Extra Credit 9

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Q19.2B

Use the table of Standard Reduction Potential to predict whether these reactions will happen spontaneously.

1. \(\mathrm{2Ag^+(aq) + Cu(s) \rightarrow 2Ag(s) + Cu^{2+}(aq)}\)
2. \(\mathrm{Fe^{3+}(aq) + Na(s) \rightarrow Fe^{2+}(aq) + Na^+(aq)}\)
3. \(\mathrm{Zn^{2+}(aq) + 2I^-(aq) \rightarrow Zn(s) + I_2(l)}\)

   S19.2B

   

   MathJax:

   First break the full reaction into two half reactions:

   Reduction : \(\mathrm{Ag^+(aq) +e^- \rightarrow Ag(s) }\) \(\mathrm E_ {reduction} =0.800V\)2Ag(s) }\) \(\mathrm E_ {reduction} =0.800V\)

   Oxidation: \(\mathrm{Cu(s) \rightarrow 2e^- + Cu(s)}\) \(\mathrm E_ {oxidation} =-0.340V\)

   The sign for the standard potential for the \(\mathrm E_ {reduction} =0.340V\) for copper has been reversed to \(\mathrm E_ {oxidation} =-0.340V\) as it is an oxidation reaction.

   Hence,

   \(\mathrm E_ {cell}=E_{reduction} + E_{oxidation}\)

   0.800V + (-0.340 V) = 0.460V

   The \(\mathrm E_ {cell}\) is positive at 0.460, making it a spontaneous reaction.

   

   MathJax:

   First break the full reaction into two half reactions in order to determine which side is being oxidized :

   Reduction : \(\mathrm{Fe^{3+} + e^- \rightarrow Fe^{2+} }\) \(\mathrm E_ {reduction} =0.771V\)

   Oxidation: \(\mathrm{Na \rightarrow e^- + Na^+ }\) \(\mathrm E_ {oxidation} =-2.713\)

   The sign for the standard potential for the \(\mathrm E_ {reduction} =-2.713V\) for sodium has been reversed to \(\mathrm E_ {oxidation} =2.713\) as it is an oxidation reaction.

   Hence,

   \(\mathrm E_ {cell}=E_{reduction} + E_{oxidation}\)

   0.771V + (2.713 V) = 3.484V

   The \(\mathrm E_ {cell}\) is positive at 3.484V, making it a spontaneous reaction.

   

   MathJax:

   First break the full reaction into two half reactions:

   Reduction : \(\mathrm{Zn^{2+} + 2e^- \rightarrow Zn^{2+} }\) \(\mathrm E_ {reduction} =-0.763V\)

   Oxidation: \(\mathrm{2I^- \rightarrow 2e^- + I_2 }\) \(\mathrm E_ {oxidation} =-0.535\)

   The sign for the standard potential for the \(\mathrm E_ {reduction} =0.535\) for iodine has been reversed to \(\mathrm E_ {oxidation} =-0.535\) as it is an oxidation reaction.

   Hence,

   \(\mathrm E_ {cell}=E_{reduction} + E_{oxidation}\)

   -0.763V + (-0.535 V) = -1.298V

   The \(\mathrm E_ c\) is negative at -1.298V, making it a not a spontaneous reaction.

   Q19.35C

   The voltaic cell in the following diagram has an \(\mathrm{E^\circ_{cell}= 0.5464\: V}\). Solve for the \(\mathrm{[Cl^-]}\) in the cell

   \(\mathrm{Ag (s)|Ag^+(0.40\:M)||Cl^- (x\:M),\,Cl_2 (0.60\:atm)|Pt(s)}\)

   

   Ignore the picture it is wrong- Max.

4. In order to solve for the \(\mathrm {[Cl^-]}\) in the cell, calculate the \(\mathrm {E _{cell}}\)

5. First break the full reaction into two half reactions and since Ag is on the right side of the cell diagram it is the anode thus being oxidize and since Cl is on the left side it is the cathode side thus being reduced:

   Reduction : \(\mathrm{Ag^+(aq) + e^- \rightarrow Ag(s) }\) \(\mathrm E_ {oxidation} =-0.800V\)

   Oxidation: \(\mathrm{Cl_2 \rightarrow 2e^- + 2Cl^- }\) \(\mathrm E_ {reduction} = 1.358V\)

   The sign for the standard potential for the \(\mathrm E_ {reduction} =0.8V\) has been reversed to \(\mathrm E_ {oxidation} =-0.8V\) as it is an oxidation reaction.

   Hence,

   \(\mathrm E^{∘}_ {cell}=E_{reduction} + E_{oxidation}\)

   (1.340 V)+(-0.8V) =0.558 V

   Using the Nernst Equation as it is non-standard conditions:

   \(\mathrm E_{cell} \) = \(\mathrm E^{∘}_{cell} \) - \(\mathrm{\frac{0.0592}{n}}\) x Log Q

   The number of electrons transferred (n)= 2

   The concentration of \(\mathrm Ag^+\) = 0.40

   The concentration of \(\mathrm Cl_2\) = 0.60

   The concentration of \(\mathrm Cl^-\) is unknown

   Q = \(\mathrm{\frac{concentration\, of\, products}{concentration\, of \,reactants}}\)

   Plugging in the known values gives us:

   \(\mathrm 0.5465 \) = \(\mathrm 0.558\) - \(\mathrm{\frac{0.0592}{2}}\) x log\(\mathrm{\frac{(0.40^2)(x^2)}{0.60}}\)

   Solving for x, which is the chloride concentration, gives us: 3.03M

   Q20.21C

   Write an equation using \(mathrm{Mn^{2+}}\) ion as an oxidizing agent in basic solution. Write an equation using Fe(s) as a reducing agent.

   

   \(\mathrm{Mn^2+} \) as an oxidizing agent. (Ignore the picture please he did it wrong -Max)

   This would mean that \(\mathrm{Mn^2+} \) is involved in a reduction half reaction in which it gains electrons. So we would have to start with the half reaction in which\(\mathrm{Mn^2+} \) is being reduced:

MnO₄^- (aq) \({\rightarrow}\) Mn²⁺ (aq)

However this half equation is unbalanced, so we have to first balance the oxygens by adding 4 water molecules to the products side.Then we have to balance the hydrogens by adding 8 hydrogen ions to the reactants side. Finally, we have to balance the charge by adding 5 electrons to the reactants side.

MnO₄^- (aq) + 8H⁺ (aq) + 5e^- \({\rightarrow}\) Mn²⁺ (aq) +4H₂O (l)

However this balanced reaction is in an acidic solution due to hydrogen ions. In order for this reaction to be in basic solution, there must be \(\mathrm{OH^+}\) ions not hydrogen ions in order to do this we have to 8 \(\mathrm{OH^+}\) to both sides. Thus resulting in

MnO₄^- (aq) + 4H₂O (l) + 5e^- \({\rightarrow}\) Mn²⁺ (aq) +8\(\mathrm{OH^+}\) (aq)

This shows that Mn goes from 7+ charge to a 2+ charge as it is gaining electrons

\(\mathrm{Fe(s)} \) as a reducing agent.

This would mean that \(\mathrm{Fe(s)} \) is involved in an oxidation reaction in which is looses electrons. Hence, \(\mathrm{Fe(s)} \) will be oxidized as shown below:

\(\mathrm{Fe(s) ^- \rightarrow Fe^+ + e^- }\)

Fe goes from 0 to +1 and is being oxidized as it is losing electrons.