn4+ E = 0.15 V Fe to Fe2+ E = –0.41 V The stronger reducing agent is Fe (to Fe2+).

The stronger oxidizing agent will be reduced.
The standard (reduction) potentials are Cl2 to Cl- E = 1.36 V MnO4- to Mn2+ E = 1.49 V The stronger oxidizing agent is MnO4-. 2

Predicting the Direction of Reaction
You can predict the direction of reaction by comparing the relative oxidizing (or reducing) strengths. The stronger oxidizing agent will be reduced. (The stronger reducing agent will be oxidized.) 2

Will dichromate ion oxidize manganese(II) ion to permanganate ion in acid solution under standard conditions? Standard potentials Cr2O72- to Cr3+ E° = 1.33 V MnO4- to Mn2+ E° = 1.49 V MnO4- has a larger reduction potential; it will oxidize Cr3+ to Cr2O72-. Cr2O72- will not oxidize Mn2+ to MnO4-.

Cd is the stronger oxidizer
What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution? Cd2+ (aq) + 2e Cd (s) E0 = V Cd is the stronger oxidizer Cd will oxidize Cr Cr3+ (aq) + 3e Cr (s) E0 = V Anode (oxidation): Cr (s) Cr3+ (1 M) + 3e- x 2 Cathode (reduction): 2e- + Cd2+ (1 M) Cd (s) x 3 2Cr (s) + 3Cd2+ (1 M) Cd (s) + 2Cr3+ (1 M) E0 = Ecathode - Eanode cell E0 = – (-0.74) cell E0 = 0.34 V cell 19.3

First we determine the oxidation numbers of N and Zn: +6 +3 0 +2
Cr2O72-(aq)  Cr3+(aq) and Zn(s)  Zn2+(aq) Cr was reduced from +6 to +3. Zn was oxidized from 0 to +2. Now we balance the half-reactions. 2

Spontaneity of Redox Reactions
DG = -nFEcell n = number of moles of electrons in reaction F = 96,500 J V • mol DG0 = -nFEcell = 96,500 C/mol DG0 = -RT ln K = -nFEcell Ecell = RT nF ln K (8.314 J/K•mol)(298 K) n (96,500 J/V•mol) ln K = = V n ln K Ecell = V n log K Ecell 19.4

Spontaneity of Redox Reactions
DG0 = -RT ln K = -nFEcell 19.4

What is the equilibrium constant for the following reaction at 250C
What is the equilibrium constant for the following reaction at 250C? Fe2+ (aq) + 2Ag (s) Fe (s) + 2Ag+ (aq) = V n ln K Ecell Oxidation: 2Ag Ag+ + 2e- n = 2 Reduction: 2e- + Fe Fe E0 = EFe /Fe – EAg /Ag 2+ + E0 = – (0.80) E0 = V V x n E0 cell exp K = V x 2 -1.24 V = exp K = 1.23 x 10-42 19.4

Physically, a voltaic cell consists of two half-cells that are electrically connected.
Each half-cell is the portion of the electrochemical cell in which a half-reaction takes place. The electrical connections allow the flow of electrons from one electrode to the other. The cells must also have an internal cell connection, such as a salt bridge, to allow the flow of ions. 2

Ions flow through the salt bridge to maintain charge balance.
The Cu2+ ions gain two electrons, forming solid copper. The zinc metal atom loses two electrons, forming Zn2+ ions. x The electrons flow through the external circuit from the zinc electrode to the copper electrode. Ions flow through the salt bridge to maintain charge balance. 2

The Effect of Concentration on Cell Emf
DG = DG0 + RT ln Q DG = -nFE DG0 = -nFE -nFE = -nFE0 + RT ln Q Nernst equation E = E0 - ln Q RT nF At 298 - V n ln Q E E = - V n log Q E E = 19.5

Dependence of Cell Potential on Concentration: Nernst Equation
DG = DG° + RT ln Q Q is the thermodynamic equilibrium constant. This equation allows us to relate E to E°. 2

Fe2+ (aq) + Cd (s) Fe (s) + Cd2+ (aq)
Will the following reaction occur spontaneously at 250C if [Fe2+] = 0.60 M and [Cd2+] = M? Fe2+ (aq) + Cd (s) Fe (s) + Cd2+ (aq) Oxidation: Cd Cd2+ + 2e- n = 2 Reduction: 2e- + Fe Fe E0 = EFe /Fe – ECd /Cd 2+ E0 = – (-0.40) - V n ln Q E E = E0 = V - V 2 ln -0.04 V E = 0.010 0.60 E = 0.013 E > 0 Spontaneous 19.5

Commercial Voltaic Cells
We next look at several commercial voltaic cells. Zinc–Carbon Dry Cell: Leclanché Anode: Zn(s)  Zn2+(aq) + 2e- Cathode: 2NH4+(aq) + 2MnO2(s) + 2e-  Mn2O3(s) + H2O(l) + 2NH3(aq) The initial voltage is about 1.5 V, but decreases and deteriorates rapidly in cold weather. 2

Batteries Dry cell Leclanché cell Anode: Zn (s) Zn2+ (aq) + 2e-
Cathode: 2NH4 (aq) + 2MnO2 (s) + 2e Mn2O3 (s) + 2NH3 (aq) + H2O (l) + Zn (s) + 2NH4 (aq) + 2MnO2 (s) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s) 19.6

Batteries Mercury Battery Anode:
Zn(Hg) + 2OH- (aq) ZnO (s) + H2O (l) + 2e- Cathode: HgO (s) + H2O (l) + 2e Hg (l) + 2OH- (aq) Zn(Hg) + HgO (s) ZnO (s) + Hg (l) 19.6

Batteries Lead storage battery Anode:
Pb (s) + SO2- (aq) PbSO4 (s) + 2e- 4 Cathode: PbO2 (s) + 4H+ (aq) + SO2- (aq) + 2e PbSO4 (s) + 2H2O (l) 4 Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO2- (aq) PbSO4 (s) + 2H2O (l) 4 19.6

Solid State Lithium Battery
Batteries Solid State Lithium Battery 19.6

Batteries A fuel cell is an electrochemical cell that requires a continuous supply of reactants to keep functioning Anode: 2H2 (g) + 4OH- (aq) H2O (l) + 4e- Cathode: O2 (g) + 2H2O (l) + 4e OH- (aq) 2H2 (g) + O2 (g) H2O (l) 19.6

Chemistry In Action: Bacteria Power
CH3COO- + 2O2 + H CO2 + 2H2O

Corrosion 19.7

Cathodic Protection of an Iron Storage Tank
19.7

Electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur. 19.8

Electrolysis of Water 19.8

Electrolysis and Mass Changes
charge (C) = current (A) x time (s) 1 mole e- = 96,500 C 19.8

How much Ca will be produced in an electrolytic cell of molten CaCl2 if a current of A is passed through the cell for 1.5 hours? Anode: 2Cl- (l) Cl2 (g) + 2e- Cathode: 2 mole e- = 1 mole Ca Ca2+ (l) + 2e Ca (s) Ca2+ (l) + 2Cl- (l) Ca (s) + Cl2 (g) mol Ca = 0.452 C s x 1.5 hr x 3600 s hr 96,500 C 1 mol e- x 2 mol e- 1 mol Ca x = mol Ca = 0.50 g Ca 19.8

What electric charge is required to plate a piece of automobile molding with 1.00 g of chromium metal using a chromium(III) ion solution? If the electrolysis current is 2.00 A, how long does the plating take?

First, we convert mass to moles Cr; then to moles e-; then, using current, to seconds.
= 2.78 × 103 s = 46.4 min

A solution of nickel salt is electrolyzed to nickel metal by a current of 2.43 A. If this current flows for 10.0 min, how many coulombs is this? How much nickel metal is deposited in the electrolysis?

We first find the charge:
Given the half-reaction: Ni2+(aq) + 2e-  Ni(s) This slide should be followed by the chapter key terms. This was originally the last slide, but I copied the logo slide from another chapter as the final slide. = g Ni

Chemistry In Action: Dental Filling Discomfort
Hg2 /Ag2Hg V 2+ Sn /Ag3Sn V 2+ Sn /Ag3Sn V 2+

Electrochemistry Chapter 18

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