give the standard line (cell) notation for a voltaic cell built on these half reactions.

Question

Consider the following half-reactions and their standard reduction potentials then give the standard line (cell) notation for a voltaic cell built on these half reactions.

Mn²⁺(aq) + 2 e^- <=> Mn(s) E° = -1.18 V

Fe³⁺(aq) + 3 e^- <=> Fe(s) E° = -0.036 V

The correct answer Mn (s) | Mn ²⁺(aq, 1.0 M) || Fe³⁺(aq, 1.0 M) | Fe(s)

Answer 1

Reaction must have a positive E.

Therefore the first half reaction must be flipped and the E becomes positive.

Therefore the Mn(s) is a reactant and the Mn+2 is the product.

And Fe+3 is the reactant and Fe(s) is the product.

Standard line notation has all solutions at 1.0 M concentrations.

Answer 2

Mn²⁺(aq) + 2 e^- <=> Mn(s) E° = -1.18 V

Fe³⁺(aq) + 3 e^- <=> Fe(s) E° = -0.036 V

From the standard reduction potentials, we know that Fe³⁺ will be reduced and therefore Mn²⁺ will be oxidized. We can write these reactions as follows:

Fe³⁺ + 3e- ==> Fe(s) Eº = -0.036 V (Cathode)

Mn(s) ===> Mn²⁺ + 2e- Eº = -1.18 V (Anode)

To write the standard line cell notation, we put the anode on the left, then salt bride, then cathode on the right.

We use 1 M solutions as standard, so we would write...

Mn(s) | Mn²⁺(aq, 1M) || Fe³⁺(aq, 1 M) | Fe(s)