The standard cell potential (E^0 cell) for the voltaic cell based on the reactio

Question

The standard cell potential (E^0 cell) for the voltaic cell based on the reaction below is_V. Sn^2+ (aq) + 2Fe^3+ (aq) right arrow 2Fe^2+ (aq) + Sn^4+ (aq) +0.617 +1.39 +1.21 -0.46 +0.46 The peroxydisulfate ion (S_2 O_8^2-) reacts with the iodide ion in aqueous solution via the reaction: S_2O_8^2- (aq) + 3I^- right arrow 2SO_4 (aq) + I3^- (aq) An aqueous solution containing 0.050 M of S_2O_8^2- ion and 0.072 M of I^- is -prepared, and the progress of the reaction folio measuring [I^-]. The data obtained is given in the table below. The concentration of S_2O_8^2- remaining at 1600 s is_M. 0.014 0.029 0.043 0.036 0.064

Explanation / Answer

Solution :-

5) Sn^2+(aq) +2Fe^3+(aq) ------ > 2Fe^2+(aq) + Sn^4+(aq)

Fe^3+ is reduced to Fe^2+ on cathode

and Sn^2+ is oxidized to Sn^4+ on anode

Eo cell = Eo cathode – Eo anode

           = 0.771 V – (0.151 V)

           = +0.62 V

So the correct answer is option is A) +0.617 V

Q6) concentration of [I-] changed from 0 to 1600 sec from 0.072 M to 0.029 M

Change in concentration = 0.072 M – 0.029 M = 0.043 M

Mole ratio of the [I-] is 3 :1 with S2O8^2-

So the moles of the S2O8^2- reacted = 0.043 M * S2O8^2- / 3 I^- = 0.0143 M

So the concentration remaining = 0.050 M – 0.0143 M = 0.036 M

So the correct answer is option D.

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