Topic 16 · page 9 of 19

Predicting feasibility

A reaction is thermodynamically feasible under standard conditions if its Ecell⊖E^\ominus_\text{cell} is positive. Find the species reduced, the species oxidised, subtract, and state the verdict.

15 min read16.10

Read first: Cell emf and cell diagrams

A positive emf means feasible 16.10

In every cell so far a positive Ecell⊖E^\ominus_\text{cell} meant that electrons flowed on their own. The same subtraction tells you whether a reaction in a single beaker will go, without building any cell. Think of zinc and copper(II) sulfate: Ecell⊖E^\ominus_\text{cell} = +1.10 V, and zinc really does coat itself with copper.

MethodPredicting whether a reaction is feasible

  1. Write the two half-equations and their E⊖E^\ominus values.
  2. Decide which couple is reduced (it contains the oxidising agent) and which is oxidised.
  3. Ecell⊖E^\ominus_\text{cell} = E⊖E^\ominus(reduced) − E⊖E^\ominus(oxidised).
  4. Positive: feasible. Negative: not feasible, but the reverse reaction is.

Worked exampleDoes iron(III) oxidise iodide?

E⊖E^\ominus(Fe³⁺/Fe²⁺) = +0.77 V and E⊖E^\ominus(I₂/I⁻) = +0.54 V.

  1. Fe³⁺ is reduced to Fe²⁺. I⁻ is oxidised to I₂.
  2. Ecell⊖E^\ominus_\text{cell} = +0.77 − (+0.54) = +0.23 V.
  3. Positive, so the reaction is feasible:
2 FeX3+(aq)+2 IX−(aq)→2 FeX2+(aq)+IX2(aq) \begin{aligned} &\ce{2Fe^3+(aq) + 2I-(aq)} \\ &\ce{-> 2Fe^2+(aq) + I2(aq)} \end{aligned}

Worked exampleDoes iron(III) oxidise bromide?

E⊖E^\ominus(Fe³⁺/Fe²⁺) = +0.77 V and E⊖E^\ominus(Br₂/Br⁻) = +1.09 V.

  1. Fe³⁺ is reduced to Fe²⁺. Br⁻ is oxidised to Br₂.
  2. Ecell⊖E^\ominus_\text{cell} = +0.77 − (+1.09) = −0.32 V.
  3. Negative, so the reaction is not feasible. The reverse is: bromine oxidises iron(II) to iron(III).
Two ladders. Left, Fe3+ with I minus: the arrow runs from the reduced form higher up to the oxidised form lower down, E cell plus 0.23 volts, feasible. Right, Fe3+ with Br minus: the arrow would run upwards, E cell minus 0.32 volts, not feasible.
The same test on a ladder. A feasible reaction runs from a reduced form higher up to an oxidised form lower down.

A shortcut for checking 16.10

In the series with the most negative at the top, a reaction is feasible when an oxidised form lower down reacts with a reduced form higher up. On the ladder this is a diagonal arrow going from the upper right to the lower left.

Use it to check your answer, but in the exam show the full method.

Worked exampleDoes chlorine oxidise iron(II)?

E⊖E^\ominus(Cl₂/Cl⁻) = +1.36 V and E⊖E^\ominus(Fe³⁺/Fe²⁺) = +0.77 V.

  1. Cl₂ is reduced to Cl⁻. Fe²⁺ is oxidised to Fe³⁺.
  2. Ecell⊖E^\ominus_\text{cell} = +1.36 − (+0.77) = +0.59 V. Feasible.
ClX2(aq)+2 FeX2+(aq)→2 ClX−(aq)+2 FeX3+(aq) \begin{aligned} &\ce{Cl2(aq) + 2Fe^2+(aq)} \\ &\ce{-> 2Cl-(aq) + 2Fe^3+(aq)} \end{aligned}

Metals and acids 16.10

For a metal reacting with dilute acid, the hydrogen ion is always the species reduced, with E⊖E^\ominus(H⁺/H₂) = 0.00 V.

Worked exampleWhich metals dissolve in dilute acid?

  1. Zinc: Ecell⊖E^\ominus_\text{cell} = 0.00 − (−0.76) = +0.76 V. Feasible: zinc fizzes in acid.
  2. Copper: Ecell⊖E^\ominus_\text{cell} = 0.00 − (+0.34) = −0.34 V. Not feasible: copper does not dissolve in dilute acid.

This matches what you see in the lab.

Check yourself

  1. Will acidified MnO₄⁻ oxidise Cl⁻ to Cl₂? Use E⊖E^\ominus(MnO₄⁻/Mn²⁺) = +1.51 V and E⊖E^\ominus(Cl₂/Cl⁻) = +1.36 V.
  2. Show that iron metal reduces Fe³⁺(aq), and write the equation. Use E⊖E^\ominus(Fe²⁺/Fe) = −0.44 V and E⊖E^\ominus(Fe³⁺/Fe²⁺) = +0.77 V.
  3. Which of Cl₂, Br₂ and I₂ will oxidise Fe²⁺ to Fe³⁺? Use +1.36, +1.09 and +0.54 V for the halogen couples and +0.77 V for Fe³⁺/Fe²⁺.
  4. Will copper reduce Zn²⁺(aq) to zinc? Use −0.76 V for Zn²⁺/Zn and +0.34 V for Cu²⁺/Cu.
  5. Will Ag⁺ oxidise Fe²⁺? Use +0.80 V for Ag⁺/Ag and +0.77 V for Fe³⁺/Fe²⁺.
  6. Explain, in terms of Ecell⊖E^\ominus_\text{cell}, why hydrochloric acid is not used to acidify manganate(VII) in a titration.

Answers cover these until you have tried the questions

  1. MnO₄⁻ is reduced, Cl⁻ is oxidised. Ecell⊖E^\ominus_\text{cell} = +1.51 − (+1.36) = +0.15 V, positive, so yes, it is feasible.
  2. Fe³⁺ is reduced and Fe is oxidised. Ecell⊖E^\ominus_\text{cell} = +0.77 − (−0.44) = +1.21 V, feasible. Fe(s)+2 FeX3+(aq)→3 FeX2+(aq)\ce{Fe(s) + 2Fe^3+(aq) -> 3Fe^2+(aq)}
  3. Each halogen is reduced and Fe²⁺ is oxidised. Cl₂: +1.36 − 0.77 = +0.59 V (yes). Br₂: +1.09 − 0.77 = +0.32 V (yes). I₂: +0.54 − 0.77 = −0.23 V (no). Cl₂ and Br₂ will, I₂ will not.
  4. Zn²⁺ is reduced and Cu is oxidised. Ecell⊖E^\ominus_\text{cell} = −0.76 − (+0.34) = −1.10 V. Negative, so not feasible.
  5. Ag⁺ is reduced and Fe²⁺ is oxidised. Ecell⊖E^\ominus_\text{cell} = +0.80 − (+0.77) = +0.03 V. Positive, so feasible, but the value is very small (see page 10).
  6. MnO₄⁻ would oxidise Cl⁻ (Ecell⊖E^\ominus_\text{cell} = +0.15 V, positive). Some of the manganate(VII) would be used up on chloride instead of the iron(II), so the titre would be too large.