10. based on the information in the table of standard reduction potentials below, what is the standard cell…

10. based on the information in the table of standard reduction potentials below, what is the standard cell potential for an electrochemical cell that has chromium, cr, and cadmium, cd, electrodes? also, identify the anode.\nstandard reduction potentials (volts) in aqueous solution\nce⁴⁺ + e⁻ → ce³⁺ +1.70\ncu⁺ + e⁻ → cu +0.520\nfe³⁺ + 3e⁻ → fe -0.036\ncd²⁺ + 2e⁻ → cd -0.400\ncr³⁺ + 3e⁻ → cr -0.73\nal³⁺ + 3e⁻ → al -1.66\na. +0.33 v with cr as the anode\nd. -0.33 v with cd as the anode\nb. -0.33 v with cr as the anode\ne. +0.33 v with cd as the anode\nc. -0.26 v with cd as the anode\n11. using the following data, determine the standard cell potential e°cell for the electrochemical cell constructed using the following reaction.\n2al(s) + 3cu²⁺(aq) → 2al³⁺(aq) + 3cu(s)\nhalf - reaction standard reduction potential (v)\ncu²⁺(aq) + 2e⁻ → cu(s) +0.34\nal³⁺(aq) + 3e⁻ → al(s) -1.66\na. +1.32 v\nd. -2.00 v\nb. -1.32 v\ne. +2.30 v\nc. +2.00 v\n12. using the following data, determine the standard cell potential e°cell for the electrochemical cell constructed using the following reaction.\n3pb(s) + 2fe³⁺(aq) → 3pb²⁺(aq) + 2fe(s)\nhalf - reaction standard reduction potential (v)\nfe³⁺(aq) + 3e⁻ → fe(s) +0.771\npb²⁺(aq) + 2e⁻ → pb(s) -0.124\na. +0.647 v\nd. -0.895 v\nb. -0.647 v\ne. -1.17 v\nc. +0.895 v\n13. if the potential of a voltaic cell is +1.20 v, what is the free - energy change when one mole of electrons is transferred in the oxidation - reduction reaction?\na. 116 kj\nd. -116 kj\nb. 1.20 kj\ne. +602 kj\nc. -1.20 kj\n14. the spontaneous redox reaction in a voltaic cell has \na. a negative value of ecell and a negative value of δg.\nb. a positive value of ecell and a positive value of δg.

10. based on the information in the table of standard reduction potentials below, what is the standard cell potential for an electrochemical cell that has chromium, cr, and cadmium, cd, electrodes? also, identify the anode.\nstandard reduction potentials (volts) in aqueous solution\nce⁴⁺ + e⁻ → ce³⁺ +1.70\ncu⁺ + e⁻ → cu +0.520\nfe³⁺ + 3e⁻ → fe -0.036\ncd²⁺ + 2e⁻ → cd -0.400\ncr³⁺ + 3e⁻ → cr -0.73\nal³⁺ + 3e⁻ → al -1.66\na. +0.33 v with cr as the anode\nd. -0.33 v with cd as the anode\nb. -0.33 v with cr as the anode\ne. +0.33 v with cd as the anode\nc. -0.26 v with cd as the anode\n11. using the following data, determine the standard cell potential e°cell for the electrochemical cell constructed using the following reaction.\n2al(s) + 3cu²⁺(aq) → 2al³⁺(aq) + 3cu(s)\nhalf - reaction standard reduction potential (v)\ncu²⁺(aq) + 2e⁻ → cu(s) +0.34\nal³⁺(aq) + 3e⁻ → al(s) -1.66\na. +1.32 v\nd. -2.00 v\nb. -1.32 v\ne. +2.30 v\nc. +2.00 v\n12. using the following data, determine the standard cell potential e°cell for the electrochemical cell constructed using the following reaction.\n3pb(s) + 2fe³⁺(aq) → 3pb²⁺(aq) + 2fe(s)\nhalf - reaction standard reduction potential (v)\nfe³⁺(aq) + 3e⁻ → fe(s) +0.771\npb²⁺(aq) + 2e⁻ → pb(s) -0.124\na. +0.647 v\nd. -0.895 v\nb. -0.647 v\ne. -1.17 v\nc. +0.895 v\n13. if the potential of a voltaic cell is +1.20 v, what is the free - energy change when one mole of electrons is transferred in the oxidation - reduction reaction?\na. 116 kj\nd. -116 kj\nb. 1.20 kj\ne. +602 kj\nc. -1.20 kj\n14. the spontaneous redox reaction in a voltaic cell has \na. a negative value of ecell and a negative value of δg.\nb. a positive value of ecell and a positive value of δg.

Answer

10.

Explanation:

Step1: Identify oxidation and reduction half - reactions

The more negative the standard reduction potential, the more likely the species is to be oxidized. Since $E^0_{Cr^{3+}/Cr}=-0.73\ V$ and $E^0_{Cd^{2+}/Cd}=-0.400\ V$, $Cr$ will be oxidized (anode) and $Cd^{2 + }$ will be reduced (cathode).

Step2: Calculate standard cell potential

The formula for standard cell potential is $E^0_{cell}=E^0_{cathode}-E^0_{anode}$. Here, $E^0_{cathode}=E^0_{Cd^{2+}/Cd}=- 0.400\ V$ and $E^0_{anode}=E^0_{Cr^{3+}/Cr}=-0.73\ V$. So, $E^0_{cell}=-0.400-(-0.73)= + 0.33\ V$.

Answer:

a. +0.33 V with Cr as the anode

11.

Explanation:

Step1: Identify oxidation and reduction half - reactions

For the reaction $2Al(s)+3Cu^{2 + }(aq)\to2Al^{3+}(aq)+3Cu(s)$, $Al$ is oxidized (anode) and $Cu^{2+}$ is reduced (cathode).

Step2: Calculate standard cell potential

Using $E^0_{cell}=E^0_{cathode}-E^0_{anode}$, where $E^0_{cathode}=E^0_{Cu^{2+}/Cu}=+0.34\ V$ and $E^0_{anode}=E^0_{Al^{3+}/Al}=-1.66\ V$. Then $E^0_{cell}=0.34-(-1.66)=+2.00\ V$.

Answer:

c. +2.00 V

12.

Explanation:

Step1: Identify oxidation and reduction half - reactions

For the reaction $3Pb(s)+2Fe^{3+}(aq)\to3Pb^{2+}(aq)+2Fe(s)$, $Pb$ is oxidized (anode) and $Fe^{3+}$ is reduced (cathode).

Step2: Calculate standard cell potential

Using $E^0_{cell}=E^0_{cathode}-E^0_{anode}$, with $E^0_{cathode}=E^0_{Fe^{3+}/Fe}=+0.771\ V$ and $E^0_{anode}=E^0_{Pb^{2+}/Pb}=-0.124\ V$. So, $E^0_{cell}=0.771-(-0.124)=+0.895\ V$.

Answer:

c. +0.895 V

13.

Explanation:

Step1: Use the formula for free - energy change

The formula for the free - energy change in an electrochemical reaction is $\Delta G=-nFE$, where $n$ is the number of moles of electrons transferred, $F = 96485\ C/mol$ (Faraday's constant), and $E$ is the cell potential. Here, $n = 1$ and $E=1.20\ V$. So, $\Delta G=-1\times96485\ C/mol\times1.20\ V=- 115782\ J/mol\approx - 116\ kJ/mol$.

Answer:

d. -116 kJ

14.

Brief Explanations:

In a spontaneous redox reaction in a voltaic cell, the cell potential $E_{cell}>0$ and the Gibbs free - energy change $\Delta G<0$ according to the relationship $\Delta G=-nFE_{cell}$.

Answer:

a. a negative value of $E_{cell}$ and a negative value of $\Delta G$ is incorrect; b. a positive value of $E_{cell}$ and a positive value of $\Delta G$ is incorrect. The correct statement is that a spontaneous redox reaction in a voltaic cell has a positive value of $E_{cell}$ and a negative value of $\Delta G$. But since the options are not correct as written, if we assume the closest correct concept, we note that for a spontaneous reaction $\Delta G<0$ and $E_{cell}>0$.