At 63.0 degree C , what is the maximum value of Q needed to produce a non-negati

Question

At 63.0 degree C , what is the maximum value of Q needed to produce a non-negative E value for the reaction SO42- (aq) + 4H+(aq) + - 2Br - (aq) Br2(aq) + - S02(g) + - 2H2O(1) In other words, what is Q when E = 0 at this temperature? Express your answer numerically using two significant figures. Express your answer numerically using two significant figures.

Explanation / Answer

sulfur is getting reduced and Br is getting oxidised reaction at cathode :- SO42- + 4?H+ + 2?e- SO2(aq) + 2?H2O .....Eo = 0.17 V and reaction at anode :- 2Br- -------> Br2 + 2e- ...Eo = 1.08 V Eo(cell) = Eo(cathode) - Eo(anode) = 0.17 - 1.08 = -0.91 V using Nernst equation ... E(cell) = Eo(cell) - 2.303RT/nF log Q given values are :- E(cell) = 0 ( according to question) Eo(cell) = -0.91 V R = 8.314 J/K/mole T = 63 + 273 = 336 K n = no.of electrons involved = 2 F = 96500 C 0 = -0.91 - 2.303 X 8.314 X 336 / 2 X 96500 log Q 0.91 = -6433.439/193000 log Q 0.91 = -0.033 log Q log Q = -0.91/0.033 = 27.576 taking antilog .... Q = 10^27.576 = 3.767 X 10^27 please check the calculations

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