Suppose a galvanic cell was constructed at 25 Suppose a galvanic cell was constr

Question

Suppose a galvanic cell was constructed at 25 Suppose a galvanic cell was constructed at 25 Suppose a galvanic cell was constructed at 25 Suppose a galvanic cell was constructed at 25 Suppose a galvanic cell was constructed at 25 degreeC using a Cu/Cu2+ half-cell (in which the molar concentration of Cu2+ was 1.00 M) and a hydrogen electrode having a partial pressure of H2 equal to 1 atm. The hydrogen electrode dipped into a solution of unknown hydrogen ion concentration, and the two half-cells were connected by a salt bridge. The precise value of EdegreeCu2+ is +0.3419 V. Choose the equation for the pH of the solution with the unknown hydrogen ion concentration, expressed in terms of Ecell and Edegreecell. If the pH of the solution were 5.68, what would be the observed potential of the cell?

Explanation / Answer

(a) Cell reaction is: Cu2+ + H2 => Cu + 2 H+

Moles of electrons transferred n = 2

Temperature T = 25 deg C = 298.15 K

Faraday constant F = 96485 C/mol

Molar gas constant R = 8.314 J/mol.K


Nernst equation:

E(cell) = Eo(cell) - 2.303RT/nF log([H+]^2/[Cu2+].P(H2))

= Eo(cell) - 8.314 x 2.303 x 298.15/(2 x 96485) x log([H+]^2/1.00 x 1.00)

= Eo(cell) - 0.0296 x log([H+])^2

= Eo(cell) - 0.0296 x 2 x log[H+]

= Eo(cell) - 0.0592 x log[H+]

= Eo(cell) + 0.0592 x (-log[H+])

= Eo(cell) + 0.0592 x pH


Rearranging gives:

pH = (E(cell) - Eo(cell))/0.0592 V


(b) E(cell) = Eo(cell) + 0.0592 x pH

= 0.3419 + 0.0592 x 5.68

= 0.6782 V = 0.678 V


(c) pH = (E(cell) - Eo(cell))/0.0592

= (0.822 - 0.3419)/0.0592

= 8.11

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