Given the following two half-reactions. write the overall reactions in the direc

Question

Given the following two half-reactions. write the overall reactions in the direction in which it is spontaneity calculate the standard cell potential. Pb^2+ (aq) + 2 e^- rightarrow Pb(s) E degree = -0.126 V Cu^2+ (aq) + 2 e^- rightarrow Cu(s) E degree = + 0.337 V Pb^2+ (aq) + Cu(s) rightarrow Pb(s) + Cu^2+(aq) E degree _cell = +0.211 V Pb(s) + Cu^2+ (aq) rightarrow Pb^2+ (aq) + Cu(s) E degree _cell = +0.926 V Pb^2+ (aq) + Cu(s) rightarrow Pb(s) + Cu^2+ (aq) E degree _cell = +0.463 V Pb(s) + Cu^2+ (aq) rightarrow Pb^2+ (aq) + Cu(s) E degree _cell = -0.211 V Pb(s) + Cu^2+ (aq) rightarrow Pb^2+ (aq) + Cu(s) E degree _cell = +0.463 V Cu^2+ is reduced to Cu(s) at an electrode. If a current of 1.25 ampere is passed for 72 hours, what mass of deposited at the electrode? Assume 100% current efficiency. 3.0 times 10^-2 g 1.1 times 10^2 g 2.90 times 10^-2 g 2.1 times 10^-2 g 4.3 times 10^-2 g Complete the following fission reaction:^1 _0 n +^235 _92 U rightarrow^236 _92 U rightarrow^141 _92 Ba + _ + 3^1 _0 n^92 _36 Kr^95 _35 Br^95 _95 Am^92 _36 U^95 _36 Kr Consider the following half-reactions: Cu^2+(aq) + 2 e^- rightarrow Cu(s) E degree = +0.34 V Sn^2+(aq) + 2 e^- rightarrow Sn(s) E degree = -0.14 V Fe^2+(aq) + 2 e^- rightarrow Fe(s) E degree = _0.44 y Al^3+(aq) + 3 e^- rightarrow Al(s) E degree =-1.66 V Mg^2+(aq) + 2 e^- rightarrow Mg(s) E degree = - 2.37 V Which of the above metals or metal ions will oxidize Fe(s)? Sn(s) and Al^3+(aq) Cu(s) and Sn(s) Cu^2+(aq) and Sn^2+(aq) Al^3+(aq) and Mg^2+(aq) Al(s) and Mg(s) A sample of magnesmm-28 is found to have an activity of 750 disintegrations per hour (dph). After 6.0 activity, has decreased to 620 dph. What is the rate constant for the decay of magnesium-28? 0.032 hr^-1 0.81 hr^-1 0.014 hr^-1 0.14 hr^-1 5.0 hr^-1

Explanation / Answer

5)
The one with greater Eo value will be reduced
So Cu2+ should be reduced to Cu
Pb should be oxidised to Pb2+
Eocell = Eo cat - EO anode
= 0.337 - (-0.126)
= 0.463 V
Answer: E

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