Chemistry · Emf of a Galvanic cell and its measurement, Nernst equation and its applications, relationship between cell potential and Gibbs' energy change
Calculate \left|\boldsymbol{H}_{2}(\boldsymbol{g})_{1 a t m}\right| \boldsymbol{
Calculate \( \boldsymbol{E}_{\text {cell}} \) \( \boldsymbol{P t}(\boldsymbol{s})\left|\boldsymbol{H}_{2}(\boldsymbol{g})_{1 a t m}\right| \boldsymbol{H} \boldsymbol{A}_{\left(\boldsymbol{K}_{a}=10^{-7}\right)} \mathbf{1} \boldsymbol{M} \| \boldsymbol{H} \boldsymbol{B}_{(\boldsymbol{k}} \)
- A. 0 .06
- B. 0.03 V
- C. \( 0.04 v \)
- D. \( 0.05 \mathrm{v} \)
Step-by-step solution
For the cell with two hydrogen electrodes, E_cell = 0.05916 (pH_anode - pH_cathode). The left side: HA (1 M, Ka=10^-7) gives [H+] = sqrt(10^-7) = 10^-3.5 M, pH = 3.5. The right side is likely HB with Ka=10^-5 (common pairing) giving [H+] = sqrt(10^-5) = 10^-2.5 M, pH = 2.5. Then E_cell = 0.05916 (3.5 - 2.5) ≈ 0.059 V ≈ 0.06 V.
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