Question

For the reaction Ni2+(aq) + 2Fe2+(aq) ? Ni(s) + 2Fe3+(aq), the standard cell potential E°cell is
A. +2.81 V.
B. +1.02 V.
C. +0.52 V.
D. -1.02 V.
E. -2.81 V.

Please show your steps, thank you :)

Answer 1

E0= E anode - E cathode
E0= Fe+2 - Ni+2
E0=0.77 -(-0.25)
E0=1.02 V

Answer 2

sol
First find which is oxidized and reduced

Oxidation: 2Fe2+--> 2Fe3+

or 2Fe2+ --> 2Fe3+ + 2e-


Reduced: Ni2+ + 2e- --> Ni



now find the E potential for each (these can be found in the appendix of your
textbook or online):

Oxidized= 0.77V

Reduced= -0.23V



Now, subtract the smaller V from the larger. 0.77-(-0.23)= 1 V = E cell

Answer 3

B. +1.02 V.

Answer 4

B. +1.02 V.

Answer 5

D. -1.02 V.

First find which is oxidized and reduced
Oxidation: 2Fe2+--> 2Fe3+
or 2Fe2+ --> 2Fe3+ + 2e-
Reduced: Ni2+ + 2e- --> Ni

now find the E potential for each (these can be found in the appendix of your textbook or online):
Oxidized= 0.77V
Reduced= -0.23V

Answer 6

B. +1.02 V.


Oxidation: 2Fe2+--> 2Fe3+
or 2Fe2+ --> 2Fe3+ + 2e-
Reduced: Ni2+ + 2e- --> Ni

now find the E potential for each (these can be found in the appendix of your textbook or online):
Oxidized= 0.77V
Reduced= -0.23V

Now, subtract the smaller V from the larger. 0.77-(-0.23)= 1 V = E cell