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Electrochemical Equilibrium and Cells

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Wize High School Grade 12 Chemistry Textbook > Electrochemistry

Reduction Potentials

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Will the following reaction proceed spontaneously or will current need to be added externally?
Cu(s)+Ca(aq)2+β†’Ca(s)+Cu(aq)2+Cu_{(s)}+Ca^{2+}_{(aq)} \to Ca_{(s)} +Cu^{2+} _{(aq)}


More Reduction Potentials Questions:
Electrochemistry: Reduction potential application
Reduction Potentials Application
Below we see a diagram showing mitochondrial cellular respiration.
If we just consider the electron carrier NADH, NADH molecules gives up their electrons to the first complex, then the electron goes to the 2nd complex, then the 3rd, then the 4th and finally oxygen (O2) is the final electron acceptor.
Electrochemistry: Reduction potential application
Reduction Potentials Application
Below we see a diagram showing mitochondrial cellular respiration.
If we just consider the electron carrier NADH, NADH molecules gives up their electrons to the first complex, then the electron goes to the 2nd complex, then the 3rd, then the 4th and finally oxygen (O2) is the final electron acceptor.
Electrochemistry: Reduction Potentials
Based on the following reduction potentials, which of these species is the strongest reducing agent?
ReactionEΒ°(V)2HCβ„“O(aq)+2H+(aq)+2eβˆ’β†’Cβ„“2(g)+2H2O(β„“)+1.63Cβ„“2(g)+2eβˆ’β†’2Cβ„“βˆ’(aq)+1.36Br2(β„“)+2eβˆ’β†’2Brβˆ’(aq)+1.08I2(s)+2eβˆ’β†’2Iβˆ’(aq)+0.535Ag+(aq)+eβˆ’β†’Ag(s)+0.799Sn4+(aq)+2eβˆ’β†’Sn2+(aq)+0.15\def\arraystretch{2} \begin{array}{c|c} \hline \rm Reaction & EΒ° (V) \\ \hline 2 HCβ„“O (aq) + 2 H^+ (aq) + 2 e^βˆ’ β†’ Cβ„“_2 (g) + 2 H_2O (β„“) & + 1.63 \\ \hline Cβ„“_2(g) + 2 e^βˆ’ β†’ 2 Cβ„“^βˆ’ (aq) &+ 1.36\\ \hline Br_2(β„“) + 2 e^βˆ’ β†’ 2 Br^βˆ’ (aq) & + 1.08\\ \hline I_2 (s) + 2 e^βˆ’ β†’ 2 I^βˆ’ (aq) & + 0.535\\ \hline Ag+ (aq) + e^βˆ’ β†’ Ag (s) &+ 0.799\\ \hline Sn^{4+} (aq) + 2 e^βˆ’ β†’ Sn^{2+} (aq) &+ 0.15\\ \hline \end{array}
Electrochemistry: Reduction potential application
Reduction Potentials Application
Below we see a diagram showing mitochondrial cellular respiration.
If we just consider the electron carrier NADH, NADH molecules gives up their electrons to the first complex, then the electron goes to the 2nd complex, then the 3rd, then the 4th and finally oxygen (O2) is the final electron acceptor.
In basic solution Se2- and SO32- ions react spontaneously:
2Se(aq)2βˆ’+2SO3(aq)2βˆ’+3H2O(l)β†’2Se(s)+6OH(aq)βˆ’+S2O(aq)2βˆ’,Ecello=0.35V2Se_{(aq)}^{2-}+2SO_{3(aq)}^{2-}+3H_2O_{(l)} \to 2Se_{(s)}+6OH_{(aq)}^-+S_2O^{2-}_{(aq)}, E^o_{cell}=0.35V
Write balanced half reactions for this process
(Duplicated)
Use the standard reduction potentials below to answer the following question: which species listed is the strongest oxidizing agent?
Reduction halfβˆ’reactionEΒ°(V)πΆπ‘Ž2+(π‘Žπ‘ž)+2π‘’βˆ’β‡ŒπΆπ‘Ž(𝑠)βˆ’2.87π‘π‘Ž+(π‘Žπ‘ž)+π‘’βˆ’β‡Œπ‘π‘Ž(𝑠)βˆ’2.71𝐴𝑙3+(π‘Žπ‘ž)+3π‘’βˆ’β‡Œπ΄π‘™(𝑠)βˆ’1.66𝑀𝑛2+(π‘Žπ‘ž)+2π‘’βˆ’β‡Œπ‘€π‘›(𝑠)βˆ’1.18𝐢𝑒2+(π‘Žπ‘ž)+2π‘’βˆ’β‡ŒπΆπ‘’(𝑠)+0.15𝐴𝑔+(π‘Žπ‘ž)+π‘’βˆ’β‡Œπ΄π‘”(𝑠)+0.80\begin{array}{c:c}\rm Reduction\ half-reaction& EΒ° (V)\\ \hline πΆπ‘Ž^{2+} (π‘Žπ‘ž)+2 𝑒^βˆ’β‡ŒπΆπ‘Ž (𝑠)& -2.87\\ \hline π‘π‘Ž^+ (π‘Žπ‘ž)+𝑒^βˆ’β‡Œπ‘π‘Ž (𝑠)&-2.71\\ \hline 𝐴𝑙^{3+} (π‘Žπ‘ž)+3 𝑒^βˆ’β‡Œπ΄π‘™(𝑠) &-1.66\\ \hline 𝑀𝑛^{2+} (π‘Žπ‘ž)+2 𝑒^βˆ’β‡Œπ‘€π‘› (𝑠)&-1.18\\ \hline 𝐢𝑒^{2+} (π‘Žπ‘ž)+2 𝑒^βˆ’ β‡ŒπΆπ‘’ (𝑠) &+0.15\\ \hline 𝐴𝑔^+ (π‘Žπ‘ž)+𝑒^βˆ’ β‡Œπ΄π‘” (𝑠)&+0.80\end{array}
Electrochemistry: Reduction potentials in a Galvanic Cell
Use the standard reduction potentials below to answer the following question: which metal could be used to create a sacrificial anode for an aluminum pipe?
Reduction halfβˆ’reactionEΒ°(V)πΆπ‘Ž2+(π‘Žπ‘ž)+2π‘’βˆ’β‡ŒπΆπ‘Ž(𝑠)βˆ’2.87π‘π‘Ž+(π‘Žπ‘ž)+π‘’βˆ’β‡Œπ‘π‘Ž(𝑠)βˆ’2.71𝐴𝑙3+(π‘Žπ‘ž)+3π‘’βˆ’β‡Œπ΄π‘™(𝑠)βˆ’1.66𝑀𝑛2+(π‘Žπ‘ž)+2π‘’βˆ’β‡Œπ‘€π‘›(𝑠)βˆ’1.18𝐢𝑒2+(π‘Žπ‘ž)+2π‘’βˆ’β‡ŒπΆπ‘’(𝑠)+0.15𝐴𝑔+(π‘Žπ‘ž)+π‘’βˆ’β‡Œπ΄π‘”(𝑠)+0.80\begin{array}{c:c}\rm Reduction\ half-reaction& EΒ° (V)\\ \hline πΆπ‘Ž^{2+} (π‘Žπ‘ž)+2 𝑒^βˆ’β‡ŒπΆπ‘Ž (𝑠)& -2.87\\ \hline π‘π‘Ž^+ (π‘Žπ‘ž)+𝑒^βˆ’β‡Œπ‘π‘Ž (𝑠)&-2.71\\ \hline 𝐴𝑙^{3+} (π‘Žπ‘ž)+3 𝑒^βˆ’β‡Œπ΄π‘™(𝑠) &-1.66\\ \hline 𝑀𝑛^{2+} (π‘Žπ‘ž)+2 𝑒^βˆ’β‡Œπ‘€π‘› (𝑠)&-1.18\\ \hline 𝐢𝑒^{2+} (π‘Žπ‘ž)+2 𝑒^βˆ’ β‡ŒπΆπ‘’ (𝑠) &+0.15\\ \hline 𝐴𝑔^+ (π‘Žπ‘ž)+𝑒^βˆ’ β‡Œπ΄π‘” (𝑠)&+0.80\end{array}
Electrochemistry: Standard Reduction Potentials
A galvanic cell is constructed from chromium and copper electrodes immersed in aqueous solutions of Cr3+and Cu2+ respectively (half-reactions shown below).
Cr3+(aq)+3eβˆ’β‡ŒCr(s)Cr^{3+}(aq)+3e^-β‡ŒCr(s) EΒ°red=βˆ’0.74VEΒ°_{red}=-0.74V
Cu2+(aq)+2eβˆ’β‡ŒCu(s)Cu^{2+}(aq)+2e^-β‡ŒCu(s) EΒ°red=+0.34VEΒ°_{red}=+0.34V
Electrochemistry: Reduction potentials
Based on the data table, which of these species is the strongest reducing agent?
Based on the following reduction potentials, which of these species is the strongest reducing agent?
ReactionEΒ°(V)2HCβ„“O(aq)+2H+(aq)+2eβˆ’β†’Cβ„“2(g)+2H2O(β„“)+1.63Cβ„“2(g)+2eβˆ’β†’2Cβ„“βˆ’(aq)+1.36Br2(β„“)+2eβˆ’β†’2Brβˆ’(aq)+1.08I2(s)+2eβˆ’β†’2Iβˆ’(aq)+0.535Ag+(aq)+eβˆ’β†’Ag(s)+0.799Sn4+(aq)+2eβˆ’β†’Sn2+(aq)+0.15\def\arraystretch{2} \begin{array}{c|c} \hline \rm Reaction & EΒ° (V) \\ \hline 2 HCβ„“O (aq) + 2 H^+ (aq) + 2 e^βˆ’ β†’ Cβ„“_2 (g) + 2 H_2O (β„“) & + 1.63 \\ \hline Cβ„“_2(g) + 2 e^βˆ’ β†’ 2 Cβ„“^βˆ’ (aq) &+ 1.36\\ \hline Br_2(β„“) + 2 e^βˆ’ β†’ 2 Br^βˆ’ (aq) & + 1.08\\ \hline I_2 (s) + 2 e^βˆ’ β†’ 2 I^βˆ’ (aq) & + 0.535\\ \hline Ag+ (aq) + e^βˆ’ β†’ Ag (s) &+ 0.799\\ \hline Sn^{4+} (aq) + 2 e^βˆ’ β†’ Sn^{2+} (aq) &+ 0.15\\ \hline \end{array}
Electrochemistry: Oxidizing agents
Which one of the species above is the best oxidizing agent?
Zn2+(aq)+2eβˆ’β‡ŒZn(s)Eo=βˆ’0.74VCu2+(aq)+2eβˆ’β‡ŒCu(s)Eo=+0.34VCo2+(aq)+2eβˆ’β‡ŒCo(s)Eo=βˆ’0.28VAg+(aq)+eβˆ’β‡ŒAg(s)Eo=+0.80V\def\arraystretch{1.5}\begin{array}{cc}Zn^{2+}(aq)+2e^-β‡ŒZn(s)&&E^o=-0.74V\\ Cu^{2+}(aq)+2e^-β‡ŒCu(s)&&E^o=+0.34V\\ Co^{2+}(aq)+2e^-β‡ŒCo(s)&&E^o=-0.28V\\ Ag^+(aq)+e^-β‡ŒAg(s)&&E^o=+0.80V\end{array}
Electrochemistry: Galvanic Cells
The cell pictured below is prepared under standard conditions. In this cell, what will be the anode?
𝐢𝑒2+(π‘Žπ‘ž)+𝐻2(𝑔)β‡ŒπΆπ‘’(𝑠)+2𝐻+(π‘Žπ‘ž)𝐸°=0.339𝑉𝐢𝑒^{2+} (π‘Žπ‘ž)+𝐻_2 (𝑔)\rightleftharpoons 𝐢𝑒 (𝑠)+2 𝐻^+ (π‘Žπ‘ž) \qquad𝐸°=0.339 𝑉
Use the standard reduction potentials below to answer the following question: which metal could be used as an anode for an aluminum pipe?
Reduction halfβˆ’reactionEΒ°(V)πΆπ‘Ž2+(π‘Žπ‘ž)+2π‘’βˆ’β‡ŒπΆπ‘Ž(𝑠)βˆ’2.87π‘π‘Ž+(π‘Žπ‘ž)+π‘’βˆ’β‡Œπ‘π‘Ž(𝑠)βˆ’2.71𝐴𝑙3+(π‘Žπ‘ž)+3π‘’βˆ’β‡Œπ΄π‘™(𝑠)βˆ’1.66𝑀𝑛2+(π‘Žπ‘ž)+2π‘’βˆ’β‡Œπ‘€π‘›(𝑠)βˆ’1.18𝐢𝑒2+(π‘Žπ‘ž)+2π‘’βˆ’β‡ŒπΆπ‘’(𝑠)+0.15𝐴𝑔+(π‘Žπ‘ž)+π‘’βˆ’β‡Œπ΄π‘”(𝑠)+0.80\begin{array}{c:c}\rm Reduction\ half-reaction& EΒ° (V)\\ \hline πΆπ‘Ž^{2+} (π‘Žπ‘ž)+2 𝑒^βˆ’β‡ŒπΆπ‘Ž (𝑠)& -2.87\\ \hline π‘π‘Ž^+ (π‘Žπ‘ž)+𝑒^βˆ’β‡Œπ‘π‘Ž (𝑠)&-2.71\\ \hline 𝐴𝑙^{3+} (π‘Žπ‘ž)+3 𝑒^βˆ’β‡Œπ΄π‘™(𝑠) &-1.66\\ \hline 𝑀𝑛^{2+} (π‘Žπ‘ž)+2 𝑒^βˆ’β‡Œπ‘€π‘› (𝑠)&-1.18\\ \hline 𝐢𝑒^{2+} (π‘Žπ‘ž)+2 𝑒^βˆ’ β‡ŒπΆπ‘’ (𝑠) &+0.15\\ \hline 𝐴𝑔^+ (π‘Žπ‘ž)+𝑒^βˆ’ β‡Œπ΄π‘” (𝑠)&+0.80\end{array}
Which one of the species above is the best oxidizing agent?
Zn2+(aq)+2eβˆ’β‡ŒZn(s)Eo=βˆ’0.74VCu2+(aq)+2eβˆ’β‡ŒCu(s)Eo=+0.34VCo2+(aq)+2eβˆ’β‡ŒCo(s)Eo=βˆ’0.28VAg+(aq)+eβˆ’β‡ŒAg(s)Eo=+0.80V\def\arraystretch{1.5}\begin{array}{cc}Zn^{2+}(aq)+2e^-β‡ŒZn(s)&&E^o=-0.74V\\ Cu^{2+}(aq)+2e^-β‡ŒCu(s)&&E^o=+0.34V\\ Co^{2+}(aq)+2e^-β‡ŒCo(s)&&E^o=-0.28V\\ Ag^+(aq)+e^-β‡ŒAg(s)&&E^o=+0.80V\end{array}
The cell pictured is prepared at standard conditions and the cell voltage measured to be +0.339 V. Which of the following changes would result in an increase to the measured cell voltage?
𝐢𝑒2+(π‘Žπ‘ž)+𝐻2(𝑔)β‡ŒπΆπ‘’(𝑠)+2𝐻+(π‘Žπ‘ž)𝐸°=0.339𝑉𝐢𝑒^{2+} (π‘Žπ‘ž)+𝐻_2 (𝑔)\rightleftharpoons 𝐢𝑒 (𝑠)+2 𝐻^+ (π‘Žπ‘ž) \qquad𝐸°=0.339 𝑉