Electrode Potentials & Cells Flashcards
State the meaning of the term electrochemical series
(List of) electrode potentials in (numerical) order
OR half cells/equations in (numerical) order of electrode potential/
Do not allow EMF in order
In an exam question, you’re shown a diagram of a cell a student set up. The left-hand electrode is copper with 0.15 mol dm^-3 CuSO4 solution. The right-hand electrode is copper with 1.0 mol dm^-3 CuSO4 solution. An initial voltage of +0.16 V at 25^C was recorded. Given that the standard electrode potential for the Cu2+/Cu electrode is Cu^2+(aq) + 2e– → Cu(s) where Eo = + 0.34 V and both electrodes contain a strip of copper metal in a solution of aqueous Cu2+ ions. State why the left-hand electrode does not have an electrode potential of +0.34 V
The +0.34 V value is for the standard electrode potential for the Cu2+/Cu electrode. As per the definition of ‘standard’, the solution concentration must be 1 mol dm^-3. The concentration is not 1 mol dm^-3 in the left-hand electrode so the value cannot be +0.34 V.
In an exam question, you’re shown a diagram of a cell a student set up. The left-hand electrode is copper with 0.15 mol dm^-3 CuSO4 solution and has an electrode potential of +0.18 V. The right-hand electrode is copper with 1.0 mol dm^-3 CuSO4 solution and has an electrode potential of +0.34 V. There’s a bulb in the circuit. The EMF of the cell decreases over time to 0 V. Explain how the concentration of copper(II) ions in the left-hand electrode changes when the bulb is alight. Give one reason why the EMF of the cell decreases to 0 V
- The left-hand electrode has a more negative electrode potential than that of the right-hand electrode so electrons in the circuit would flow clockwise (from left-hand electrode to right-hand through the wire not through the salt bridge)
- So left-hand electrode is where oxidation is occurring so Cu -> Cu^2+ + 2e- which shows Cu^2+ as product so Cu^2+ concentration increases in left-hand electrode
- Naturally, this means that reduction is occurring at right-hand electrode so Cu^2+ + 2e- -> Cu so Cu^2+ concentration decreases
- Electrons will continue to flow around the circuit while the 2 above points occur simultaneously until of course the [Cu^2+] in the two solutions become equal which is when the current would stop flowing and there would no longer be any difference in voltage so EMF would be 0 V (as EMF = Eo r - Eo l)
Explain what Electrode Potentials and Electrochemical Cells are
Electrode potentials can demonstrate simple equilibria and how to measure them. Remember that simple equilibria involve reversible reactions that are reaching or have reached equilibrium at standard conditions. In this case, electrode potentials arise from redox reactions in simple equilibria. This electron flow can create a potential difference, which lays the basis of an electrochemical circuit or electric currency; exactly like the process we see in common batteries.
What’s an electrode?
An electrode is a conductor of electricity
Explain what happens if you have two electrodes made of two metals with different reactivity
An electrode potential will be created and the electron flow from the more reactive to the less reactive metal can generate electricity. This two-electrode electricity generator is also called a galvanic cell, a simple battery system
What’s a half-cell?
A half-cell is one of the two electrodes in a galvanic cell. It is a conductive electrode that is bathed in an electrolyte
In an electrochemical cell, what must the electrolytes be, depending on the electrodes used?
It can be a solution of any electron-accepting molecule, such as sulphate (SO4^2- ) or nitrate (NO3^2-)
Explain using theory how and why a galvanic cell works?
In 2 half-cells there’s an electrolyte (either sulphate or nitrate molecules). The metal in each half-cell will react with the aqueous solution and ionise into their respective ions, forming either a metal sulphate or metal nitrate in equilibrium solutions. One of the metals, we’ll call metal A, will lose its electrons more readily than the metal in the other half-cell, metal B; so the equilibrium of A will move farther to the left than that of B. The reactivity of metals is empirically determined and it is known that A is a more reactive metal with higher reducing power compared to B. Each half-cell will produce a potential energy called the energy potential, which can be measured in voltage. Because there is a difference in energy potentials, 2 half-cells can be combined to create a circuit of energy flow that can be measured.
Standard electrode potentials are measured by comparison with the
standard hydrogen electrode. State the substances and conditions needed in a standard hydrogen electrode.
- H2(g) AND 100kPa
- 1 mol dm^−3 AND H+
- Pt electrode AND temperature of 298 K (25^C)
2H+(aq) + 2e− → H2(g) has Eo of 0.00 V and Cu^2+(aq) + 2e− → Cu(s) has Eo of +0.34 V
Use the data to explain why copper does not react with most acids
Eo H+/H2 (or the hydrogen electrode) less +ve than Eo Cu2+/Cu (or the copper
electrode) so H+ cannot oxidise Cu to Cu2+ / H+ poorer oxidising agent
