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Electrochemistry: Standard Reduction Potentials

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Wize High School Grade 12 Chemistry Textbook > Electrochemistry

Redox Reactions

7 Activities

Wize High School Grade 12 Chemistry Textbook > Electrochemistry

Galvanic Cells

7 Activities

Wize High School Grade 12 Chemistry Textbook > Electrochemistry

Reduction Potentials

5 Activities

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
Which of the following is true?
More Redox Reactions 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: Redox
Write the half-reactions for the following redox reaction. Which is the oxidizing agent and which is the reducing agent?
Au2+(aq) + 2I-(aq) <----> Au(s) + I2(s)
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.
Practice: Fill in the Blanks Review Activity (Oxidation vs Reduction)
Review of Oxidation/Reduction:
Redox reaction: involve changes in oxidation states or numbers.
Oxidation AND reduction must occur in order for it to be called a redox reaction!
A voltaic cell is constructed with an Sn/Sn2+ half-cell and a Zn/Zn2+ half-cell. The zinc electrode is negative.
Write a balanced overall reaction for this cell
Draw a diagram of this cell, labeling electrodes with their charges and showing the direction of electron flow in the wire and anion/cation flow in the salt bridge
Consider the following balanced redox equation:
16H(aq)++2MnO4(aq)βˆ’+10Cl(aq)βˆ’β†’2Mn(aq)2++5Cl2(g)+8H2O(l)16H^+_{(aq)}+2MnO^-_{4(aq)}+10Cl^-_{(aq)} \to 2Mn^{2+}_{(aq)}+5Cl_{2(g)}+8H_2O_{(l)}
Which species is being oxidized?
Electrochemistry: Redox
Consider the following balanced redox equation:
16H(aq)++2MnO4(aq)βˆ’+10Cl(aq)βˆ’β†’2Mn(aq)2++5Cl2(g)+8H2O(l)16H^+_{(aq)}+2MnO^-_{4(aq)}+10Cl^-_{(aq)} \to 2Mn^{2+}_{(aq)}+5Cl_{2(g)}+8H_2O_{(l)}
Which species is being oxidized?
Electrochemistry: Redox
Consider the following balanced redox equation:
16H(aq)++2MnO4(aq)βˆ’+10Cl(aq)βˆ’β†’2Mn(aq)2++5Cl2(g)+8H2O(l)16H^+_{(aq)}+2MnO^-_{4(aq)}+10Cl^-_{(aq)} \to 2Mn^{2+}_{(aq)}+5Cl_{2(g)}+8H_2O_{(l)}
Which species is the reducing agent?
Consider the following balanced redox equation:
16H(aq)++2MnO4(aq)βˆ’+10Cl(aq)βˆ’β†’2Mn(aq)2++5Cl2(g)+8H2O(l)16H^+_{(aq)}+2MnO^-_{4(aq)}+10Cl^-_{(aq)} \to 2Mn^{2+}_{(aq)}+5Cl_{2(g)}+8H_2O_{(l)}
Which species is being reduced?
More Galvanic Cells Questions:
Practice: Galvanic Cells
Galvanic Cell Practice
Construct a galvanic cell with the highest voltage given the following half reactions. Determine the voltage produced and express it in standard cell notation.
Na(aq)++eβˆ’β†’Na(s)Eo=βˆ’2.71VNa^+_{(aq)} +e^- \to Na_{(s)} \hspace{40pt} E^o=-2.71V
Practice: Galvanic Cells
Construct a galvanic cell with the highest voltage given the following half reactions. Determine the voltage produced and express it in standard cell notation.
Na(aq)++eβˆ’β†’Na(s)Eo=βˆ’2.71VNa^+_{(aq)} +e^- \to Na_{(s)} \hspace{40pt} E^o=-2.71V
S2O8(aq)2βˆ’+2eβˆ’β†’2SO4(aq)2βˆ’Eo=2.01VS_2O_{8(aq)}^{2-} +2e^- \to 2SO_{4(aq)}^{2-} \hspace{20pt} E^o=2.01V
Practice: Galvanic Cells
Galvanic Cell Practice
Construct a galvanic cell with the highest voltage given the following half reactions. Determine the voltage produced and express it in standard cell notation.
Na(aq)++eβˆ’β†’Na(s)Eo=βˆ’2.71VNa^+_{(aq)} +e^- \to Na_{(s)} \hspace{40pt} E^o=-2.71V
Electrochemistry: Galvanic Cells
In a galvanic cell, oxidation occurs at the:
Electrochemistry: Galvanic Cell
In a zinc-lead cell the reaction is:
Pb2+(aq)+Zn(s)β†’Pb(s)+Zn2+(aq)       EΒ°=0.637VPb^{2+}\left(aq\right)+Zn\left(s\right)\to Pb\left(s\right)+Zn^{2+}\left(aq\right)\ \ \ \ \ \ \ E\degree=0.637V
Which of the following statements about this cell is FALSE?
Electrochemistry: Galvanic Cells
Construct a galvanic cell with the highest voltage given the following half reactions. Determine the voltage produced and express it in standard cell notation.
Na(aq)++eβˆ’β†’Na(s)Eo=βˆ’2.71VNa^+_{(aq)} +e^- \to Na_{(s)} \hspace{40pt} E^o=-2.71V
S2O8(aq)2βˆ’+2eβˆ’β†’2SO4(aq)2βˆ’Eo=2.01VS_2O_{8(aq)}^{2-} +2e^- \to 2SO_{4(aq)}^{2-} \hspace{20pt} E^o=2.01V
A voltaic cell is constructed with an Sn/Sn2+ half-cell and a Zn/Zn2+ half-cell. The zinc electrode is negative.
Write a balanced overall reaction for this cell
Draw a diagram of this cell, labeling electrodes with their charges and showing the direction of electron flow in the wire and anion/cation flow in the salt bridge
Choose the INCORRECT statement
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}
A voltaic cell is constructed with an Sn/Sn2+ half-cell and a Zn/Zn2+ half-cell. The zinc electrode is negative.
Draw a diagram of this cell, labeling electrodes with their charges and showing the direction of electron flow in the wire and anion/cation flow in the salt bridge

Electrochemistry: Battery
Given the following half cells, what is the maximum voltage of a battery that could be constructed?
Li(aq)++eβˆ’β†’Li(s)Eo=βˆ’3.04 VCe(aq)4++eβˆ’β†’Ce(s)3+Eo=1.44 VCo(aq)3++eβˆ’β†’Co(aq)2+Eo=1.82 V\begin{array}{llll}&Li^+_{(aq)}&+&e^{-}&\rightarrow&Li_{(s)}&&E^o=-3.04\ V\\ \\&Ce^{4+}_{(aq)}&+&e^{-}&\rightarrow&Ce^{3+}_{(s)}&&E^o=1.44 \ V \\ \\&Co^{3+}_{(aq)}&+&e^{-}&\rightarrow&Co^{2+}_{(aq)}&&E^o=1.82 \ V \end{array}
Electrochemistry: Ksp calculations
Use the following reduction potentials to calculate the Ksp of La(OH)3(s)
La(OH)3(s)+3eβˆ’β†’La(s)+3 OH(aq)βˆ’Eo=βˆ’2.90VLa(aq)3++3eβˆ’β†’La(s)Eo=βˆ’2.379 V\begin{array}{llllll}&La(OH)_{3(s)}&+&3e-&\rightarrow& La_{(s)}&+&3\ OH^-_{(aq)}& &&E^o=-2.90 V \\ &La^{3+}_{(aq)}&+&3e^-&\rightarrow& La_{(s)}&&&&& E^o=-2.379\ V \end{array}
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.
Electrochemical Equilibrium and Cells
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)}
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: 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