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Question

A hydrogen electrode in a solution with PH_₂ = 2 atm and [H⁺] = 0.002 M operates at 298 K. What is its potential? (Given: E°H⁺/H_₂ = 0.00 V )

Options

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Explanation

E = E° - (0.059/2) log (PH_₂/[H⁺]²) . E = 0 - 0.0295 log (2/0.000004) = 0 - 0.0295 × 5.699 = -0.1681 V .