Given that,

Cu2+(aq) + 2e– → Cu(s) E° = +0.34 V

NO3–(aq) + 4H+(aq) + 3e– → NO(g) + 2H2O(l) E° = +0.96 V

calculate the value of E°cell, in volts, for the reaction,

3Cu(s) +2NO3–(aq) +8H+(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O(l)

Question

Given that,

Cu2+(aq) + 2e– → Cu(s)                                           E° = +0.34 V

NO3–(aq) + 4H+(aq) + 3e– → NO(g) + 2H2O(l)     E° = +0.96 V

calculate the value of E°cell, in volts, for the reaction,

3Cu(s) +2NO3–(aq) +8H+(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O(l)

BlackTom AI · Accepted Answer

First, restating the problem helps ensure we’re evaluating the right species and potentials. We’re given two half-reactions with their standard reduction potentials and asked for the E°cell of the overall reaction: 3 Cu(s) + 2 NO3– + 8 H+ → 3 Cu2+ + 2 NO + 4 H2O.

Option by option analysis:
Option A: –0.62 V. To get the overall E°cell, we combine the cathode (reduction) and anode (oxidation) potentials as E°cell = E°cathode − E°anode. The reduction half-reaction for NO3– to NO has E° = +0.96 V. The oxidation half-reaction for Cu to Cu2+ has E°oxidation = −0.34 V (since the corresponding reduction potential is +0.34 V). Therefore, E°cell = 0.96 − 0.34 = +0.62 V, not –0.62 V. So this option contradicts the correct sign and value.
Option B: +0.62 V. This value matches the proper combination: the cathodic reduction (NO3– to NO) uses +0.96 V, the anodic oxidation (Cu to Cu2+) effectively contributes −0.34 V when using the E°cell = E°cathode − E°anode formula, giving 0.96 − 0.34 = +0.62 V. The stoichiometry (3 Cu and 2 NO3– with the required H+ and H2O balanced) supports using these two half-rereactions, so +0.62 V is consistent.
Option C: –0.90 V. This would imply a very unfavorable overall energy, which contradicts the standard potentials provided, since we add a relatively strong oxidizing ability (NO3– reduction at +0.96 V) with copper’s oxidation (−0.34 V) to yield a positive cell potential. The math does not support such a negative result.
Option D: +0.90 V. Although positive, this is not equal to the calculated difference between the two given half-reactions. The correct subtraction is 0.96 − 0.34 = 0.62, not 0.90, so this option overestimates the cell potential.

Additional reasoning tied to balancing electrons: in the overall reaction, Cu(s) is oxidized by losing 6 electrons (3 Cu → 3 Cu2+ + 6 e−), while NO3– is reduced by gaining 6 electrons (2 NO3– + 8 H+ + 6 e− → 2 NO + 4 H2O). The electrons balance, validating that we should combine E°cathode = +0.96 V with E°anode (as a reduction potential) = +0.34 V, leading to E°cell = 0.96 − 0.34 = +0.62 V. The sign and magnitude align with standard electrochemical conventions for a spontaneous reaction under standard conditions.

Given that, Cu2+(aq) + 2e– → Cu(s) E° = +0.34 V NO3–(aq) + 4H+(aq) + 3e– → NO(g) + 2H2O(l) E° = +0.96 V calculate the value of E°cell, in volts, for the reaction, 3Cu(s) +2NO3–(aq) +8H+(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O(l)单项选择题

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