If E^$\omicron$for the following cell is 0.36 V at 25^$\omicron$C
Pb(s) |PbSO₄(s) |SO₄²^-(aq,0.60 M) M H⁺(aq,0.70 M) |H₂(g,192.5 kPa) |Pt
How is the Nernst equation for the cell properly expressed at this temperature?

Question

Quizplus · Accepted Answer

The Nernst equation is:E = E° - (RT/nF)lnQwhere:E is the cell potentialE° is the standard cell potentialR is the gas constant (8.314 J/mol·K)T is the temperature in Kelvinn is the number of moles of electrons transferredF is the Faraday constant (96,485 C/mol)Q is the reaction quotientIn this case, n = 2, and Q = [SO42-]/[H+]2[Pb2+]. Substituting these values into the Nernst equation, we get:E = E° - (RT/2F)ln[SO42-]/[H+]2[Pb2+]At 25°C, T = 298 K. Substituting this value and the given concentrations into the equation, we get:E = E° - (0.02569 V)ln[0.60/{(0.70)2(1.90)}]E = E° - 0.02569 Vln[1.90/{(0.70)2(0.60)}]Therefore, the correct answer is A.

If E ο\omicronο for the Following Cell Is 0 ο\omicronο C Pb(s)|PbSO4(s)|SO42-(aq,0

If E^$\omicron$for the Following Cell Is 0 $\omicron$ C
Pb(s)|PbSO₄(s)|SO₄²^-(aq,0