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natima [27]
3 years ago
13

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
Chemistry
2 answers:
cupoosta [38]3 years ago
6 0

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.

Natalija [7]3 years ago
4 0
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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Explanation:

Step 1: Data given

Volume of the Ca(OH)2 = 100.0 mL = 0.100 L

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Step 2: The balanced equation

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Step 3: Calculate molarity of Ca(OH) 2

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⇒with b = the coefficient of HBr = 2

⇒with Va = the volume of Ca(OH)2 = 0.100 L

⇒with ca = the concentration of Ca(OH)2 = TO BE DETERMINED

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⇒with Vb = the volume of HBr = 0.0365 L

⇒with Cb = the concentration of HBr = 5.00 * 10^-2 = 0.05 M

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Ca = (0.0365*0.05) / 0.200

Ca = 0.009125 M

Step 4: Calculate moles HBr

Moles HBr = concentration HBr * volume HBr

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Moles HBr = 0.001825 moles

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