Consider the following E^ values: E^_Fe^3+/Fe^2+ = 0.77\ V E^_Sn^2+/Sn = -0.14\ V Under standard conditions the potential for the reaction: Sn(s) + 2F - Chemistry MCQ AIEEE 2004 2026 | ExamTest.live

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Consider the following E^{\circ} values: E_{Fe^{3 +} / Fe^{2 +}}^{\circ} = 0.77 V E_{Sn^{2 +} / Sn}^{\circ} = - 0.14 V Under standard conditions the potential for the reaction: Sn (s) + 2 Fe^{3 +} (a q) \to 2 Fe^{2 +} (a q) + Sn^{2 +} (a q) is:

Solution & Step-by-step Explanation

In the reaction Sn (s) + 2 Fe^{3 +} (a q) \to 2 Fe^{2 +} (a q) + Sn^{2 +} (a q) : Sn is oxidized to Sn^{2 +} (Anode). Fe^{3 +} is reduced to Fe^{2 +} (Cathode). E_{ce ll}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ} . E_{ce ll}^{\circ} = E_{Fe^{3 +} / Fe^{2 +}}^{\circ} - E_{Sn^{2 +} / Sn}^{\circ} . E_{ce ll}^{\circ} = 0.77 - (- 0.14) = 0.77 + 0.14 = 0.91 V .

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Consider the following E^{\circ} values: E_{Fe^{3 +} / Fe^{2 +}}^{\circ} = 0.77 V E_{Sn^{2 +} / Sn}^{\circ} = - 0.14 V Under standard conditions the potential for the reaction: Sn (s) + 2 Fe^{3 +} (a q) \to 2 Fe^{2 +} (a q) + Sn^{2 +} (a q) is:

1.68 V

0.63 V

0.91 V

1.40 V

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