Electrochemical Cells (College Board AP® Chemistry): Flashcards
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Define voltaic (galvanic) cell.
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Define voltaic (galvanic) cell.
An electrochemical cell that converts the chemical energy of a spontaneous redox reaction into electrical energy, driving a current through an external circuit.
Why does a voltaic cell require a salt bridge?
Without a salt bridge, charge would build up in each half-cell and the current would stop flowing. The salt bridge provides a pathway for ions to migrate between the two solutions, maintaining electrical neutrality in both half-cells and completing the circuit.
True or False?
In a voltaic cell, oxidation occurs at the anode and the anode is the negative electrode.
True.
In voltaic cells, the anode is where oxidation occurs and carries a negative charge.
Define electrolyte (in an electrolytic cell).
The ionic substance (molten or in aqueous solution) through which current passes during electrolysis, supplying the ions that are oxidized or reduced at the electrodes.
True or False?
In an electrolytic cell, the cathode is the positive electrode where reduction occurs.
False.
In an electrolytic cell, the cathode is the negative electrode. Reduction (gain of electrons) does still occur at the cathode, but it carries a negative charge — the opposite polarity to a voltaic cell.
In a zinc–copper voltaic cell, which electrode acts as the anode and which as the cathode? What reactions occur at each?
The zinc electrode is the anode where oxidation occurs: Zn(s) → Zn2+(aq) + 2e−. The copper electrode is the cathode where reduction occurs: Cu2+(aq) + 2e− → Cu(s). Electrons flow through the external circuit from the zinc anode to the copper cathode.
In both voltaic and electrolytic cells, oxidation occurs at the .......... and reduction occurs at the .......... .
In both voltaic and electrolytic cells, oxidation occurs at the anode and reduction occurs at the cathode.
Define standard electrode potential (E°).
The potential difference measured when a standard half-cell is connected to a standard hydrogen electrode (E° = 0.00 V) under standard conditions (1 M ion concentration, 298 K, 100 kPa).
True or False?
A reaction is thermodynamically feasible when E°cell is positive.
True.
A positive E°cell means the forward reaction is spontaneous (ΔG° is negative). If E°cell is negative, the reverse reaction is favored and the forward reaction requires an external energy input.
How do you calculate E°cell from the standard electrode potentials of two half-cells?
E°cell = E°cathode − E°anode (equivalently, E°reduction − E°oxidation). The half-cell with the more positive E° value is assigned as the cathode (right-hand side in the cell diagram), and the half-cell with the less positive E° value is the anode (left-hand side).
Define Gibbs free energy change (ΔG°) in the context of electrochemical cells.
ΔG° = −nFE°cell, where n is the number of electrons transferred and F is the Faraday constant (96,485 C mol-1). A negative ΔG° confirms the cell reaction is spontaneous under standard conditions.
True or False?
For an electrolytic cell, the applied external voltage must be greater than the magnitude of E°cell for the non-spontaneous reaction to proceed.
True.
Because the cell reaction is non-spontaneous (E°cell is negative for the desired direction), an external voltage larger than |E°cell| must be applied. In practice, the required voltage is even higher due to overpotentials and resistance losses.
Why does a more positive E° value for a half-reaction indicate a greater tendency for reduction?
E° is measured relative to the standard hydrogen electrode, so a more positive E° means the species has a stronger thermodynamic drive to gain electrons — it is a stronger oxidising agent than H+/H2. The more positive the value, the further the half-reaction equilibrium lies toward the reduced form.
The standard Gibbs free energy change is calculated using .......... = −nFE°cell, where n is the number of .......... transferred.
The standard Gibbs free energy change is calculated using ΔG° = −nFE°cell, where n is the number of electrons transferred.
Define reaction quotient (Q) as used in electrochemistry.
Q = [oxidised species] / [reduced species]. It describes the ratio of product to reactant concentrations at any given moment, and its value relative to K determines the direction of spontaneous change in the cell.
True or False?
When a galvanic cell reaches equilibrium, Ecell = 0 V and the cell is 'dead'.
True.
At equilibrium, Q = K and there is no net driving force for the redox reaction. With no potential difference, no current flows and the cell can no longer do electrical work.
Why does increasing the concentration of the oxidised species in a galvanic cell increase Ecell above E°cell?
The oxidised species (e.g. Cu2+) is reduced at the cathode and appears in the denominator of Q for the cell reaction. Increasing its concentration decreases Q, which reduces the log Q term in the Nernst equation (Ecell = E°cell − (0.0592/n) log Q) and raises Ecell above E°cell.
Define concentration cell.
An electrochemical cell in which both half-cells contain the same redox couple but at different concentrations. The voltage is generated by the tendency of concentrations to equalise; Ecell → 0 as the concentrations converge.
True or False?
In a concentration cell, reduction occurs in the more concentrated half-cell.
True.
The more concentrated half-cell has a higher ion concentration and therefore a more positive electrode potential, so it acts as the cathode (reduction occurs there), while the less concentrated half-cell is the anode where oxidation releases ions to raise its concentration.
As a galvanic cell reaches equilibrium, Q approaches .......... and Ecell decreases toward .......... , meaning the cell is 'dead'.
As a galvanic cell reaches equilibrium, Q approaches K and Ecell decreases toward 0 V, meaning the cell is 'dead'.
Define the Nernst equation.
E = E° − (RT/nF) ln Q, or at 25 °C: E = E° − (0.0592/n) log10 Q. It gives the electrode potential under nonstandard conditions by accounting for the effect of temperature and ion concentration on the standard electrode potential.
True or False?
Pure solids and pure liquids are excluded from Q in the Nernst equation, but gases are included via their partial pressures.
True.
Pure solids and pure liquids have fixed concentrations (set to 1) and are omitted from Q. Gases contribute to Q through their partial pressures — the Nernst equation accounts for all species in solution or in the gas phase.
How does the Nernst equation predict E will change if the concentration of the oxidised species decreases?
Decreasing [oxidised species] lowers Q, so log Q decreases (becomes more negative), making the term −(0.0592/n) log Q less negative and causing E to rise above E°. This is because a lower concentration of the species being reduced places the half-cell further from equilibrium, increasing the thermodynamic driving force for reduction.
In the Nernst equation E = E° − (0.0592/n) log Q, what does n represent and why does it matter?
n is the number of moles of electrons transferred in the balanced half-reaction. It scales the correction term: the larger n is, the smaller the correction (0.0592/n) applied per unit change in log Q. This means half-reactions involving more electrons are less sensitive to concentration changes than those involving fewer electrons.
True or False?
The simplified Nernst equation E = E° − (0.0592/n) log Q is valid at any temperature.
False.
The 0.0592 value is derived from RT/F at 298 K (25 °C). At any other temperature, the full equation E = E° − (RT/nF) ln Q must be used with the actual temperature in kelvin.
At 25 °C, the simplified Nernst equation is E = E° − (.......... /n) log10 Q, where n is the number of .......... transferred.
At 25 °C, the simplified Nernst equation is E = E° − (0.0592/n) log10 Q, where n is the number of electrons transferred.
Define the faraday (F).
The faraday is the amount of electric charge carried by 1 mole of electrons: F = 96,485 C mol-1. It equals Avogadro's number multiplied by the charge on one electron.
True or False?
The mass of substance deposited during electrolysis is proportional to the current and the time for which it flows.
True.
Charge Q = I × t, and the moles of electrons (and therefore moles of product) are directly proportional to Q. Doubling current or doubling time both double the charge and the mass deposited.
Why does depositing 1 mol of Mg2+ require twice the charge needed to deposit 1 mol of Na+?
Each Mg2+ ion requires 2 electrons for reduction (Mg2+ + 2e- → Mg), whereas each Na+ requires only 1 electron (Na+ + e- → Na). Depositing 1 mol of Mg therefore requires 2 × 96,485 C = 192,970 C, compared to 96,485 C for 1 mol of Na.
Define electrolysis.
Electrolysis is the non-spontaneous decomposition of an ionic compound (electrolyte) using electrical energy. An external current drives oxidation at the anode and reduction at the cathode.
True or False?
During electrolysis of a molten salt, the metal is always deposited at the anode.
False.
Metal cations are attracted to the cathode (negative electrode), where they gain electrons and are reduced to the metal. Non-metal anions migrate to the anode, where they are oxidised.
How do you calculate the mass of a metal deposited during electrolysis from current, time, and molar mass data?
Step 1: Calculate charge — Q = I × t (in coulombs). Step 2: Calculate moles of electrons — mol e- = Q / 96,485. Step 3: Use the half-equation to find moles of metal — mol metal = mol e- / n (where n = electrons per ion). Step 4: Convert to mass — mass = mol metal × molar mass.
The charge passed during electrolysis is calculated using Q = .......... × t, where Q is in .......... and t is in seconds.
The charge passed during electrolysis is calculated using Q = I × t, where Q is in coulombs and t is in seconds.
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