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Question
The cell in which the following reaction occurs:
`2Fe^(3+) (aq) + 2I^(-) (aq) ---> 2Fe^(2+) (aq) + I_2 (s)` has `E_"cell"^@` = 0.236 V at 298 K. Calculate the standard Gibbs energy of the cell reaction. (Given : 1 F = 96,500 C mol−1)
Solution
Here n = 2, `E_"cell"^@ = 0.236 V`, T = 298 K
We know that
`triangle G^@ = -nFE_"cell"^@`
= −2 × 96500 × 0.236
= −45548 J mol−1
= −45.548 kJ mol−1
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RELATED QUESTIONS
The cell in which the following reactions occurs: \[\ce{2Fe^{3+}_{( aq)} + 2I^-_{( aq)} -> 2Fe^{2+}_{( aq)} + I2_{(s)}}\] has \[\ce{E^Θ_{cell}}\] = 0.236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
Following reaction takes place in the cell:
`Zn(s) + Ag_2O(s)+H_2O(l) -> Zn^{2+}(aq) + 2Ag (s) + 2OH^- (aq)`
Calculate `Delta_r G^0` of the reaction
[Given ; `E^0_(Zn^{2+}//Zn)` = -0.76V
`E_((Zn^{2+}//Zn)) = 0.76V`
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When a reversible work is done by a galvanic cell, the relationship between Gibbs free energy and emf of the cell is given by the relationship as ____________.
Match the items of Column I and Column II.
| Column I | Column II |
| (i) Lechlanche cell | (a) cell reaction \[\ce{2H2 + O2 -> 2H2O}\] |
| (ii) Ni–Cd cell | (b) does not involve any ion in solution and is used in hearing aids. |
| (iii) Fuel cell | (c) rechargeable |
| (iv) Mercury cell | (d) reaction at anode, \[\ce{Zn -> Zn^{2+} + 2e^{-}}\] |
| (e) converts energy of combustion into electrical energy |
What is the relationship between Gibbs free energy of the cell reaction in a galvanic cell and the emf of the cell? When will the maximum work be obtained from a galvanic cell?
The correct order of first ionization energy for the following elements, Hydrogen (H), Helium (He), Lithium (Li), Borm (B) is:
Calculate the ΔrG0 and log Kc, for the given reaction at 298 K:
\[\ce{Ni_{(s)} + 2Ag^+_{( aq)} <=> Ni^{2+}_{( aq)} + 2Ag_{(s)}}\]
Given: `"E"_("Ni"^(2+)//"Ni")^0` = −0.25 V, `"E"_("Ag"^+//"Ag")^0` = +0.80 V, 1F = 96500 C mol−1.
