Homework 15D: Advanced Voltaic Cells and Electrochemical Potentials

Procedures for Balancing Cell Reactions and Determining $E^{\circ}$

• Criteria for Standard Cell Potential ($E^{\circ}_{cell}$):

  • For a voltaic (galvanic) cell to be functional and spontaneous, the standard cell potential ($E^{\circ}_{cell}$) must be a positive value.

  • The formula used is: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$, where both values are standard reduction potentials.

• Balancing Half-Reactions:

  • Identify which species is being reduced (the one with the more positive reduction potential) and which is being oxidized (the one with the more negative reduction potential).

  • The number of electrons lost in the oxidation half-reaction must equal the number of electrons gained in the reduction half-reaction.

  • If the electron counts do not match, multiply the entire half-reaction by a necessary coefficient. Note: Multiplying a half-reaction by a coefficient does not change the value of $E^{\circ}$.

• Example 1: Copper and Gold Voltaic Cell

  • Half-reactions provided:

    1. $Cu^{2+} + 1e^{-} \rightarrow Cu^{+}$ ($E^{\circ}_{red} \approx +0.16\,V$)

    2. $Au^{3+} + 3e^{-} \rightarrow Au$ ($E^{\circ}_{red} \approx +1.50\,V$)

  • Determination of Anode and Cathode:

    • Gold ($Au^{3+}$) has a higher reduction potential, so it serves as the cathode (reduction).

    • Copper ($Cu^{+}$) must be oxidized, serving as the anode ($Cu^{+} \rightarrow Cu^{2+} + 1e^{-}$).

  • Balancing: Multiply the copper half-reaction by 3 to equalize electrons.

    • Oxidation: $3(Cu^{+} \rightarrow Cu^{2+} + 1e^{-})$

    • Reduction: $Au^{3+} + 3e^{-} \rightarrow Au$

  • Balanced Overall Reaction: $3Cu^{+} + Au^{3+} \rightarrow 3Cu^{2+} + Au$

  • Calculation of $E^{\circ}$: $E^{\circ}_{cell} = 1.50\,V - 0.16\,V = 1.34\,V$

Sketching and Components of Voltaic Cells

• Defining Cell Components:

  • Anode: The electrode where oxidation occurs. It is the source of electrons and is typically labeled as the negative terminal in a voltaic cell.

  • Cathode: The electrode where reduction occurs. It is the destination for electrons and is typically labeled as the positive terminal.

  • Electron Flow: Electrons always flow through the external circuit from the anode to the cathode.

  • Salt Bridge: Essential for maintaining electrical neutrality; ions flow through it to balance the charge build-up in the half-cells.

• Example 2: Copper and Magnesium Cell

  • Materials: Copper electrode in $1\,M\,Cu^{2+}$ solution; Magnesium electrode in $1\,M\,Mg^{2+}$ solution.

  • Standard Reduction Potentials:

    • $Cu^{2+} + 2e^{-} \rightarrow Cu$ ($E^{\circ}_{red} = +0.34\,V$)

    • $Mg^{2+} + 2e^{-} \rightarrow Mg$ ($E^{\circ}_{red} = -2.37\,V$)

  • Anode Identification: Magnesium is the anode because it has the lower reduction potential ($-2.37\,V$).

  • Cathode Identification: Copper is the cathode because it has the higher reduction potential ($+0.34\,V$).

  • Overall Reaction: $Mg(s) + Cu^{2+}(aq) \rightarrow Mg^{2+}(aq) + Cu(s)$

  • Calculation of $E^{\circ}$: $E^{\circ}_{cell} = 0.34\,V - (-2.37\,V) = 2.71\,V$

  • Direction of Flow: Electrons flow from the $Mg$ electrode to the $Cu$ electrode.

Electrochemical Cell Shorthand Notation

• General Format: Anode | Anode Ion (Concentration) || Cathode Ion (Concentration) | Cathode

  • A single vertical line ($|$) represents a phase boundary (e.g., solid metal to aqueous solution).

  • A double vertical line ($||$) represents the salt bridge.

• Shorthand for Magnesium/Copper Cell: $Mg(s) | Mg^{2+} (1\,M) || Cu^{2+} (1\,M) | Cu(s)$

• Shorthand for Copper/Gold Cell: $Pt | Cu^{+}(aq), Cu^{2+}(aq) || Au^{3+}(aq) | Au(s)$

  • Note: In cases where both species in a half-cell are aqueous, an inert electrode like Platinum ($Pt$) may be used.

Determining Spontaneity of Redox Reactions

• Rule of Spontaneity: A reaction is spontaneous as written if the calculated $E^{\circ}_{cell}$ is greater than zero (E^{\circ} > 0).

• Problem 1: Silver and Cadmium

  • Reaction: $Ag(s) + Cd^{2+}(aq) \rightarrow Ag^{+}(aq) + Cd(s)$

  • Half-reactions:

    • Oxidation: $Ag \rightarrow Ag^{+} + 1e^{-}$ ($E^{\circ}_{ox} = -0.80\,V$)

    • Reduction: $Cd^{2+} + 2e^{-} \rightarrow Cd$ ($E^{\circ}_{red} = -0.40\,V$)

  • Calculation: $E^{\circ}_{cell} = -0.40\,V - (+0.80\,V) = -1.20\,V$

  • Spontaneous? No, because $E^{\circ}$ is negative.

• Problem 2: Chromium and Platinum

  • Reaction: $Cr + Pt^{2+} \rightarrow Cr^{3+} + Pt$

  • Half-reactions:

    • Oxidation: $Cr \rightarrow Cr^{3+} + 3e^{-}$ ($E^{\circ}_{red} \text{ for } Cr^{3+} = -0.74\,V$\text{, so }$E^{\circ}_{ox} = +0.74\,V$)

    • Reduction: $Pt^{2+} + 2e^{-} \rightarrow Pt$ ($E^{\circ}_{red} = +1.18\,V$)

  • Calculation: $E^{\circ}_{cell} = 1.18\,V - (-0.74\,V) = 1.92\,V$

  • Spontaneous? Yes, because $E^{\circ}$ is positive.

Reactions Involving the Standard Hydrogen Electrode (SHE)

• Problem 3: Hydrogen and Aluminum

  • Half-reactions:

    1. $2H^{+} + 2e^{-} \rightarrow H_2$ ($E^{\circ} = 0.00\,V$)

    2. $Al^{3+} + 3e^{-} \rightarrow Al$ ($E^{\circ} = -1.66\,V$)

  • Balancing:

    • The aluminum half-reaction must be reversed and multiplied by 2 ($2Al \rightarrow 2Al^{3+} + 6e^{-}$).

    • The hydrogen half-reaction must be multiplied by 3 ($6H^{+} + 6e^{-} \rightarrow 3H_2$).

  • Balanced Overall Reaction: $2Al(s) + 6H^{+}(aq) \rightarrow 2Al^{3+}(aq) + 3H_2(g)$

  • Calculation of $E^{\circ}$: $E^{\circ}_{cell} = 0.00\,V - (-1.66\,V) = 1.66\,V$

Effects of Non-Standard Concentrations (Nernst Equation)

• Principle: The cell potential ($E$) will deviate from the standard cell potential ($E^{\circ}$) if the concentrations are not $1\,M$. This is predicted by the Nernst Equation:

  • $E = E^{\circ} - \frac{0.0592}{n} \log Q$

  • Where $Q$ is the reaction quotient (ratio of products to reactants).

• Case 4: Zinc and Silver Cell

  • Cell Logic: $Zn + 2Ag^{+} \rightarrow Zn^{2+} + 2Ag$

  • Conditions: $[Zn^{2+}] = 1\,M$; $[Ag^{+}] = 5.0\,M$

  • Analysis: The concentration of the reactant ($Ag^{+}$) is significantly higher than the standard $1\,M$. According to Le Chatelier’s Principle, increasing reactant concentration drives the reaction forward, increasing the driving force.

  • Result: Voltage will be higher than $E^{\circ}$.

• Case 5: Manganese and Iron Cell

  • Cell Logic: $3Mn + 2Fe^{3+} \rightarrow 3Mn^{2+} + 2Fe$

  • Conditions: $[Mn^{2+}] = 0.5\,M$; $[Fe^{3+}] = 1\,M$

  • Analysis: The concentration of the product ($Mn^{2+}$) is lower than the standard $1\,M$. Decreasing product concentration shifts the equilibrium toward the products.

  • Result: Voltage will be higher than $E^{\circ}$.