Homework Answers
Cell potential can be calculated using Nernst equation :
Ecell = E°cell - 0.059/n log([Fe2+]/[Ag+]2)
Here, E°cell = E°cathode - E°anode
At cathode, reduction takes place. In given reaction reduction of Ag+ takes place (because it's oxidation state decreases from +1 to 0) hence Ag+ is at cathode whereas Fe is at anode. Thus,
E°cell = 0.80 -(-044) = 1.24 V
Number of electrons lost or gained, n = 2
(Because oxidation state of Fe changes from 0 to +2 thus 2 electrons are lost)
[Fe2+] = 1.50
[Ag+] = 0.050
Substituting values, we get :
Ecell = 1.24 - 0.059/2 log(1.50/(0.05)2)
= 1.24 - 0.0295×log(600)
= 1.24 - 0.0295×2.778
= 1.24 - 0.082
= 1.16 V
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would be helpful if you could explain it in words too. thanks
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Which combination of metals, if used to create an
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could someone please answer and explain the questions please!! #’s 7,8,9,13,14,17,18,19,20 please and thank you What...
could someone please answer and explain the questions please!!
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please and thank you
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Using the table below:
19. Three combinations of metals are listed below, which
combination would produce the largest voltage if they were used to
construct an electrochemical cell?
Copper (Cu) with zinc (Zn)
Lead (Pb) with zinc (Zn)
Lead (Pb) with cadmium (Cd)
Liu lur the reaction between Zn and Cu2+ ions is 1.1030 V, we can use the known value for the half-cell potential for zinc to determine the half-cell potential for copper: Zn(s) → Zn2+(aq) + 2e +...
Using the information in the table:
Which combination of metals, if used to create an
electrochemical cell, would produce the largest voltage?
Liu lur the reaction between Zn and Cu2+ ions is 1.1030 V, we can use the known value for the half-cell potential for zinc to determine the half-cell potential for copper: Zn(s) → Zn2+(aq) + 2e + Cu2+(aq) + 2e → Cu(s) Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) E half-cell = 0.7628 V Eºhalf-cell =...
could someone please answer and explain the questions please!!
#’s 7,8,9,13,14,17,18,19,20
please and thank you
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