Spontaneity from E°_cell and ΔG = −nFE: Exercises

This page is a self-test ladder for the parent note Spontaneity from E°_cell and ΔG = −nFE. Every problem hides its worked solution in a collapsible callout — read the problem, try it on paper, THEN reveal.

Before we start, let us pin down the three numbers we will reuse every single time, so no symbol is ever a mystery.

Below, "cathode" always means the electrode where reduction (electron gain) happens, "anode" where oxidation (electron loss) happens — see Galvanic vs Electrolytic Cells.

A reference table used across problems (all are standard reduction potentials, $E^{\circ}$ in volts):

Half-reaction	$E^{\circ}$ (V)
$F_{2} + 2 e^{-} \to 2 F^{-}$	$+ 2.87$
$A g^{+} + e^{-} \to A g$	$+ 0.80$
$C u^{2 +} + 2 e^{-} \to C u$	$+ 0.34$
$2 H^{+} + 2 e^{-} \to H_{2}$	$0.00$
$N i^{2 +} + 2 e^{-} \to N i$	$- 0.25$
$F e^{2 +} + 2 e^{-} \to F e$	$- 0.44$
$Z n^{2 +} + 2 e^{-} \to Z n$	$- 0.76$
$A l^{3 +} + 3 e^{-} \to A l$	$- 1.66$

Figure — Spontaneity from E°_cell and ΔG = −nFE

The figure above is the mental picture behind every exercise: the higher a couple sits, the more it "wants" the electrons. The gap between the cathode line and anode line is $E_{ce l l}^{\circ}$ .

Level 1 — Recognition

Problem 1.1

State the sign of $Δ G^{\circ}$ when $E_{ce l l}^{\circ} = + 0.50 V$ , and say whether the reaction is spontaneous.

Recall Solution 1.1

$Δ G^{\circ} = - n F E_{ce l l}^{\circ}$ . Here $E_{ce l l}^{\circ}$ is positive, so $- n F \times (positive)$ is negative. $Δ G^{\circ} < 0 \Rightarrow$ spontaneous. (We didn't even need $n$ — the sign alone decides it.)

Problem 1.2

For $E_{ce l l}^{\circ} = - 0.30 V$ , is the forward reaction spontaneous?

Recall Solution 1.2

Negative $E_{ce l l}^{\circ}$ → $Δ G^{\circ} = - n F (- 0.30) = + something > 0$ . $Δ G^{\circ} > 0 \Rightarrow$ non-spontaneous (needs external energy, e.g. a battery — that's an electrolytic cell).

Problem 1.3

In the cell $Z n (s) + C u^{2 +} \to Z n^{2 +} + C u (s)$ , which electrode is the cathode?

Recall Solution 1.3

Cathode = where reduction happens = where electrons are gained. $C u^{2 +}$ gains 2 electrons to become $C u$ , so copper is the cathode. Zinc is oxidised → anode.

Level 2 — Application

Problem 2.1

Compute $E_{ce l l}^{\circ}$ for $Z n (s) + C u^{2 +} \to Z n^{2 +} + C u (s)$ using the table.

Recall Solution 2.1

Cathode (reduction) = $C u^{2 +} / C u = + 0.34 V$ . Anode (oxidation) = $Z n^{2 +} / Z n = - 0.76 V$ . $E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ} = (+ 0.34) - (- 0.76) = + 1.10 V$ Positive → spontaneous, as the boulder-downhill picture promised.

Problem 2.2

Find $Δ G^{\circ}$ (in kJ) for the Zn/Cu cell above. Take $n = 2$ , $F = 96485 C/mol$ .

Recall Solution 2.2

$Δ G^{\circ} = - n F E_{ce l l}^{\circ} = - (2) (96485) (1.10) = - 212267 J$ Convert to kJ (divide by 1000): $Δ G^{\circ} \approx - 212.3 kJ$ per mole of reaction. Negative → spontaneous.

Problem 2.3

For the silver–copper cell $C u (s) + 2 A g^{+} \to C u^{2 +} + 2 A g (s)$ , find $E_{ce l l}^{\circ}$ .

Recall Solution 2.3

$A g^{+}$ is reduced (cathode, $+ 0.80 V$ ); $C u$ is oxidised (anode, $+ 0.34 V$ ). $E_{ce l l}^{\circ} = 0.80 - 0.34 = + 0.46 V$ Note: even though we wrote $2 A g^{+}$ , we do not multiply the voltage. $E^{\circ}$ is an intensive property (voltage per charge), so stoichiometric coefficients never scale it. Only $n$ (and thus $Δ G$ ) scales.

Level 3 — Analysis

Problem 3.1

For $2 A l (s) + 3 C u^{2 +} \to 2 A l^{3 +} + 3 C u (s)$ , determine $n$ carefully, then compute $E_{ce l l}^{\circ}$ and $Δ G^{\circ}$ (kJ).

Recall Solution 3.1

Count electrons. Each Al loses 3 e⁻ → $2 Al$ lose $6$ e⁻. Each $C u^{2 +}$ gains 2 e⁻ → $3 C u^{2 +}$ gain $6$ e⁻. They match: $n = 6$ . Voltage: cathode $C u^{2 +} / C u = + 0.34$ , anode $A l^{3 +} / A l = - 1.66$ . $E_{ce l l}^{\circ} = 0.34 - (- 1.66) = + 2.00 V$ Free energy: $Δ G^{\circ} = - (6) (96485) (2.00) = - 1157820 J \approx - 1158 kJ$

Problem 3.2

The reaction $C u (s) + Z n^{2 +} \to C u^{2 +} + Z n (s)$ is the reverse of Problem 2.1's. Find $E_{ce l l}^{\circ}$ and $Δ G^{\circ}$ without redoing all the work.

Recall Solution 3.2

Reversing a reaction flips the sign of both $E^{\circ}$ and $Δ G^{\circ}$ . $E_{ce l l}^{\circ} = - 1.10 V, Δ G^{\circ} = + 212.3 kJ$ Positive $Δ G^{\circ}$ → non-spontaneous (you'd force it in an electrolytic cell).

Problem 3.3

A student computes $Δ G^{\circ} = - 88.8 kJ$ for a 2-electron cell. What was $E_{ce l l}^{\circ}$ ?

Recall Solution 3.3

Rearrange $Δ G^{\circ} = - n F E_{ce l l}^{\circ}$ : $E_{ce l l}^{\circ} = \frac{- Δ G ^{\circ}}{n F} = \frac{- ( - 88 800 J )}{( 2 ) ( 96 485 )} = \frac{88 800}{192 970} \approx + 0.46 V$ (Convert kJ → J first: $- 88.8 kJ = - 88800 J$ .)

Level 4 — Synthesis

Problem 4.1

Link to equilibrium: given $Δ G^{\circ} = - R T ln K$ and $Δ G^{\circ} = - n F E_{ce l l}^{\circ}$ , derive the relation between $E_{ce l l}^{\circ}$ and $K$ , then find $K$ for a cell with $E_{ce l l}^{\circ} = + 0.46 V$ , $n = 2$ , $T = 298 K$ . Use $R = 8.314 J/(mol\cdotK)$ .

Recall Solution 4.1

Set the two expressions for $Δ G^{\circ}$ equal: $- n F E_{ce l l}^{\circ} = - R T ln K \Rightarrow ln K = \frac{n F E _{ce l l}^{\circ}}{R T}$ Plug in: $ln K = \frac{( 2 ) ( 96 485 ) ( 0.46 )}{( 8.314 ) ( 298 )} = \frac{88 766.2}{2477.6} \approx 35.83$ . $K = e^{35.83} \approx 3.6 \times 1 0^{15}$ A large positive $E^{\circ}$ → huge $K$ → reaction goes essentially to completion. See Relationship between K_eq and ΔG°.

Problem 4.2

You are told $E_{ce l l}^{\circ} = + 1.10 V$ but the cell runs at non-standard concentrations. Which vault tool corrects the voltage, and does diluting the product ion $Z n^{2 +}$ make the cell more or less spontaneous?

Recall Solution 4.2

The correction tool is the Nernst Equation: $E = E^{\circ} - \frac{R T}{n F} ln Q$ , where $Q$ is the reaction quotient. For $Z n + C u^{2 +} \to Z n^{2 +} + C u$ , $Q = \frac{[ Z n ^{2 +} ]}{[ C u ^{2 +} ]}$ . Diluting the product $Z n^{2 +}$ lowers $Q$ , so $ln Q$ becomes more negative, so $- \frac{R T}{n F} ln Q$ is positive → $E$ rises above $E^{\circ}$ . The cell becomes more spontaneous. (Le Chatelier agrees: removing product pushes forward.)

Level 5 — Mastery

Problem 5.1

An electrolysis (forced, non-spontaneous) deposits copper: $C u^{2 +} + 2 e^{-} \to C u$ . If the cell's spontaneous version had $Δ G^{\circ} = - 63.6 kJ$ ( $n = 2$ ), what minimum voltage must an external supply exceed to reverse it and plate copper? Then, using Faraday's Laws of Electrolysis, how many grams of Cu are deposited by $2.0 A$ flowing for $30$ minutes? ( $M_{C u} = 63.5 g/mol$ .)

Recall Solution 5.1

Part A — voltage to reverse. First recover the spontaneous $E_{ce l l}^{\circ}$ : $E_{ce l l}^{\circ} = \frac{- Δ G ^{\circ}}{n F} = \frac{63 600}{( 2 ) ( 96 485 )} \approx + 0.330 V$ To drive it backward you must supply a voltage exceeding this in the opposite sense, i.e. more than $0.33 V$ (ignoring overpotential). Below that, nothing plates; above it, deposition proceeds.

Part B — mass deposited. Charge passed: $q = I t = (2.0 A) (30 \times 60 s) = 3600 C$ . Moles of electrons: $\frac{q}{F} = \frac{3600}{96 485} = 0.03731 mol e^{-}$ . Each $C u$ needs 2 e⁻, so moles of Cu $= 0.03731/2 = 0.018655 mol$ . Mass $= 0.018655 \times 63.5 \approx 1.18 g$ .

Problem 5.2

Rank the spontaneity (most to least, by $Δ G^{\circ}$ ) of these three cells, without computing $Δ G^{\circ}$ for cell C:

A: $Z n + C u^{2 +}$ ( $E^{\circ} = + 1.10$ , $n = 2$ )
B: $2 A l + 3 C u^{2 +}$ ( $E^{\circ} = + 2.00$ , $n = 6$ )
C: $C u + 2 A g^{+}$ ( $E^{\circ} = + 0.46$ , $n = 2$ )

Recall Solution 5.2

Spontaneity by $Δ G^{\circ}$ depends on both $n$ and $E^{\circ}$ (their product), not voltage alone.

A: $n E^{\circ} = 2 \times 1.10 = 2.20$
B: $n E^{\circ} = 6 \times 2.00 = 12.0$
C: $n E^{\circ} = 2 \times 0.46 = 0.92$ Since $Δ G^{\circ} = - n F E^{\circ}$ and $F$ is a shared positive constant, the most negative $Δ G^{\circ}$ has the largest $n E^{\circ}$ . Ranking (most spontaneous → least): $B > A > C$ . $Δ G^{\circ}$ values: B $= - 6 (96485) (2.00) = - 1158 kJ$ , A $= - 212.3 kJ$ , C $= - 88.8 kJ$ .

Recall Quick self-check clozes

$E_{ce l l}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$ ; spontaneous means == $E^{\circ} > 0$ and $Δ G^{\circ} < 0$ ==. The bridge equation ::: $Δ G^{\circ} = - n F E_{ce l l}^{\circ}$ What links $E^{\circ}$ to the equilibrium constant $K$ ? ::: $ln K = \frac{n F E _{ce l l}^{\circ}}{R T}$ Does doubling stoichiometric coefficients change $E_{ce l l}^{\circ}$ ? ::: No — voltage is intensive; only $n$ (and $Δ G$ ) scale.