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Answer: \(E^0_{cell} = +0.02\) V

Zn is the anode (left) and Cr the cathode (right):

\(E^0_{cell} = E^0_{cathode} - E^0_{anode} = E^0_{Cr} - E^0_{Zn} = (-0.74) - (-0.76) = +0.02\) V

(If the actual potential at the given concentrations is wanted: cell reaction 3Zn + 2Cr3+ → 3Zn2+ + 2Cr, \(n = 6\); \(E_{cell} = 0.02 - \dfrac{0.0592}{6}\log\dfrac{(0.08)^3}{(0.1)^2} = 0.02 - 0.00987\times(-1.291) \approx 0.033\) V.)

Doubtly AI · 29 Oct 2026 AI-verified · Nov 2026 Report
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