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A voltaic cell is constructed with an Sn/Sn2+ half-cell and a Zn/Zn2+ half-cell…

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Wize High School Grade 12 Chemistry Textbook > Electrochemistry

Galvanic Cells

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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
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)++e−→Na(s)Eo=−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)++e−→Na(s)Eo=−2.71VNa^+_{(aq)} +e^- \to Na_{(s)} \hspace{40pt} E^o=-2.71V
S2O8(aq)2−+2e−→2SO4(aq)2−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)++e−→Na(s)Eo=−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)→Pb(s)+Zn2+(aq)       E°=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)++e−→Na(s)Eo=−2.71VNa^+_{(aq)} +e^- \to Na_{(s)} \hspace{40pt} E^o=-2.71V
S2O8(aq)2−+2e−→2SO4(aq)2−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 half−reactionE°(V)𝐶𝑎2+(𝑎𝑞)+2𝑒−⇌𝐶𝑎(𝑠)−2.87𝑁𝑎+(𝑎𝑞)+𝑒−⇌𝑁𝑎(𝑠)−2.71𝐴𝑙3+(𝑎𝑞)+3𝑒−⇌𝐴𝑙(𝑠)−1.66𝑀𝑛2+(𝑎𝑞)+2𝑒−⇌𝑀𝑛(𝑠)−1.18𝐶𝑢2+(𝑎𝑞)+2𝑒−⇌𝐶𝑢(𝑠)+0.15𝐴𝑔+(𝑎𝑞)+𝑒−⇌𝐴𝑔(𝑠)+0.80\begin{array}{c:c}\rm Reduction\ half-reaction& E° (V)\\ \hline 𝐶𝑎^{2+} (𝑎𝑞)+2 𝑒^−⇌𝐶𝑎 (𝑠)& -2.87\\ \hline 𝑁𝑎^+ (𝑎𝑞)+𝑒^−⇌𝑁𝑎 (𝑠)&-2.71\\ \hline 𝐴𝑙^{3+} (𝑎𝑞)+3 𝑒^−⇌𝐴𝑙(𝑠) &-1.66\\ \hline 𝑀𝑛^{2+} (𝑎𝑞)+2 𝑒^−⇌𝑀𝑛 (𝑠)&-1.18\\ \hline 𝐶𝑢^{2+} (𝑎𝑞)+2 𝑒^− ⇌𝐶𝑢 (𝑠) &+0.15\\ \hline 𝐴𝑔^+ (𝑎𝑞)+𝑒^− ⇌𝐴𝑔 (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)+3e−⇌Cr(s)Cr^{3+}(aq)+3e^-⇌Cr(s) E°red=−0.74VE°_{red}=-0.74V
Cu2+(aq)+2e−⇌Cu(s)Cu^{2+}(aq)+2e^-⇌Cu(s) E°red=+0.34VE°_{red}=+0.34V

Electrochemistry: Battery
Given the following half cells, what is the maximum voltage of a battery that could be constructed?
Li(aq)++e−→Li(s)Eo=−3.04 VCe(aq)4++e−→Ce(s)3+Eo=1.44 VCo(aq)3++e−→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)+3e−→La(s)+3 OH(aq)−Eo=−2.90VLa(aq)3++3e−→La(s)Eo=−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}