The equation below represents the overall cell reaction for an electrochemical cell that undergoes a spontaneous reaction. Fe(s)
+ Pb2+(aq) mc011-1.jpg Fe2+(aq) + Pb(s) Emc011-2.jpgcell = –0.31 V Which reaction is the spontaneous reduction reaction? Fe(s) mc011-3.jpg Fe2+(aq) + 2e– Emc011-4.jpg = –0.44 V Fe(s) mc011-5.jpg Fe2+(aq) + 2e– Emc011-6.jpg = +0.44 V Pb2+(aq) + 2e– mc011-7.jpg Pb(s) Emc011-8.jpg = –0.13 V Pb2+(aq) + 2e– mc011-9.jpg Pb(s) Emc011-10.jpg = +0.13 V
2 answers:
Answer:
Pb2+(aq) + 2e– ------------>Pb(s) = –0.13 V
Explanation:
Given the question asked, we have to examine the reduction potentials of Fe2+ and Pb2+.
Fe2+= -0.44V
Pb2+= -0.13V
Recall that the moire negative the reduction potential of a species, the less likely it is to participate in spontaneous reduction reactions.
From the data before us, Pb2+ will undergo a spontaneous reduction reaction as shown in the answer.
Its b Fe(s) <span> Fe</span>2+(aq) + 2e– <span><span> </span>E</span><span> = </span><span>+0.44 V</span>
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