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Find the electrochemical potential of a galvanic cell using Zn/Zn² and Cu/Cu² electrodes at 25°C if [Zn² ] = 0.01 M and [Cu² ] = 1.0 M.
n = 2 (number of electrons transferred). Q = [Zn² ] / [Cu² ] = 0.01 M / 1.0 M = 0.01 E_cell = E°_cell - (0.0592/n)logQ E_cell = 1.10 V - (0.0592/2)log(0.01) E_cell = 1.10 V - (0.0296)(-2) E_cell = 1.10 V + 0.0592 V E_cell = 1.1592 V. The electrochemical potential of the galvanic cell is approximately 1.16 V.
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Asked 29 days ago|2/6/2026 12:16:26 PM
Updated 29 days ago|2/6/2026 12:32:30 PM
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Added 29 days ago|2/6/2026 12:32:30 PM
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n = 2 (number of electrons transferred).

Q = [Zn² ] / [Cu² ] = 0.01 M / 1.0 M = 0.01
E_cell = E°_cell - (0.0592/n)logQ
E_cell = 1.10 V - (0.0592/2)log(0.01)
E_cell = 1.10 V - (0.0296)(-2)
E_cell = 1.10 V + 0.0592 V
E_cell = 1.1592 V.

The electrochemical potential of the galvanic cell is approximately 1.16 V.
Added 29 days ago|2/6/2026 12:32:10 PM
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