Cell potential, the Nernst correction, and copper on the cathode
SCH4U Grade 12 Chemistry · Electrochemistry
A technician builds a copper–zinc cell from tabulated half-cells, +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 is 0.0100 mol/L and the copper half-cell 0.400 mol/L, giving Q = 0.0250, at 25.0 °C. A separate plating bath then runs 1.50 A for 25.0 min through a copper solution. Find the standard cell potential, the actual potential, the charge passed, and the mass of copper deposited, taking copper as 63.55 g/mol.
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.
Carried onward at full precision, not this rounded figure.
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.
Carried onward at full precision, not this rounded figure.
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.
Carried onward at full precision, not this rounded figure.
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.
Carried onward at full precision, not this rounded figure.
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.