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🧪 NEET Chemistry❓ Question

E°cell from standard reduction potentials: why not multiply by the coefficients? And does Cu reduce Fe³⁺?

Ishita GoyalDropper Sample

29 Sept 2026 · 1 min read

I like having a clean template for every type of question, and electrochemistry is the only chapter where I can't make one. Let me show where my template breaks.

Given:

Fe3++e−→Fe2+E∘=+0.77 V\text{Fe}^{3+} + e^- \to \text{Fe}^{2+}\quad E^\circ = +0.77\ \text{V} Cu2++2e−→CuE∘=+0.34 V\text{Cu}^{2+} + 2e^- \to \text{Cu}\quad E^\circ = +0.34\ \text{V}

Question: will copper metal reduce Fe³⁺ to Fe²⁺ under standard conditions?

My template says: write the cell as Cu∣Cu2+∥Fe3+,Fe2+∣Pt\text{Cu} | \text{Cu}^{2+} \| \text{Fe}^{3+}, \text{Fe}^{2+} | \text{Pt}. Ecell∘=Ecathode∘−Eanode∘E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}. But the balanced reaction needs two Fe³⁺ for one Cu, so I multiplied the Fe half-reaction by 2 and got Ecell∘=2(0.77)−0.34=1.20E^\circ_{cell} = 2(0.77) - 0.34 = 1.20 V. The answer key says 0.43 V.

So (1) why can't I multiply E° by 2? (2) Also, I always forget which electrode is positive and which is negative in a galvanic cell. (3) What is the job of the salt bridge, exactly? (4) How do I connect E° with ΔG?

A tidy explanation, if someone has one!

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6 answers & replies

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  • [+]Tushar PawarDropperSample· 21d ago

    Electrode potential is an intensive property, like temperature or density, it doesn't depend on how much you have. Doubling the reaction doesn't double the voltage; it doubles the energy (ΔG\Delta G), not the potential. Think two identical 1.5 V cells side by side in parallel: still 1.5 V, the 'energy per coulomb' stays the same, only the number of coulombs you can push doubles.

    So Ecell∘=0.77−0.34=0.43E^\circ_{cell} = 0.77 - 0.34 = 0.43 V. Positive, so spontaneous: yes Cu reduces Fe³⁺.

    • [+]Ishita GoyalDropperSample· 20d agoAuthor

      ahh E° as intensive, that was my gap. I was mixing it up with ΔG. Updating my template: subtract, never multiply, then ΔG = −nFE. Thank you!

  • [+]Rohan Sharma11thSample· 20d ago

    if you must multiply, multiply both then, E°cell = 2(0.77) − 2(0.34) = 0.86 V, that is what the reaction needs for 2 electrons. consistent no? at least both halves are scaled the same way, so it feels more fair than scaling just one. I don't have a book to check, going by logic

    • [+]Neel Chatterjee12thSample· 20d ago

      No, 0.86 V would be wrong too. E∘E^\circ is never multiplied by stoichiometric coefficients at all. The thing that changes with coefficients is nn (the number of electrons) and ΔG∘\Delta G^\circ:

      ΔG∘=−nFEcell∘\Delta G^\circ = -nFE^\circ_{cell}

      Here n=2n = 2, so ΔG∘=−2×96500×0.43≈−83,000\Delta G^\circ = -2\times96500\times0.43 \approx -83{,}000 J = −83 kJ per mole of Cu. If you wrote the equation for one electron, nn would be 1 and ΔG\Delta G would be half, but E° is still 0.43 V.

  • [+]Kavya NaikDropperSample· 20d ago

    Signs: in a galvanic cell the anode is where oxidation happens and it's the negative electrode (electrons are produced there, they flow through the wire to the cathode). The cathode, where reduction occurs, is positive. Remember 'An Ox, Red Cat'. In an electrolytic cell the signs are flipped because the battery imposes them (anode positive), but the definition (anode = oxidation) stays.

    For your cell: Cu is the anode (−), Pt in the Fe³⁺/Fe²⁺ half cell is the cathode (+).

  • [+]Vivek Mishra12thSample· 20d ago

    Salt bridge: it has an inert electrolyte (like KCl or KNO₃ in agar). As Cu oxidises, positive charge builds up in the anode half cell; at the cathode, cations are used up so negative charge builds up. The bridge lets ions move so each half cell stays neutral, and completes the circuit. Without it, the cell stops almost immediately. Also, it stops the two solutions from mixing directly.

    For the Daniell cell: E∘=0.34−(−0.76)=1.10E^\circ = 0.34 - (-0.76) = 1.10 V and ΔG∘=−2×96500×1.10≈−212\Delta G^\circ = -2\times 96500\times1.10 \approx -212 kJ/mol.

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