Predicting the Direction of Redox Reactions
This lesson covers:
- What the electrochemical series is
- Using the electrochemical series to calculate standard cell potentials
- Predicting whether reactions will occur using electrode potentials
The electrochemical series ranks redox half-reactions
The electrochemical series is a list of reduction half-equations in order of their standard electrode potentials (E⦵).
- The half-equation with the most positive E⦵ value is placed at the bottom. The oxidised species in this half-equation is the strongest oxidising agent as it most readily accepts electron to become reduced.
- The half-equation with the most negative E⦵ value is at the top. This reduced species in this half-equation is the strongest reducing agent as it most readily loses electrons to becom
For example, the half-equation for fluorine gas gaining electrons has a very positive E⦵ value (+2.87 V):
F2(g) + 2e- ➔ 2F-(aq) E⦵ = +2.87 V
This tells us that fluorine has a great tendency to be reduced, so is a powerful oxidising agent.
By contrast, the half-equation for lithium metal losing an electron has a very negative E⦵ value (-3.04 V):
Li+(aq) + e- ➔ Li(s) E⦵ = -3.04 V
This negative E⦵ value indicates that the forward reaction is not favoured. Rather, lithium metal very readily undergoes oxidation to form lithium ions, so is a strong reducing agent.
Calculating cell potentials from E⦵ values
Standard electrode potentials can be used to calculate the standard cell potential (E⦵cell) for an electrochemical cell:
- Write the reduction half-reactions for both species.
- Determine which reaction occurs in the oxidation direction (more negative E⦵) and which occurs in the reduction direction (more positive E⦵).
- Combine the half-reactions to give the overall redox reaction.
- Calculate E⦵cell using the equation:
Ecell⊖=Ereduced⊖−Eoxidised⊖
Worked example 1 - Calculating E⦵cell for an electrochemical cell
Calculate E⦵cell for a cell comprising copper and silver ions, given the standard electrode potential data below. The overall reaction is:
Cu(s) + 2Ag+(aq) ➔ Cu2+(aq) + 2Ag(s)
| Half reaction | E⦵ (V) |
|---|---|
| Cu2+(aq) + 2e- ⇌ Cu(s) | +0.34 |
| Ag+(aq) + e- ⇌ Ag(s) | +0.80 |
Step 1: Identify oxidation and reduction half-reactions
The copper half-reaction proceeds in the oxidation direction (Cu(s) ➔ Cu2+(aq) + 2e-), as indicated by its lower positive E⦵ value relative to silver's half-reaction.
The silver half-reaction proceeds in the reduction direction (Ag+(aq) + e- ➔ Ag(s)), as indicated by its higher E⦵ value.
Step 2: Equation
Ecell⊖=Ereduced⊖−Eoxidised⊖
Step 3: Substitution and correct evaluation
Ecell⊖=+0.80−(+0.34)=+0.46 V
Worked example 2 - Calculating E⦵cell for an electrochemical cell
Calculate E⦵cell for a cell comprising sodium metal and chlorine gas, given the standard electrode potential data below. The overall reaction is:
2Na(s) + Cl2(g) ➔ 2Na+(aq) + 2Cl-(aq)
| Half reaction | E⦵ (V) |
|---|---|
| Na+(aq) + e- ⇌ Na(s) | -2.71 |
| Cl2(g) + 2e- ⇌ 2Cl-(aq) | +1.36 |
Step 1: Identify oxidation and reduction half-reactions
The sodium half-reaction proceeds in the oxidation direction (Na(s) ➔ Na+(aq) + e-), as indicated by its lower (i.e. negative) E⦵ value relative to chlorine's half-reaction.
The chlorine half-reaction proceeds in the reduction direction (Cl2(g) + 2e- ➔ 2Cl-(aq)), as indicated by its higher (i.e. positive) E⦵ value.
Step 2: Equation
Ecell⊖=Ereduced⊖−Eoxidised⊖
Step 3: Substitution and correct evaluation
Ecell⊖=+1.36−(−2.71)=+4.07 V
Predicting reaction feasibility using electrode potentials
To predict if a redox reaction will occur spontaneously, compare the standard reduction potentials of the two half-reactions.
The half-reaction with the more negative E⦵ will proceed in the oxidation direction, while the half-reaction with the more positive E⦵ will proceed in the reduction direction.
If the resulting E⦵cell is positive, the reaction is feasible under standard conditions. The more positive the value of E⦵cell, the more feasible the redox reaction is.
Worked example 3 - Predicting the feasibility of a redox reaction
Calculate E⦵cell for the cell comprising of MnO4- ions and Fe2+ ions, given the following standard electrode potentials.
Determine whether the reaction is spontaneous under standard conditions.
| Half reaction | E⦵ (V) |
|---|---|
| MnO4-(aq) + 8H+(aq) + 5e- ➔ Mn2+(aq) + 4H2O(l) | +1.51 |
| Fe3+(aq) + e- ➔ Fe2+(aq) | +0.77 |
Step 1: Compare standard reduction potentials
The half-reaction involving manganese has a more positive E⦵ value, indicating it proceeds in the reduction direction:
MnO4-(aq) + 8H+(aq) + 5e- ➔ Mn2+(aq) + 4H2O(l)
The half-reaction involving iron has a less positive E⦵ value, indicating it proceeds in the oxidation direction:
Fe2+(aq) ➔ Fe3+(aq) + e-
Step 2: Equation
Ecell⊖=Ereduced⊖−Eoxidised⊖
Step 3: Substitution and correct evaluation
Ecell⊖=+1.51−(+0.77)=+0.74 V
Step 4: Predict feasibility
As E⦵cell is positive, the reaction between MnO4- and Fe2+ ions is spontaneous under standard conditions.