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Gibbs energy change of a cell reaction from its EMF

This type of question, asked 3 times: 26 Jun 2022, 27 Jun 2022, 23 Jan 2025. All 3 questions below, oldest first.

Try each one first, then open its answer.

  1. Question 1

    26 Jun 2022

    147
    Cu(s)+Sn2+(0.001M)→Cu2+(0.01M)+Sn(s)Cu(s) +Sn^{2 +}(0.001M) \rightarrow Cu^{2 +}(0.01M) + Sn(s) The Gibbs free energy change for the above reaction at 298 K298\text{ }K is x×10−1 kJ mol−1x \times10^{- 1}\text{ }kJ{\text{ }mol}^{- 1}. The value of xx is ____\_\_\_\_. [nearest integer] [Given: ECu2+/Cu⊖=0.34 V;ESn2+/Sn⊖=−0.14 V;F=96500Cmol−1E_{Cu^{2 +}/Cu}^{\ominus}= 0.34\text{ }V;E_{Sn^{2 +}/Sn}^{\ominus}= - 0.14\text{ }V;F = 96500Cmol^{- 1} ]

  2. Question 2

    27 Jun 2022

    147
    For the reaction taking place in the cell: PtPt (s) ∣H2\mid H_{2} (g) ∣H+\mid H^{+}(aq) ∥Ag+\parallel Ag^{+}(aq) ∣Ag(s)\mid Ag(s) ECell o=+0.5332 VE_{\text{Cell~}}^{o}= + 0.5332\text{ }V. The value of ΔfG0\Delta_{f}G^{0} is ____ kJ mol−1\_\_\_\_\ kJ\ mol^{- 1}. (in nearest integer)

  3. Question 3

    23 Jan 2025

    102
    Standard electrode potentials for a few half cells are mentioned below:
    ECu2+/Cu∘=0.34 V,EZn2+/Zn∘=−0.76 VEAg+/Ag∘=0.80 V,EMg2+/Mg∘=−2.37 V\begin{matrix} & E_{Cu^{2 +}/Cu}^{\circ}= 0.34\text{ }V,E_{Zn^{2 +}/Zn}^{\circ}= - 0.76\text{ }V \\ & E_{Ag^{+}/Ag}^{\circ}= 0.80\text{ }V,E_{Mg^{2 +}/Mg}^{\circ}= - 2.37\text{ }V \end{matrix}
    Which one of the following cells gives the most negative value of ΔG∘\Delta G^{\circ} ?

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