EO = Eº 1.500V = 3+ / Au is more and hence it will reduce. Eº 1.500V Cathode Eº 0.440V = Anode - 0.440V Fe²+ (aq)

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Chapter19: Electrochemistry
Section: Chapter Questions
Problem 19.86QP: What is the cell potential of the following cell at 25C? Ni(s)Ni2+(1.0M)Sn2(1.5104M)Sn(s)
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please answer questions d, e, f
12:43
Step2
b)
Calculations:
E⁰
Eº = 1.500V
3+
/ Au is more and hence it will reduce.
EO = 1.500V Cathode
Eº = 0.440V
Anode
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?
√x
- 0.440V
Fe²+ (aq)
94V
Do
8
(
Transcribed Image Text:12:43 Step2 b) Calculations: E⁰ Eº = 1.500V 3+ / Au is more and hence it will reduce. EO = 1.500V Cathode Eº = 0.440V Anode Was this solution helpful? ? √x - 0.440V Fe²+ (aq) 94V Do 8 (
12:43
3. A voltaic cell was made from Au/Au" and Fe/Fe: From the given standard reduction potential,
determine: (a) overall cell reaction; (b) value of n; (c) emf of the cell; (d) value of AG; (e) Would
you expect K to be > or < 1? (f) Give the shorthand notation of the cell diagram. Write the answers
for (a) and (f) below and draw a box around it.
Fe (aq) + 2e
Fe(s) -0.440
Au" (aq) + 3e
Au (s) +1.500
(
Transcribed Image Text:12:43 3. A voltaic cell was made from Au/Au" and Fe/Fe: From the given standard reduction potential, determine: (a) overall cell reaction; (b) value of n; (c) emf of the cell; (d) value of AG; (e) Would you expect K to be > or < 1? (f) Give the shorthand notation of the cell diagram. Write the answers for (a) and (f) below and draw a box around it. Fe (aq) + 2e Fe(s) -0.440 Au" (aq) + 3e Au (s) +1.500 (
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