If E $\circ$ for the following cell is 0.36 V at 25 $\circ$ 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

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The Nernst equation for the cell 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 transferred in the cell reactionF is the Faraday constant (96,485 C/mol)Q is the reaction quotientIn this case, the cell reaction is:Pb(s) + SO₄²^-(aq) + 2H⁺(aq) → PbSO₄(s) + H₂(g)The standard cell potential is 0.36 V. The temperature is 25°C, which is 298 K. The number of moles of electrons transferred is 2. The reaction quotient is:Q = [PbSO₄(s)]/[Pb(s)][SO₄²^-(aq)]²[H⁺(aq)]²Substituting these values into the Nernst equation, we get:E = 0.36 - (0.02569 V)ln[192.5/{(0.70)²(0.60)}]Therefore, the correct answer is B.

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

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