Extra Credit 35
- Page ID
- 83006
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Q17.5.3
Consider a battery made from one half-cell that consists of a copper electrode in 1 M \(CuSO_4\) solution and another half-cell that consists of a lead electrode in 1 M \(Pb(NO_3)_2Pb(NO_3)_2\) solution.
- What are the reactions at the anode, cathode, and the overall reaction?
- What is the standard cell potential for the battery?
- Most devices designed to use dry-cell batteries can operate between 1.0 and 1.5 V. Could this cell be used to make a battery that could replace a dry-cell battery? Why or why not.
- Suppose sulfuric acid is added to the half-cell with the lead electrode and some \(PbSO_4(s)PbSO_4(s)\) forms. Would the cell potential increase, decrease, or remain the same?
S17.5.3
1.) What are the reactions at the anode, cathode, and overall reaction?
$$Cu^2+2e^-\to Cu \qquad E^\circ=0.34V$$
$$Pb^2+2e^-\to Pb \qquad E^\circ=-0.13V$$
The standard potential of Pb is less then Cu so it undergoes oxidation and Cu under goes reduction, or sometimes you could look at the oxidation states.
Edit: The standard potential refers to the potential difference between the anode and the cathode. It is under standard conditions. See the attached reduction potential table for the necessary values.
At anode $$Pb^2+2e^-\to Pb \qquad E^\circ=-0.13V$$
At cathode $$Cu^2+2e^-\to Cu \qquad E^\circ=0.34V$$
2. The Standard Potential for the battery is:
$$E^\circ=cathode-anode$$
$$E^\circ=0.34-(-0.13)$$
$$E^\circ=0.34+0.13$$
$$E^\circ=0.47V$$
3. Most devices designed to use dry-cell batteries can operate between 1.0 and 1.5 V and yes this cell be used to make a battery that could replace a dry-cell battery. Because when 3 cells are connected in a series the standard battery potential becomes 3 times the original amount $$E^\circ_{battery}=E^\circ_{cell}\times{3}$$$$E^\circ_{battery}=3\times0.47$$$$E^\circ=1.41$$ So yes this cell could be used to make a battery that could replace a dry battery cell because the new \(E^\circ_{battery}\) is in between 1.0V and 1.5V.
4. If Lead(II) Sulfate forms the concentration of Lead(II) decreases because lead (II) ions appear as products Q<1, so we can use the Nernst Equation $$ E_{cell}=E^\circ_{cell}-\frac{0.0592V}{n}logQ$$ Subsitute Q for $$Q=\frac{Pb^{2+}}{Cu^{2+}}$$ And just based off this information we can see that, since the log of a number less then one will be negative and since the value of \(Pb^{2+}\) increases that mean overall \(E_{cell}\) will increase
Edit: The purpose of the Nernst Equation is to aid in determining the cell potential under non-standard conditions. Overall, each component of this question is correct.
12.2.2
Go to the PhET Reactions & Rates interactive. Use the Single Collision tab to represent how the collision between monatomic oxygen, \(O\) ,and carbon monoxide, \(CO\) ,results in the breaking of one bond and the formation of another. Pull back on the red plunger to release the atom and observe the results. Then, click on “Reload Launcher” and change to “Angled shot” to see the difference. What happens when the angle of collision changes? Explain how this is relevant to the rate of reaction.
S12.2.2
When the collision was straight on and the red puller was pulled back to the max, the selected atom broke the bond of the other atom instantly and then reformed a bond when hit again. Now when the angle changed, even in the slightest, the bond did NOT break, and when the angle was drastically changed, the two molecules took several seconds to minutes (and sometimes never) to even collide with one another, and when they did the bond still did not always break. So when the angle was changed, the selected atom did not reach the other atom for a good amount of time and then when it did, angle would no longer be at the appropriate orientation for a reaction to occur.
The Collision Theory states that particles must hit each other with energy greater then or equal to the activation energy(the minimum energy in which reactants must collide), \(E_a\) and at they hit at the right orientation for there to be a reaction to occur. Reactions must have sufficient energy in order to break and form new bonds.
Edit: Molecules must collide in order to react. For an effective reaction to occur, collisions must obtain the required energy or activation energy to break/form the bonds. As the temperature rises, molecules move more quickly increasing the chances of bonds to break when colliding.
12.5.7
How much faster does the reaction proceed at \(45^\circ\) then \(25^\circ\) degrees? How much faster does the reaction proceed at \(95^\circ\) then \(25^\circ\)?
S12.5.7
The rate doubles for every \(10^\circ\) rise, so if we start at \(25^\circ\) degrees then go to \(45^\circ\), the reaction has gone up \(25^\circ\) so it has doubled twice. $$2\times2=4x$$ so the reaction goes 4 times faster then what we originally started with.
For the reaction going from \(95^\circ\) to \(25^\circ\)
So since the reactions goes from \(25^\circ\) to \(95^\circ\), so lets go through in increments of \(10^\circ\) to find out how much it doubles $$25\, to \,35 = 2x\,rate$$ $$35\, to \,45 =4x\,rate$$ $$45\,to\, 55= 8x\,rate$$ $$ 55\,to\,65=16x\, rate$$ $$65\,to\,75=32x\, rate$$ $$75\,to\, 85=64x\, rate$$ $$85\,to \,95=128\,rate$$ so it doubles 7 times for every \(10^\circ\) and is 128 times faster at \(95^\circ\) then \(25^\circ\)
Edit: This question is correct. In general, \(25^\circ\) is the standard temperature in Celsius. It can be converted to 298 Kelvin.
Q21.4.2
What changes occur to the atomic number and mass of a nucleus during each of the following decay scenarios?
- an \(\alpha\) particle is emitted
- a \(\beta\) particle is emitted
- \(\gamma\) radiation is emitted
- a positron is emitted
- an electron is captured
S21.4.2
A alpha particle consists of two protons and two neutrons bound together into a particle identical to a helium nucleus. When alpha particle is emitted: A 2 protons will be lost and mass will decrease by 4 $$_2^4\alpha$$ $$Example: _{106}^{263}Sg\rightarrow_{104}^{259}Rf+_2^4He$$
Beta decay occurs when a proton or neutron is transformed into the other because there are to many in the nucleus. There is beta minus decay and beta plus decay. In beta minus decay a neutron decays into a proton, electron and antineutrino. When a beta particle is emitted: The proton will be gained and the mass will not be effected $$_{-1}^0\beta$$$$Example:\,_{29}^{61}Cu\rightarrow_{30}^{61}Zn+_{-1}^0\beta$$
When gamma radiation is emitted: The mass and proton are not effected $$_0^0\gamma$$
A positron is a subtype of beta emission. When a positron is emitted: A proton is lost and mass is not effected $$_1^0\beta$$$$Example:\,_{25}^{50}Mn\rightarrow_{24}^{50}Cr+_1^0e$$
During electron capture the electron in the atoms most inner shell is pulled toward the nucleus and combines with a proton. This forms a neutron and a neutrino which is ejected from the nucleus. When a electron is captured: A proton is lost and the mass is not effected $$_1^0e$$ $$Example:\,_{30}^{63}+_{-1}^0e\rightarrow_{29}^{63}Cu$$
Edit: The provided explanations are correct.
20.2.6
Of these elements, which would you expect to be easiest to reduce: \(Se\), \(Sr\), or \(Ni\)? Explain your reasoning.
S20.2.6
Out of the elements \(Se\), \(Sr\), and \(Ni\), \(Sr\) is the easiest to reduce. Based on Periodic Trends, ionization energy increases as you go to the right and decreases as you go down (increases as you go up). Ionization energy is the energy required to remove an electron from a atom. Reduction is related to Ionization energy and has the same trends. So based on that Periodic Trend, \(Se\) is the most far right element out of the 3 and therefore the easiest to reduce.
Edit: There are different "levels" of ionization energies. The first ionization energy is the energy needed to remove the outermost electron from a neutral atom in the gas phase. When releasing energy, the removal of the following electrons after the first, leads to less energy being released each time.
Q20.5.1
State whether you agree or disagree with this reasoning and explain your answer: Standard electrode potentials arise from the number of electrons transferred. The greater the number of electrons transferred, the greater the measured potential difference. If 1 mol of a substance produces 0.76 V when 2 mol of electrons are transferred—as in \(Zn(s)\rightarrow\, Zn^2+(aq) + 2e^-\,\)—then 0.5 mol of the substance will produce 0.76/2 V because only 1 mol of electrons is transferred.
S20.5.1
I would disagree because standard reduction potential is independent of size and amount of matter in a system. So, the standard reduction potential remains the same for a given metal-metal ion couple. It doesn't matter if there was one electron transfer or two electrons transferred. The standard reduction potential of $$Zn(s)\rightarrow\,Zn^2+2e^-$$ is the same.
Edit: I disagree as well because the quantity of electrons does not cause an effect. The term "standard" can be applied to standard conditions.
Q20.9.10
Electrolysis of \(Cr^{3+}\) produces \(Cr^{2+}\). If we had 500mL of a 0.15M solution of \(Cr^{3+}\) how long would it take to reduce the \(Cr^{3+}\) to \(Cr^{2+}\) using a 0.158A current?
S.20.9.10
Start by writing the equation and formula you will need
$$Cr^{3+}+e^-\rightarrow\,Cr^{2+}$$
$$m=\frac{IT}F \cdot\frac{M}{n}$$
n= number of moles transferred
F= Faradays constant
M= molarity
I=current
t=time
That equation can be reduced to \(Q=IT\)
First we find n, so we are going from \(Cr^{3+}\) to \(Cr^{2+}\), so there is only one electron that is transferred.
$$n=1mole\,e^-\, (going\, from \,Cr^{3+}\, to \,Cr^{2+})$$
Next, we need to get volume into Liters to isolate moles from the molarity
$$Volume=500mL\times\frac{10^-3L}{1mL}$$
$$Volume=0.5L$$
$$moles\, of\, Cr^{3+}=.15\frac{mol}{L}\times0.5L$$
$$moles\, of\, Cr^{3+}=0.075mol$$
Now that we have moles, we can use the number of moles, faradays constant \(F\), and the number of electrons transferred \(n\) to find \(Q\)
$$F=96487\frac{C}{mol\,e^-}$$
$$F=96487\frac{C}{mol\,e^-}\times1mol\,e^-$$
$$F=96487C$$
$$Q=0.075mol Cr^3\times96487{C}=7236.5C$$
Now we can plug everything into the equation. Don't forget to convert \(C\) to \(A\) conversion \(1A=1C\)
$$Q=It$$
$$t=\frac{Q}{I}$$
$$t=\frac{7236.5C}{0.158A}\times\frac{A}{C}$$
$$t=45800.8s$$
So it would have taken 45800.8 seconds
Edit: Everything for this question appears to be correct.
Q14.6.8
Nitramide \(O_2NNH_2\) decomposes in aqueous solution to \(N_2O\) and \(H_2O\). What is the experimental rate law \(\frac{\nabla[N_2O]}{\nabla\,t}\) for the decomposition of nitramide if the mechanism for the decomposition is as follows? $$O_2NNH_2\Leftrightarrow^{k_1}_{k_{-1}}\,O_2NNH^-+H^+(fast)$$
$$O_2NNH^-\rightarrow^{k_2}\,N_2O+OH^-(slow)$$
$$H^++OH^-\rightarrow^{k_3}\,H2O(fast)$$
S14.6.8
So, the rate for the overall reaction is equal to the rate law for the slow step $$rate=rate_2=k_2[O_2NNH^-]$$
But, Nitramide is a intermediate so we have to substitute it for the equivalent expression from the equilibrium step.
$$rate_1=rate_{-1}$$
For the first fast step the reaction is going forward and reverse so \(k_1\, and\, k_{-1}\) rates equal each other
$$k_1[O_2NNH_2]=k_{-1}[O_2NNH^-][H^+]$$
Now we want to get \(O_2NNH^-\) alone because it's the intermediate
$$[O_2NNH^-]=\frac{k_1[O_2NNH_2]}{k_{-1}[H^+]}$$
The second reaction in the mechanism is the slow step and the rate of the reaction depends on the slow step. So
$$rate=rate_2=k_2[O_2NNH^-]$$
Now we can just plug in \(O_2NNH^-\) into the slow step to get the overall rate of reaction
$$rate=k_2\times\frac{k_1[O_2NNH_2]}{k_{-1}[H^+]}$$
In terms of just k, the rate would be
$$rate=k=\frac{k_1k_2}{k_{-1}}$$
The overall rate of the reaction is
$$rate=k\frac{[O_2NNH_2]}{[H^+]}$$
Edit: Notice that the overall rate of a reaction doesn't include the intermediate product in the final rate law answer. Intermediates can be isolated because of their fully formed bonds.