Cell EMF E°_cell = E°_cathode − E°_anode: Exercises

This page is a self-testing ladder. Each problem sits inside a collapsible solution — try it first, then reveal. We climb five levels: L1 Recognition (just spot the right value), L2 Application (plug into the formula), L3 Analysis (reason about signs and direction), L4 Synthesis (combine EMF with Gibbs energy / Nernst), and L5 Mastery (multi-step, real design thinking).

Everything here rests on the parent formula from the Cell EMF topic:

Before we start, a tiny reminder of two words we'll use in every problem:

Standard potentials used throughout (all reduction, at 25 °C, 1 M):

Half-reaction	$E °$ (V)
$F_{2} + 2 e^{-} \to 2 F^{-}$	$+ 2.87$
$Ag^{+} + e^{-} \to Ag$	$+ 0.80$
$Fe^{3 +} + e^{-} \to Fe^{2 +}$	$+ 0.77$
$Cu^{2 +} + 2 e^{-} \to Cu$	$+ 0.34$
$2 H^{+} + 2 e^{-} \to H_{2}$	$0.00$
$Ni^{2 +} + 2 e^{-} \to Ni$	$- 0.25$
$Fe^{2 +} + 2 e^{-} \to Fe$	$- 0.44$
$Zn^{2 +} + 2 e^{-} \to Zn$	$- 0.76$
$Al^{3 +} + 3 e^{-} \to Al$	$- 1.66$
$Li^{+} + e^{-} \to Li$	$- 3.04$

The "number line" figure below is your compass for the whole page — up = more positive = wants electrons = cathode.

Figure — Cell EMF E°_cell = E°_cathode − E°_anode

Level 1 — Recognition

Goal: read the table and spot which electrode is which. No arithmetic beyond a subtraction.

Problem L1.1

Given $E ° (Ag^{+} / Ag) = + 0.80 V$ and $E ° (Cu^{2 +} / Cu) = + 0.34 V$ , which metal is the cathode in a spontaneous cell, and which is the anode?

Recall Solution L1.1

What we do: compare the two reduction potentials. Why: the half-cell that "wants electrons more" (higher $E °$ ) is where reduction wins — that is the cathode.

$+ 0.80 > + 0.34$ , so Ag is the cathode (reduction: $Ag^{+} + e^{-} \to Ag$ ).
Cu is the anode (oxidation: $Cu \to Cu^{2 +} + 2 e^{-}$ ).

Look at the accent point on figure s01: the higher dot pulls electrons in. Answer: cathode = Ag, anode = Cu.

Problem L1.2

For a Zn–Ag cell, name the electrode that loses mass as the cell runs.

Recall Solution L1.2

The anode is oxidized; solid metal dissolves into ions, so the anode loses mass. Zn has the lower $E °$ ( $- 0.76 < + 0.80$ ), so Zn is the anode. Answer: the zinc electrode loses mass.

Level 2 — Application

Goal: plug numbers into $E °_{cell} = E °_{cathode} - E °_{anode}$ correctly, minding signs.

Problem L2.1

Compute $E °_{cell}$ for the Zn–Ag cell ( $E °_{Zn} = - 0.76$ , $E °_{Ag} = + 0.80$ ).

Recall Solution L2.1

Step 1 — identify: higher $E °$ is cathode → Ag ( $+ 0.80$ ); anode → Zn ( $- 0.76$ ). Step 2 — subtract: $E °_{cell} = E °_{cathode} - E °_{anode} = 0.80 - (- 0.76) = 1.56 V$ Why the double-minus: subtracting a negative adds. Zn's low potential means it pushes electrons hard, boosting the total drive. Answer: $1.56 V$ .

Problem L2.2

Compute $E °_{cell}$ for a cell built from $Ni^{2 +} / Ni$ ( $- 0.25$ ) and $Fe^{2 +} / Fe$ ( $- 0.44$ ). Identify anode and cathode.

Recall Solution L2.2

Step 1: compare: $- 0.25 > - 0.44$ . Ni is higher, so Ni = cathode, Fe = anode. Step 2: $E °_{cell} = (- 0.25) - (- 0.44) = 0.19 V$ Positive → spontaneous as written. Answer: $0.19 V$ , Ni cathode, Fe anode.

Problem L2.3

A cell reports $E °_{cell} = + 2.71 V$ with $Li$ as the anode ( $E ° = - 3.04$ ). What is the cathode's $E °$ ?

Recall Solution L2.3

Rearrange the formula: $E °_{cathode} = E °_{cell} + E °_{anode}$ . $E °_{cathode} = 2.71 + (- 3.04) = - 0.33 V$ Answer: $E °_{cathode} = - 0.33 V$ (close to Ni).

Level 3 — Analysis

Goal: reason about direction, spontaneity, and what a sign means physically.

Problem L3.1

You wire Cu ( $+ 0.34$ ) as cathode and Ag ( $+ 0.80$ ) as anode, by choice. Compute $E °_{cell}$ for this assignment. What does the sign tell you?

Recall Solution L3.1

$E °_{cell} = E °_{cathode} - E °_{anode} = 0.34 - 0.80 = - 0.46 V$ Negative → this configuration is non-spontaneous. Nature refuses to run it; you'd need an external power source (an electrolytic cell) to force it. The spontaneous cell would instead make Ag the cathode, giving $+ 0.46 V$ . Answer: $- 0.46 V$ , non-spontaneous as assigned.

Problem L3.2

Which single ion in the table is the strongest oxidizing agent, and which neutral metal is the strongest reducing agent? Justify using $E °$ .

Recall Solution L3.2

Oxidizing agent = the species that gets reduced (it takes electrons). The strongest one has the most positive $E °$ : that's $F_{2}$ at $+ 2.87 V$ . Reducing agent = the species that gets oxidized (it gives electrons). The strongest is the reduced form of the most negative $E °$ : that's $Li$ at $- 3.04 V$ . On figure s01: strong oxidizers live at the top, strong reducers at the bottom. Answer: $F_{2}$ strongest oxidizer; Li strongest reducer.

Problem L3.3

Will $Cu$ metal spontaneously reduce $Fe^{3 +}$ to $Fe^{2 +}$ in solution? Use $E ° (Fe^{3 +} / Fe^{2 +}) = + 0.77$ and $E ° (Cu^{2 +} / Cu) = + 0.34$ .

Recall Solution L3.3

The proposed reaction: $Cu$ is oxidized (anode, $+ 0.34$ ), $Fe^{3 +}$ is reduced (cathode, $+ 0.77$ ). $E °_{cell} = 0.77 - 0.34 = + 0.43 V$ Positive → yes, spontaneous. Cu will dissolve while Fe³⁺ becomes Fe²⁺. Answer: yes, $E °_{cell} = + 0.43 V$ .

Level 4 — Synthesis

Goal: connect EMF to Gibbs free energy and the Nernst equation.

Two bridges we'll cross here:

Problem L4.1

For the Daniel cell ( $E °_{cell} = 1.10 V$ , $n = 2$ ), compute $Δ G °$ in kJ/mol.

Recall Solution L4.1

$Δ G ° = - n F E °_{cell} = - (2) (96485) (1.10) = - 212, 267 J/mol$ Convert: $\approx - 212.3 kJ/mol$ . Negative → spontaneous, consistent with the positive EMF. Answer: $\approx - 212.3 kJ/mol$ .

Problem L4.2

A Cu concentration cell has $[Cu^{2 +}]_{cathode} = 1.0 M$ and $[Cu^{2 +}]_{anode} = 0.010 M$ , with $n = 2$ and $E °_{cell} = 0$ . Find $E_{cell}$ .

Recall Solution L4.2

Reaction quotient: for $Cu_{(anode)}^{2 +} \to Cu_{(cathode)}^{2 +}$ effectively driving from dilute to concentrated, $Q = \frac{[ Cu ^{2 +} ] _{anode}}{[ Cu ^{2 +} ] _{cathode}} = \frac{0.010}{1.0} = 0.010$ . $E_{cell} = 0 - \frac{0.0592}{2} lo g (0.010) = - 0.0296 \times (- 2) = + 0.0592 V$ Positive → the dilute side is the anode; nature tries to equalize concentrations. Answer: $+ 0.0592 V$ .

Problem L4.3

Using $Δ G ° = - R T ln K$ and $Δ G ° = - n F E °_{cell}$ , find the equilibrium constant $K$ for the Daniel cell at 25 °C. (Shortcut: $lo g K = \frac{n E ° _{cell}}{0.0592}$ .)

Recall Solution L4.3

$lo g K = \frac{n E ° _{cell}}{0.0592} = \frac{2 \times 1.10}{0.0592} = 37.16$ $K = 1 0^{37.16} \approx 1.4 \times 1 0^{37}$ Enormous — the reaction goes essentially to completion. Answer: $lo g K \approx 37.2$ , $K \approx 1.4 \times 1 0^{37}$ .

Level 5 — Mastery

Goal: multi-step design and comparison problems combining everything.

Problem L5.1

You want a battery of at least $3.5 V$ from two half-cells in the table. Pick the pair with the largest $E °_{cell}$ , state which is cathode/anode, and give the voltage.

Recall Solution L5.1

Largest spread = highest minus lowest: cathode $F_{2}$ ( $+ 2.87$ ), anode $Li$ ( $- 3.04$ ). $E °_{cell} = 2.87 - (- 3.04) = 5.91 V$ This exceeds $3.5 V$ . Answer: F₂ cathode, Li anode, $5.91 V$ . (This is why lithium chemistries dominate high-voltage cells.)

Problem L5.2

An iron object corrodes when $O_{2}$ acts as cathode. Suppose instead you connect it to a metal so that iron becomes the cathode (protected). From the table, which metals could serve as the sacrificial anode for $Fe^{2 +} / Fe$ ( $E ° = - 0.44$ ), and why?

Recall Solution L5.2

To make Fe the cathode, the attached metal must be the anode, i.e. have a more negative $E °$ than $- 0.44 V$ so it oxidizes preferentially. From the table: Zn ( $- 0.76$ ), Al ( $- 1.66$ ), Li ( $- 3.04$ ) all qualify. Zn and Al are the practical choices (Li reacts with water). Check with Zn: $E °_{cell} = E °_{Fe} - E °_{Zn} = - 0.44 - (- 0.76) = + 0.32 V$ (spontaneous, Zn dissolves, Fe spared). Answer: Zn, Al (and in principle Li).

Problem L5.3

A student measures $E °_{cell} = 0.46 V$ for a Cu–Ag cell and claims $n = 2$ . Compute $Δ G °$ , then explain whether the value of $Δ G °$ changes if the real electron count is $n = 1$ for the silver half.

Recall Solution L5.3

Balance electrons: $Cu \to Cu^{2 +} + 2 e^{-}$ needs two Ag: $2 Ag^{+} + 2 e^{-} \to 2 Ag$ . So the overall balanced reaction transfers $n = 2$ electrons. $Δ G ° = - n F E °_{cell} = - (2) (96485) (0.46) = - 88, 766 J/mol \approx - 88.8 kJ/mol$ The silver half-reaction alone shows $n = 1$ , but to balance charge with Cu we double it. $E °_{cell}$ stays $0.46 V$ (intensive), yet $Δ G °$ uses the balanced $n = 2$ . Answer: $Δ G ° \approx - 88.8 kJ/mol$ , with $n = 2$ .

[!recall]- Self-Check Cloze Review

Cathode gets the more positive reduction potential

because reduction wins where electrons are most wanted.

The formula is

E °_{cell} = E °_{cathode} - E °_{anode}

A negative

E °_{cell}

means

the reaction is non-spontaneous as written (electrolytic).

Δ G °

relates to EMF by

Δ G ° = - n F E °_{cell}

E °

does not scale with

n

because

it is an intensive property (volts per unit charge).

Connections

Parent topic (Hinglish)
Standard Reduction Potentials (Table) — source of every $E °$ used here
Gibbs Free Energy and Spontaneity — the $Δ G ° = - n F E °$ bridge
Nernst Equation — non-standard concentrations (L4.2)
Galvanic vs Electrolytic Cells — sign of $E °_{cell}$ (L3.1)
Daniel Cell (Detailed Mechanism) — worked in L4.1, L4.3
Faraday's Laws of Electrolysis — where $F$ comes from