Cell EMF E°_cell = E°_cathode − E°_anode: Question bank

Reminders you should carry into every item:

$E °_{cell} = E °_{cathode} - E °_{anode}$ , and both $E °$ values are looked up as reduction potentials from the Standard Reduction Potentials (Table).
$Δ G ° = - n F E °_{cell}$ — sign of $E °_{cell}$ tells you spontaneity (see Gibbs Free Energy and Spontaneity).
" $E °$ " (with the little circle) means standard conditions: 1 M, 1 atm, 25°C. Drop the circle and you need the Nernst Equation.

Formula Two symbols you'll need for the edge cases — defined right here

This page is self-contained: whenever a question drops the little circle and talks about non-standard conditions, it uses these two things, so pin them down now.

The reaction quotient $Q$ is "how far along" the reaction currently sits. Take the balanced reaction and write: $Q = \frac{[ products ] ^{coeff}}{[ reactants ] ^{coeff}}$ where each concentration is raised to its coefficient in the balanced equation (pure solids and liquids count as $1$ ). At the exact moment concentrations are all $1 M$ , every bracket is $1$ , so $Q = 1$ .

The Nernst equation turns the standard voltage $E °_{cell}$ into the actual voltage $E_{cell}$ once concentrations drift away from $1 M$ (at $25° C$ ): $E_{cell} = E °_{cell} - \frac{0.0592}{n} lo g Q$ Read it as: start from the standard value, then subtract a correction that grows when there is more product ( $Q > 1$ , correction positive, voltage drops) and adds voltage when there is more reactant ( $Q < 1$ , $lo g Q < 0$ , correction negative). Here $E_{cell}$ (no circle) is the real measured voltage and $n$ is moles of electrons — same $n$ as in $Δ G ° = - n F E °_{cell}$ .

True or false — justify

E °_{cell}

is found by adding the two half-cell reduction potentials.

False — voltage is a difference in electron "height", not a sum; you subtract anode from cathode, like a hill 100 m minus a hill 50 m gives a 50 m drop, not 150 m (see the electron-hill figure above).

A negative

E °_{cell}

means you did the arithmetic wrong.

False — a negative value is a real, useful answer: it says that particular cathode/anode assignment is non-spontaneous, so the reaction actually runs the other way (or needs external power).

Reversing a half-reaction (reduction → oxidation) flips the sign of its

E °

before you plug it in.

False — you never manually flip

E °

in this formula; the subtraction already accounts for the anode running backwards, so you always insert the tabulated reduction value.

E °_{cell}

depends on how many moles of electrons (

n

) are transferred.

False —

E °

is an intensive property (electron "pressure per charge"), so doubling the reaction doesn't change it; only

Δ G ° = - n F E °

scales with

n

If you double every coefficient in the balanced cell reaction,

E °_{cell}

doubles.

False — potential is per-electron and stays the same; you just moved twice as much charge, so

Δ G °

doubles while

E °

is unchanged.

A cell with

E °_{cell} = 0

can never produce any voltage.

False under non-standard conditions — a concentration cell has

E °_{cell} = 0

yet the Nernst equation

E_{cell} = - \frac{0.0592}{n} lo g Q

gives a real

E_{cell} \neq = 0

once concentrations differ.

The electrode made of the more reactive metal is always the anode in a galvanic cell.

True for a galvanic cell — the more reactive (more easily oxidised, lower reduction potential) metal loses electrons, making it the anode; but only because the cell is spontaneous, not by definition of "reactive".

In a galvanic cell the cathode is the positive terminal; in an electrolytic cell the cathode is the negative terminal.

True — the chemistry label (cathode = reduction) never changes, but the sign flips because in electrolysis an external supply pushes electrons into the cathode, making it the negative terminal.

Standard reduction potentials are absolute voltages you could measure with one electrode alone.

False — a single half-cell has no measurable voltage; all

E °

values are relative to the standard hydrogen electrode defined as exactly

0 V

Spot the error

"Zn/Cu cell:

E °_{cell} = E °_{Zn} - E °_{Cu} = - 0.76 - 0.34 = - 1.10

V, so it's non-spontaneous."

Error is swapping cathode and anode — Cu has the higher

E °

so Cu is the cathode; correct is

0.34 - (- 0.76) = + 1.10

V, spontaneous.

"At the anode, reduction potential is

- 0.76

V, so I flip it to

+ 0.76

V, then

E °_{cell} = 0.34 + 0.76 = 1.10

V."

The number is right but the method is a trap — flipping the sign and adding double-counts the direction; use the tabulated

- 0.76

and subtract:

0.34 - (- 0.76)

"Cu–Ag cell: cathode Cu (+0.34), anode Ag (+0.80),

E °_{cell} = 0.34 - 0.80 = - 0.46

V — so this cell doesn't work."

The assignment is wrong, not the cell — Ag has the higher

E °

so Ag is really the cathode; the true galvanic cell gives

0.80 - 0.34 = + 0.46

V and works fine.

"Because

E °_{cell} = 0.46

V, the reaction releases

0.46

J of energy."

Volts are energy per coulomb, not joules; total energy is

Δ G ° = - n F E °

, so you must multiply by

n

and Faraday's constant

F = 96, 485

C/mol.

"The concentration cell has identical electrodes, so

Q = 1

always and it can never generate voltage."

Q = 1

only when the two concentrations are equal; recall

Q = \frac{[ products ]}{[ reactants ]}

, so make them unequal and

Q \neq = 1

, giving a nonzero

E_{cell}

through the Nernst equation.

E °_{cell}

was positive, so I'll reverse the anode reaction's

E °

sign to double-check."

No second sign-flip is needed or allowed — the subtraction already handled the reversal; flipping again would give the wrong magnitude.

Why questions

Why do we always tabulate potentials as reduction potentials rather than oxidation potentials?

A single, consistent convention lets one table serve both electrodes — the subtraction formula converts a reduction value into anode behaviour automatically, so we never need a second table.

Why does the electrode with the higher reduction potential become the cathode in a galvanic cell?

A higher reduction potential means that species "wants" electrons more strongly, so electrons flow toward it and reduction (the cathode process) happens there.

Why is

E °_{cell}

independent of

n

while

Δ G °

is not?

E °

measures energy per unit charge (an intensive ratio), whereas

Δ G ° = - n F E °

measures total energy, which grows with the amount of charge

n

moved.

Why does a positive

E °_{cell}

guarantee spontaneity?

From

Δ G ° = - n F E °_{cell}

, a positive

E °

makes

Δ G °

negative, and a negative free-energy change is the definition of a spontaneous process.

Why can't we just measure the voltage of one half-cell directly?

Voltage is always a difference between two points; a lone electrode has nothing to compare against, so we reference everything to the standard hydrogen electrode set to

0 V

Why does an electrolytic cell have

E °_{cell} < 0

for its forced direction, yet still run?

Left alone it would run backwards; an external power source supplies more than that missing voltage, pushing electrons uphill against the natural tendency (see Galvanic vs Electrolytic Cells).

Why does the Daniel cell give exactly

+ 1.10

V and not, say, the sum

0.34 + 0.76 = 1.10

that also happens to look equal?

The coincidence is because

E °_{anode}

is negative; the real operation is

0.34 - (- 0.76)

, a subtraction — the "looks like addition" is just the double-negative, not a rule you can generalise.

Edge cases

What is

E °_{cell}

when both half-cells are chemically identical at 1 M?

Exactly

0 V

— cathode and anode potentials cancel, so at standard conditions there is no driving force.

Two identical Cu electrodes, but the cathode side has more concentrated

Cu^{2 +}

than the anode side — is there voltage?

Yes —

E °_{cell} = 0

, but with

Q = \frac{[ Cu ^{2 +} ] _{anode}}{[ Cu ^{2 +} ] _{cathode}} < 1

we get

lo g Q < 0

, and the Nernst equation

E_{cell} = - \frac{0.0592}{n} lo g Q

gives

E_{cell} > 0

; this is how concentration cells and biological ion sensors work.

If you compute

E °_{cell}

and get exactly

0 V

for two different metals, what does it mean?

The two half-cells sit at the same reduction potential, so neither has a driving edge — at standard conditions no net reaction and no current, a genuine equilibrium.

What happens to spontaneity if

E °_{cell}

is a tiny positive number like

+ 0.01

Still spontaneous (

Δ G ° < 0

), just barely — the reaction proceeds but with a very small driving force, so in practice it may be slow or easily reversed by small concentration changes.

Can two half-reactions with both positive

E °

still form a working galvanic cell?

Yes — spontaneity depends only on which is larger; the higher one is cathode, the lower one is forced to run as anode, and

E °_{cell}

= (higher − lower) > 0.

If you physically swap which beaker you call the cathode, does the measured cell voltage change sign?

The magnitude is fixed by chemistry; only your bookkeeping sign flips — nature still drives electrons the same real direction regardless of your labels.

Recall One-line self-test

Cathode minus anode, both as reduction values, never re-flip a sign — say it once more. ::: $E °_{cell} = E °_{cathode} - E °_{anode}$ , both looked up as reductions, positive means spontaneous.

Connections

2.7.03 Cell EMF E°_cell = E°_cathode − E°_anode (Hinglish) — the parent topic
Standard Reduction Potentials (Table) — the values these traps quote
Gibbs Free Energy and Spontaneity — why sign of $E °_{cell}$ decides direction
Nernst Equation — for every non-standard / concentration edge case above
Galvanic vs Electrolytic Cells — the cathode-sign trap
Daniel Cell (Detailed Mechanism) — the $+ 1.10$ V worked example
Faraday's Laws of Electrolysis — where $n$ and $F$ turn voltage into energy