Chem 131 exam 2 practice ch 20 only

Separate the redox reaction into its component half reactions.

O2 + 2Cu => 2CuO

Oxidation: Cu(s) => Cu2+(aq) + 2e- Reduction: O2(g)+2e- => 2O2-(aq)

the element with the higher reduction potential acts as the strong ____ agent.

oxidizing

the element with the lower reduction potential acts as the strong ____ agent.

reducing

Which of the following reagents would oxidize r to Cr3+, but not Fe to Fe3+?

Br- (EoBr-/Br2 = 1.09V)
Co2+ (EoCo2+/Co = -0.28V)
Ca2+ (EoCa2+/Ca = -0.286V)
Br2 (EoBr2/Br- = I messed up)
Co (EoCo/Co2+ = -0.28V)

*EoCr/Cr3+ = -0.74)
*EoFe/Fe3+ -0.04)

Co2+, Co

Which of he following reagents would oxidize Ag to Ag+, but not F- to F2?

Ca (EoCa/Ca2+ = -2.87V)
Ca2+ (EoCa2+/Ca = -2.87V)
Co (EoCo/Co2+ = -0.28V)
Co2+ (EoCo2+’Co = -0.28V)
Br- (EoBr-/Br2 = 1.09V)
Br2 (EoBr2/Br- = 1.09V)

*EoAg/Ag+ = 0.80V
*EoF-/F2 = 2.87

Br-, Br2

define Cell Potential (Ecell)

measure of potential difference in Volts (V) between two half cells in an electrochemical cell; The potential difference is caused by the ability of electrons to flow from one half cell to the other, electrons are able to move between electrodes because the chemical reaction is a redox reaction.

Formula for Ecell

Eocell = (Ered Anode V) – (Ered Cathode V)

Based on the sign of the Ecell, classify the reaction as spontaneous/nonspontaneous as written (assume std conditions)

Pb+2(aw) + H2(g) => Pb(s) + 2H+(aq)

Reduction Rxn: Pb2+ (aq) + 2e- => Pb(s)
Ecell = -0.126 V

Oxidation Rxn: H2(g) => 2H+(aq) + 2e-
Ecell = +0.00 V

Nonspontaneous (-0.126V – 0.000V = -0.126 V)

Based on the sign of the Ecell, classify the reaction as spontaneous/nonspontaneous as written (assume std conditions)

Al+3(aq) + 3Na(s) => Al(s) + 3Na+(aq)

Reduction Rxn: Al+3(aq) + 3e- => Al(s)
Ecell = -1.66 V

Oxidation Rxn: Na(s) => Na+(aq) +e-
*Ecell = -2.71 V

Spontaneous (-1.66V – -2.71V = 1.05V)

Based on the sign of the Ecell, classify the reaction as spontaneous/nonspontaneous as written (assume std conditions) (i = iodine)

i2(s) + Cu(s) => 2i-(aq) + Cu+2(aq)

Reduction Rxn: i2(s) + 2e- => 2i-
Ecell = +0.54V

Oxidation Rxn: Cu(s) => Cu+2(aq) + 2e-
Ecell = +0.34V

Spontaneous (0.54V – 0.34V=0.20V)

A positive value for E°cell indicates a _____ reaction. Calculate E°cell for each reaction. (Keep in mind that the coefficients in the overall reaction do not affect the cell potential.)

spontaneous

Write the half-reactions as they occur at each electrode and the net cell reaction for this electrochemical cell containing copper and silver:
Cu(s)|Cu+2 (aq)||Ag+ (aq)|Ag(s)

Anode:
Cathode:
Net Rxn:

Anode: Cu => Cu+2 +2e- Cathode: Ag+ + e- =>Ag Net Rxn: Cu + 2Ag+ => Cu+2 + 2Ag

Calculate the standard free-energy change for the following reaction at 25 °C.

2Au3+(aq) + 3Zn(s) <=> 2Au(s) + 3 Zn+2(aq)

Anode/Reduction: Au+3(aq) + 3e- <=> Au(s)
Cathode/Oxidation: Zn(s) <=> Zn+2(aw_ + 2e-

deltaG = -nFEcell

<b>F = 96485J/V</b>mol e-

-1307 kJ

Give the formula for gibbs energy (deltaG)

deltaG = -nFEcell

What does F stand for in the gibbs energy equation
deltaG = -nFEcell

F = 96485J/V*mol e-

When K&lt;1, Ecell is ____ and deltaG is ____.

neg, pos

When K = 1, Ecell is ____ and delta G is ____.

0, 0

When K&gt;1, Ecell is _____ and delta G is _____.

pos, neg

For the following reaction

Cr(s)+Sn+2(aq)&lt;–&gt;Cr+2(aq)+Sn(s)

Which element is probably being oxidized? At which node?

Cr, anode

For the following reaction

Cr(s)+Sn+2(aq)&lt;–&gt;Cr+2(aq)+Sn(s)

Which element is probably being reduced? At which node?

Sn+2, cathode

Eo cell = Eo______ – Eo_______

cathode, anode

In the Nernst Equation:
E = Eo – [(RT)/(nF)]*ln(Q)

What is Q?

Q = [C]^c*[D]^c / [A]^a[B]^b (cathode/anode)

In the Nernst Equation:
E = Eo – [(RT)/(nF)]*ln(Q)

What is Eo?

Eo = Cell Standard Potential

In the Nernst Equation:
E = Eo – [(RT)/(nF)]*ln(Q)

What is R?

R =Gas Constant = 8.314J/Kmol

In the Nernst Equation:
E = Eo – [(RT)/(nF)]*ln(Q)

What is T?

T = temp in K

In the Nernst Equation:
E = Eo – [(RT)/(nF)]*ln(Q)

What is F?

F = Faraday Constant = 96485 C/mol

In the Nernst Equation:
E = Eo – [(RT)/(nF)]*ln(Q)

What is n?

n = # of mols

What is a concentration cell?

A cell holding one type of element that produces a voltage because of the two different concentrations of that element as ion(s)

How can concentration cells produce energy if only one type of element + its ion are present?

The element and its ion are at different concentrations, creating a potential difference

For the reaction:
Zn(s)|Zn+2 (aq, 0.100 M)||Zn+2 (aq, ? M)|Zn(s)

Which is the product and reactant at the CATHODE end?

Zn+2, Zn

The voltage generated by the zinc concentration cell described by,
Zn(s)|Zn2 (aq, 0.100 M)||Zn2 (aq, ? M)|Zn(s)
is 19.0 mV at 25 °C. Calculate the concentration of the Zn2 (aq) ion at the cathode.

0.439M ( 19.0×10^-3=0-[(8.314)(25.0+273.15)/(2)(96480)]ln(0.100/?M) )

Define sacrificial anode/cathodic protection

a metal used to protect the metal from corosion, is more active (i.e. has more neg reduction protential) than the metal being protected

What is the reduction potential of H2O?

-1.78

Group ___ elements (eg: lithium, sodium, potassium, rubidium and caesium) react with water

1

Which of the following metals would act as a sacrificial anode/cathodic protection for iron?

Cr (Ered(Cr+3/Cr)= -0.740V)
Ni(Ered(Ni+2/Ni)= -0.260V)
Mg(Ered(Mg+2/Mg)= -2.38V)
Na(Ered(Na+/Na)= -2.71V)
Pb(Ered(Pb+2/Pb)= -0.130V)

(Ered(Fe+2/Fe)=-0.440V)

Cr, Mg, Na

Despite its ability to act as a sacrificial anode for iron, which metal\’s reactivity with water makes it unsuitable to attach to the hull of a steel ship?

Pb
Na
Ni
Cr
Mg

Na (it’s the only G1 element

define electrolysis

the decomposition of a substance, either in the molten state or in an electrolyte soln, by means of electric current

_____ ___ = charge(C)/Faraday Constant(C/mol)

mols e-

_______ = Current(A)*Time(sec)

charge

give the formula for charge

charge =current(A)*time(sec)

Gallium is produced by the electrolysis of a solution obtained by dissolving gallium oxide in concentrated NaOH(aq). Calculate the amount of Ga(s) that can be deposited from a Ga(III) solution by a current of 0.330 A that flows for 90.0 min.

Use the formulas:
Charge(C)= Current(A)*Time(sec)
Mols e- = Charge(C)/Faraday Constant(mols/C)

Molar Mass of Ga = 69.72g/mol

0.429g Ga

Electrons are transfered in ____ ____ reactions

ozidation reduction

Reduction is the _______ of electrons

gain

Oxidation is the ______ of electrons

loss

A ____ bridge maintains electrical contact between two half cells

salt

A ___ ______ is a symbolic way to show cell components

cell diagram

An anode is where ____ occurs,

oxidation

A cathode is where _____ occurs

reduction

The ____ is on the left in a cell diagram

anode

The _____ available in a cell = -nFEcell

work

_______ = -nFEdegcell

DeltaGdeg

When Ecell is &gt; 0 the reaction is _______

spontaneous

When Ecell = 0 is the reaction is at _____

equilibrium

______ a cell reaction changes the sign of Ecell

Reversing

Calculate the Ksp of lead iodide from the following standard electrode potentials, at 25degC:

2e- + PbI2(s) -&gt; Pb(s) + 2I-(aq) Edeg=-0.365V
2e- + Pb+2(aq) -&gt; Pb(s) Edeg=-0.126V

…

Electromotive force (emf), or cell potential, is represented by the symbol _____

Ecell

Define electrode potential

potential difference between electrode and solution; the difference between the charge on an electrode and the charge in the solution

Calculate the Ksp for AgI. Ecell is 0.417V.
Ag(s)|Ag+(satd AgI)||Ag+(0.100M)|Ag
AgI(s) &lt;=&gt; Ag+(aq) + I-(aq)

8.3*10^-17 ( in saturated AgI solution, the concentrations of Ag+ and I- are equal Ksp = [Ag+][I-] = (9.110^-9M)(9.110^-9M) = 8.310^-17)

Calculate the Ksp of lead iodide from the following standard electrode potentials, at 25 °C:
2 e- + PbI2(S) Pb(s) + 2 I-(aq) E° = -0.365 V
2 e- + Pb2+(aq) Pb(s) E° = -0.126 V

A) 8 × 10 -9
B) 9 × 10 – 5
C) 2 × 10 – 17
D) 5 × 10 – 13
E) 6 × 10 -5

a

A concentration cell consists of half cells with identical ________ but different ____ __________.

electrodes, ion concentrations

The cell reactions of _______ batteries cannot be reversed.

primary

The cell reactions of ______ batteries can be reversed.

secondary

Cells be joined in series to increase the total ______

voltage

T/F a primary batter cannot be recharged

T

1 mole of Hydrogen ions will have a positive charge of ____ faraday

1

In the electrolysis of a chloride solution, one Faraday of charge will release ______ g of a Cl- solution

35.45

One mole of electrons has a charge of ___ Faraday

1

One mole of H2 gas requires __ Faradays.

2

What is the difference between a voltaic (galvanic) cell and an electrolytic cell?

A Voltaic (galvanic) cell uses chemical change to produce electricity. An electrochemical cell uses electricity to produce a nonspontaneous reaction driven by electrolysis.

An electrolyic cell applies elecric energy by a process called _____. This process uses electricity to produce a nonspontaneous reaction.

electrolysis

1 mol e- = ______C = 1 Faraday

96485

An electrolysis is carried out by passing an electric current through a solution of copper sulfate using inert electrodes. This causes:

A) Cu to plate out on the negative electrode
B) Cu to plate out on the anode
C) Oxygen to form on the negative electrode
D) H2 to form at the negative electrode
E) Sulfur to form on the positive electrode

a

Write the net redox reaction that occurs in the galvanic cell.

Zn(s) | Zn2+(aq) (1.00 M) || Pb2+(aq) (1.00 M)|Pb(s)

A) Pb(s) + Zn(s) -&gt; Pb2+(aq) + Zn2+(aq) + 4 e-
B) Pb(s) + Zn2+(aq) -&gt; Pb2+(aq) + Zn(s)
C) Pb2+(s) + Zn2+(s) -&gt; Pb(s) + Zn(s) + 4 e-
D) Pb2+(aq) + Zn(s) -&gt; Zn2+(aq) + Pb(s)

d

For the reaction: Mg(s) + AgNO3(aq) Ag(s) + Mg(NO3)2(aq)
Ag+(aq) + e- -&gt; Ag(s) E° = 0.800 V
Mg2+(aq) + 2 e- -? Mg(s) E° = -2.356 V
For the reaction, determine E° for the cell.

A) 1.556 V
B) -3.156 V
C) 3.156 V
D) 0.800 V
E) 2.356 V

C (Eocell = Eo(Ag+/Ag) – Eo(Mg+2/Mg) =(0.800) – (-2.356V) =3.156V)

Determine E°cell for the reaction: I2 + Cu | 2 I- + Cu2+, the half reactions are:
I2(s) + 2 e- -&gt; 2 I- (aq) E° = +0.535 V
Cu2+ + 2 e- -&gt; Cu E° = 0.340 V

A) 0.730 V
B) -0.875 V
C) 0.875 V
D) 0.195 V
E) -0.195 V

d (Eocell = Eo(I2/I-) – Eo(Cu+2/Cu) =0.535V – 0.340V =0.195V)

The following galvanic cell has a measured cell potential E°cell = 0.646 V.
Pt(s) | Sn4+(aq), Sn2+(aq) | Ag+(aq) | Ag(s)
Given that the standard reduction potential for the reduction of Ag+ to Ag(s) is +0.800 V, what is the standard reduction potential for the Sn4+/Sn2+ half-reaction?

A) +0.154 V
B) -0.154 V
C) +1.45 V
D) -1.45 V
E) +0.308 V

A

What is the cell diagram for the spontaneous cell involving the Fe3+ | Fe2 + (0.771V) half cell and the Zn2+ | Zn (-0.763 V) half cell?

A) Fe2+(aq) | Fe3+(aq) || Zn2+(aq) | Zn(s)
B) Pt(s) | Fe2+(aq), Fe3+(aq) | Zn2+(aq) | Zn(s)
C) Zn(s) | Zn2+(aq) || Fe3+(aq), Fe2+(aq) | Pt(s)
D) Zn(s) | Zn2+(aq) || Fe3+(aq) | Fe2+(aq)
E) Zn(s) | Zn2+(aq) || Fe3+(aq), Fe2+(aq) | Zn(s)

C

What is the cell reaction of the spontaneous cell made from the Zn/Zn2+ (0.76 V) and 2Br-/Br2 ( -1.07 V) half cells?

A) 2 Br- + Zn2+ -&gt; Zn + Br2
B) Br2 + Zn -&gt; Zn2+ + 2 Br-
C) 2 Br- + Zn -&gt; Br2 + Zn2+
D) Br2 + Zn -&gt; Zn2+ + Br-
E) Br2 + Zn2+ -&gt; Zn + 2Br-

B

The net reaction in a voltaic cell with E°cell = +0.726 V is,
2 Fe3+(aq) + 3 Zn(s) 2 Fe(s) + 3 Zn2+(aq)
What is ∆G° for this reaction at 25°C?

A) -210 kJ
B) -140 kJ
C) -700 kJ
D) -463 kJ
E) -420 kJ

E ( Overall Rxn: 2 Fe3+(aq) + 3 Zn(s) 2 Fe(s) + 3 Zn2+(aq Oxidation: 3Zn -> 3Zn+2 + 6e- Reduction: 2Fe+3 + 6e- -> 2Fe mols. e- transferred: n = 6 Faraday Constant: F = 96485C/mol Eo(cell) = +0.726V DeltaGo = -nFEo(cell) =-(6mol)(96485C/mol)(0.726V) =-420288.7 J/mol =-420 kj/mol)

The standard free energy change for the following voltaic cell is ∆G° = -89.3 kJ at 25°C. What is E°cell? Cu(s) | Cu2+(aq) || Ag+(aq) | Ag (s)

A) +0.463 V
B) -0.926 V
C) -0.463 V
D) +0.926 V
E) +0.231 V

A ( Oxidation: Cu -> Cu+2 + 2e- Reduction: {Ag+ +1e- -> Ag }*2 Mols e- transferred: n=2 Faraday Constant: F = 96485 C/mol DeltaGo = -89.3kJ = -89300J DeltaGo = -nFEo(cell) (-89300J) = -(2)(96485C/mol)(Eo(cell)) Eo(cell) = +0.463V

What is Keq at 25°C of the spontaneous cell made from Ag+/Ag (0.80 V) and Cl2/Cl- (1.36 V) half cells?

A) 1073
B) 10-10
C) 109
D) 1019
E) 10-19

D

Find Ecell for the following voltaic cell at 25°C:
Ti(s) | Ti2+(aq) (0.550M) || Sn2+(aq) (0.005 M) | Sn(s)
The half-reactions are:
Ti2+(aq) + 2 e- Ti(s) E° = -1.630 V
Sn2+(aq) + 2 e- Sn(s) E° = -0.137 V

A) 1.372 V
B) 1.707 V
C) 1.646 V
D) 1.433 V
E) -1.646 V

D

What is the [H+] in a concentration cell at 25°C if the cell potential is 0.0451 V and cell diagram:

Pt | H2 (g, 1 atm), H+ (? M)|| H+ (1 M) | H2 (g, 1 atm) |Pt
2 H+(aq) + 2 e- H2(g) E° = 0. V

A) 0.51 M
B) 0.42 M
C) 0.030 M
D) 0.17 M
E) 5.7 M

C

Calculate the mass of I2 produced at the anode if a current of 2.85 A is passed through a solution of KI for 56 minutes.

A) 0.42 g
B) 0.13 g
C) 13 g
D) 25 g
E) 4.3 g

C (2I—->I2+2e- 2.85A56min(60s/min)=9.576*10^3C 9.57610^3C+(1mole e-/96485C)(1mol I2/2mole-)*(245g I2/1mol I2)= 12.59 I2)

The cathode is on the ______ in a cell diagram

right

Oxidation occurs at the _____.

Anode

Reduction occurs at the ______.

cathode

Choose the correct statement.

a. a concentration cell consists of half cells with identical electrodes but different ion concentrations
b. the cell reaction for a secondary battery can be reversed
c. cells can be joined in series to increase the total voltage
d. all of the above

d

A concentration cell consists of half cells with identical _____ but different __ __________.

electrodes, ion concentrations

The cell reaction for a ______ battery can be reversed.

secondary

Cells can be joined in series to increase the total _____.

voltage

_____(C) = current(c/s)*time(s)

Charge

1 ampere =____

C/s

The boundary between half-cell compartments, commonly a salt bride, is represented by a ___ ___ ___. Different species within the same solution are separated from each other by a ___.

double vertical line, comma

The properly connected combination of two half-cells is called an __________ ____.

electrochemical cell

The ____ _____/______________ (emf)/______ ________ is the potential difference between the half cells.

cell voltage/electromotive force/cell potential

An electrode is often a strip of _______.

metal

An electrode in a solution of its ions is a _____ _______.

half cell

define half cell

an electrode in a solution of its ions

the cell voltage/electromotive force (emf)/cell potential is the ______ __________ between the half cells.

cell potential

What is the cell reaction of the spontaneous cell made from the Zn+2/Zn and Cl2/Cl- ?

Zn+2 + Cl- -> Cl2 + Zn

Corrosion of _____ is an oxidation reduction process

metals

Metals can be protected by ______ _______

cathodic protection

Cathodic protection is attaching a more _____ metal to the protected metal.

active

In cathodic protection, the active metal is called a _____ _______.

sacrificial anode

Give the general calculation for determining the moles of electrons involved in an electrolysis reaction.

mol e- = current(C/s)time(s)1mol e-/96485C

one mole of electrons has a charge of ______

96485C

Write the net redox reaction that occurs in the following galvanic cell.

Ti / Ti+3 // Se-2 / Se

2Ti + 3Se -> 2Ti+3 + 3Se-2

Determine the Eocell for the reaction: Pb+2 + Zn -&gt; Pb + Zn+2, the half reactions are:
Pb+2(aq) + 2e- -&gt; Pb(s) Eo = -0.125V
Zn+2(aq) + 2e- -&gt; Zn(s) Eo = -0.763V

0.888V (Eocell = Eo(Ox) + Eo(Red) Eocell = (0.763V) + (-0.125V) Eocell = 0.638V)

The following galvanic cell has a measured cell potential Eocell = 0.646V.

Pt / Sn+4(aq) , Sn+2(aq) // Ag+(aq) / Ag(s)

Given that the standard reduction potential for the reduction of Ag+ to Ag(s) is +0.800 V, what is the standard reduction potential for the Sn+4/Sn+2 half-reaction?

+0.154 V (Eocell = Eo(Ox) + Eo(Red) (0.646V) = Eo(Ox) + (+0.800V) Eo(Ox) = -0.154V)

Will magnesium metal displace Al3+ ion from an aqueous solution?

Mg2+(aq) + 2 e- Mg(s) E° = -2.356 V
Al3+(aq) + 3 e- Al(s) E° = -1.676 V

A) No, the system is at equilibrium.
B) Yes, since E°cell is negative.
C) No, since E°cell is negative.
D) No, the reverse reaction is spontaneous.
E) Yes, since E°cell is positive.

E (Eocell = Eo(Ox) + Eo(Red) Eocell = (2.356V) + (-1.676V) Eocell = 0.680V reverse the Eo for Al)

Determine E°cell for the reaction: 2 Al + 3 Zn2+ 2 Al3+ + 3 Zn.
The half reactions are:
Al3+(aq) + 3 e- Al(s) E° = -1.676 V
Zn2+(aq) + 2 e- Zn(s) E° = -0.763 V
A) 0.913 V
B) 2.439 V
C) -0.913 V
D)-1.063 V
E) -2.439 V

A

9) What is the cell diagram for the spontaneous cell involving the Fe3+ / Fe2+ (0.771V) half cell and the Zn2+ / Zn (-0.763 V) half cell?

A) Fe2+(aq) / Fe3+(aq) // Zn2+(aq) / Zn(s)
B) Zn(s) / Zn2+(aq) // Fe3+(aq) / Fe2+(aq) / Pt(s)
C) Pt(s) / Fe2+(aq), Fe3+(aq) // Zn2+(aq) / Zn(s)
D) Zn(s) / Zn2+(aq) // Fe3+(aq) / Fe2+(aq)
E) Zn(s) / Zn2+(aq) // Fe3+(aq) / Fe2+(aq) / Zn(s)

B

What is the cell diagram for the spontaneous cell involving the Fe3+ A Fe2+ (0.771V) half cell and the Zn2+ A Zn (-0.763 V) half cell?

A) Fe2+(aq) A Fe3+(aq) ? Zn2+(aq) A Zn(s)
B) Zn(s) A Zn2+(aq) ? Fe3+(aq) A Fe2+(aq) A Pt(s)
C) Pt(s) A Fe2+(aq), Fe3+(aq) ? Zn2+(aq) A Zn(s)
D) Zn(s) A Zn2+(aq) ? Fe3+(aq) A Fe2+(aq)
E) Zn(s) A Zn2+(aq) ? Fe3+(aq) A Fe2+(aq) A Zn(s)

B

What is the cell diagram for the spontaneous cell involving the Cl2 / Cl- (1.358V) and the Sn+2 / Sn (-0.137V) half cells?

Cl2(g) / Cl-(aq) // Sn+2(aq) / Sn(s)

What is the cell diagram for the spontaneous cell involving the Cl2 / Cl (1.358 V) and the Sn2+ / Sn (-0.137V) half cells in STP condition?

A) Pt A Cl2(g) / Cl(aq) // Sn2+(aq) / Sn(s)
B) Sn(s) / Cl(aq) ? Sn2+(aq) / Sn(s)
C) Sn(s) / Sn2+(aq) // Cl(aq) / Cl2(g) / Pt
D) Sn(s) // Sn2+(aq) // Cl(aq) / Cl2(g)
E) Cl2(g) / Cl(aq) ? Sn2+(aq) / Sn(s)

C

Determine E°cell for the reaction:
2 Al + 3 Zn2+ 2 Al3+ + 3 Zn.

The half reactions are:
Al3+(aq) + 3 e- Al(s) E° = -1.676 V
Zn2+(aq) + 2 e- Zn(s) E° = -0.763 V

A) 1.227 V
B) 1.611 V
C) 1.487 V
D) 1.467 V
E) 1.562 V

D (Eocell = Eo(reduction half-cell) – Eo(oxidation half-cell) ignore the moles of electrons! Eocell =

In a cell diagram, the components to the left of the two vertical lines comprise the ________ reaction, while the components to the right of the two vertical lines comprise the _________ reaction.

oxidation, reduction

Consider the cell:
Ni / Ni+2 (?M) // Cu+2 (0.136M) / Cu

Ni+21/Ni: Eo= -0.257
Cu+2/Cu: Eo = 0.340V

The measured potential of the cell is 0.612V. What is [Ni+2] at 25degC?

0.05M (Ni +Cu2+ –>Ni2+ + Cu…… (no. of electrons transferred =z=2) (A)Eocell=0.257+0.34= 0.597 (B)Ecell=Eocell-(0.059/z)x(log[Ni2+]/[Cu2+]) 0.612=0.597-(0.059/2)xlog([Ni2+]/0.136) -0.508+log(0.136)=log([Ni2+]=-1.374 [Ni2+]=0.042 M)

A solution of AuCl3 is electrolyzed by passing 85.0 A of current for 17.6 min. What mass of gold plates out?

A. 61.1g
B. 1.02 g
C. 183 g
D. 0.483 g

A ([17.6min][60sec/1min][85.0C/1sec][1mol e-/96485C][1mol/96485C][1mol Au/3mol e-][186.967g/1mol Au] = 58.0g close enough lol)

Anions flow through the salt bridge into the _______- compartment while cations flow into
the __________ compartment to maintain electroneutrality

anode, cathode

The anode of a voltaic cell is __________ (like an anion), and the cathode is ________ (like a
cation). Electrons move from the negative anode through an external circuit to the
positive cathode.

negative, positive

Given the following half-reactions occurring in the silver-zinc hearing aid battery, calculate the deltaGo at 25degC of this battery.

ZnO(s) + H2O(l) + 2e- —&gt; Zn(s) + 2OH- Eo = -1.260V
Ag2O(s) + H2O(l) + 2e- —&gt; 2Ag(s) + 2OH- Eo = +0.243V

-290 kJ (The element with the lower reduction potential acts as a strong reducing agent. A reducing agent is oxidized. The element with a higher reduction potential acts as a strong oxidizing agent. An oxidizing agent is reduced. DeltaGo = -zFEocell DeltaGo = -(2)(96500)(0.243-[-1.260]) Delta Go = -290000J Delta Go = -290 kJ)

What is the [H+] in a concentration cell at 25degC if the cell potential is 00451 V and the cell diagram is:

Pt / H2 (g, 1atm) / H+ (?M) // H+(1M) / H2 (g, 1atm) / Pt

2H+(aq) + 2e- —&gt; H2(g) Eo = 0.V

…

A solution of AuCl3 is electrolyzed by passing 85.0 A of current for 17.6 min. What mass of gold plates out?

A. 61.1g
B. 1.02 g
C. 183 g
D. 0.483 g

c

A 0.235 g sample of an alloy containing lead is dissolved in nitric acid to produce an aqueous solution of lead (II) ions. This acidic solution is titrated with a 0.0123 M solution of potassium permanganate, foming lead (IV) and manganese (II) ions. If the titration requires 14.36 mL of the potassium permanganate to reach the equivalence point,

a. How many moles of permanganate were used?

b. How many moles of lead ions were titrated?

c. How many grams of lead are present?

d. What is the percent lead by mass in the alloy?

(Reaction of Lead (Pb) with nitric goes by the equation Pb(s) + 4HNO3(aq) —–> Pb(NO3)2(aq) + 2H2O(l) + 2NO2(g) 14.26 mL of KMnO4 solution having concentration of 0.0123 M was used So the moles of KMnO4 used were 0.175 mmoles of KMnO4 were used As per the data provided Lead gets converted to Pb(IV) and Manganese gets reduced to Mn(II) In KMnO4 Mn is in +7 state So the equation with formal oxidation state would be Pb2+ —> Pb4+ + 2e- Mn7+ + 5e- —-> Mn2+ balancing is possible when we multiply 1st reaction with 5 and second reaction with 2 So we get 5 moles of Pb2+ reacts with 2 moles on Mn7+ Since 0.175 mmoles of KMnO4 were used we get 0.4375 mmoles of lead ions were titrated. Lead Atomic mass = 207 So lead present in the alloy is 0.4375 mmol or 0.0905 g of lead. Percentage lead by mass is 38.5 %.)