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''Easy'' Nernst potential half cell

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    Q) The standard electrode potential for the Pt Sn4+, Sn2+ half cell is +0.15 volts. What is the value for E for this half cell when [Sn4+] = 0.3 mol dm-3 and [Sn2+] 0.1 mol dm-3 (T = 298 K).

    keq= sn4+/sn2+

    My answer:

    Using E0= E - (RT/NF) x (ln keq) .... Textbook confirms this is the correct formula and order and that values are correct, but not the answer.

    E0= 0.15 - ((8.314*298 / 2*96500) x (ln (0.1/0.3)) = 0.164

    Textbook answer:

    Using the Nernst equation: E = E° – RT/nF´ln([Sn2+]/[Sn4+] We use [Sn2+]/[Sn4+] ratherthan [Sn4+]/[Sn2+], because the half cell is written as Sn4+ + e-→ Sn2+ and products are divided byreactants in the Nernst equation (and others related to it).

    E = (+0.15 V) – (8.31J K-1mol-1´ 298 K)/(2´ 96500JV-1mol-1) ´ ln(0.1 M/0.3 M)E= (+0.15 V) – (-0.69 V) = +0.84 V Hereit doesn’t matter what units of concentration you use as they cancel out butyou must use M where this isn’t the case and it would be wisest to do so as amatter of course.

    Anyone know where I'm going wrong?
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