Cell potential, the Nernst correction, and copper on the cathode

SCH4U Grade 12 Chemistry · Electrochemistry

At one end of the electrochemistry bench a technician assembles a copper–zinc cell: two beakers joined by a salt bridge, a polished zinc strip in one and a copper strip in the other, and a high-impedance voltmeter across the terminals. The tabulated half-cells give +0.342 V for Cu²⁺/Cu and −0.762 V for Zn²⁺/Zn, but the beakers are not at standard concentrations: the zinc half-cell was made up at 0.0100 mol/L and the copper half-cell at 0.400 mol/L, giving Q = 0.0250, with the water bath holding everything at 25.0 °C. At the other end of the bench a separate plating bath is wired through an ammeter to a DC supply, a weighed cathode hanging in a copper solution. The supply is trimmed to hold 1.50 A and the run is timed at 25.0 min, the deposit reddening visibly on the cathode as the minutes pass. Find the standard cell potential, the actual potential at these concentrations, the charge passed, and the mass of copper deposited, taking copper as 63.55 g/mol.

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given

E°_Cu = 0.342 V — Cu²⁺/Cu reduction potential
E°_Zn = -0.762 V — Zn²⁺/Zn reduction potential
Q = 0.025 — — Reaction quotient, [Zn²⁺]/[Cu²⁺]
T = 25 °C — Cell temperature
I = 1.5 A — Plating-bath current
t = 25 min — Plating time
M = 63.55 g/mol — Molar mass of copper

Determine

(a)the standard cell potential
(b)the actual potential at these concentrations
(c)the charge the plating bath passes
(d)the mass of copper deposited

Step 1 of 4(a) · solve for Standard cell potential

Copper is reduced, so it is the cathode; zinc is oxidised, so it is the anode. Subtract the anode's tabulated reduction potential — the minus sign in the formula already handles the reversal.

Rearranged for E°cell

E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}

E°cathode

E°anode

Your values, in your units

E^{\circ}_{\text{cell}} = \left( 0.342\ \text{V} \right) - \left( -0.762\ \text{V} \right)

Answer

E^{\circ}_{\text{cell}} = 1.104\ \text{V}

in

Carried onward at full precision, not this rounded figure.

Open the Standard Cell Potential from Half-Cells solver →

Step 2 of 4(b) · solve for Cell potential

Tabulated potentials assume every solution is 1 mol/L. Here the products are scarce and the reactants plentiful, so Q < 1 and the cell reads high — 1.151 V instead of 1.104 V.

Rearranged for E

E = E^{\circ} - \frac{RT}{nF}\ln Q

E°1.104 Vcarried from step 1

n

Q

T

Your values, in your units

E = \left( 1.104\ \text{V} \right) - \frac{\left( 8.31446\ \text{J/(mol}{\cdot}\text{K)} \right) \, \left( 25\ ^{\circ}\text{C} \right)}{\left( 2 \right) \left( 96{,}485.3\ \text{C/mol} \right)}\ln\left(\left( 0.025 \right)\right)

Answer

E = 1.1514\ \text{V}

in

Carried onward at full precision, not this rounded figure.

Open the Nernst Equation solver →

Step 3 of 4(c) · solve for Charge

The plating bath is a different question with a different currency. Voltage says whether the reaction runs; charge says how much of it runs, and charge is current multiplied by time.

Rearranged for Q

Q = I t

I

t

Your values, in your units

Q = \left( 1.5\ \text{A} \right) \, \left( 25\ \text{min} \right)

Converted to base units

Q = \left( 1.5\ \text{A} \right) \, \left( 1{,}500\ \text{s} \right)

Answer

Q = 2.25\ \text{kC}

in

Carried onward at full precision, not this rounded figure.

Open the Electric Charge (Q = It) solver →

Step 4 of 4(d) · solve for Mass deposited or dissolved

Faraday's law converts coulombs into grams: divide by F for moles of electrons, by n = 2 for moles of copper, then multiply by the molar mass.

Rearranged for m

m = \frac{Q M}{n F}

Q2.25 kCcarried from step 3

M

n

Your values, in your units

m = \frac{\left( 2{,}250\ \text{C} \right) \times \left( 63.55\ \text{g/mol} \right)}{\left( 2 \right) \times \left( 96{,}485.3\ \text{C/mol} \right)}

Answer

m = 740.98\ \text{mg}

in

Carried onward at full precision, not this rounded figure.

Open the Faraday's Law of Electrolysis (m = QM/nF) solver →

Why this order

This chain deliberately runs in two halves, because electrochemistry answers two different questions with two different quantities and students constantly fuse them. Steps 1 and 2 are about potential — an intensive quantity, energy per coulomb, which tells you whether electrons will move and how hard they are pushed. Steps 3 and 4 are about charge — an extensive quantity, which tells you how many electrons actually went and therefore how much metal appeared. A higher voltage does not plate more copper. Only more coulombs do that, which is why the answer from step 2 correctly feeds nothing at all.

Three sign traps live in step 1. Do not negate the anode value before subtracting: 0.342 − (−0.762) = 1.104 V, and flipping it twice cancels the cell. Do not scale a half-cell potential by its coefficient — potential is per coulomb and is not extensive, so balancing the electrons never changes it. And in step 2, Q is products over reactants, [Zn²⁺]/[Cu²⁺]; written upside down it moves the voltage the wrong way by twice the correction. The correction itself is small, about 47 mV, and that smallness is worth internalising: the Nernst factor at 25 °C is 59 mV per decade divided by n, so even a fortyfold concentration mismatch shifts a cell less than a twentieth of a volt.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.