Question

Write net equation for the spontaneous redox reaction and determine the standard cell potential that results when each of the following redox couples are connected inan electrochemical cell.

Instructions:

Enter all substances in the order listed at the top of the column.
Use a carot to indicate a superscript, but do nothing for subscripts.
Use a hyphen + greater than (->) for yields.
Click on the eye symbol to check your formatting.

Couples Rxn: Ox(1) + Red(2) + other -> Red(1) + Ox(2) + other
Cu/Cu2+ and Cr3+/Cr
Hg2+/Hg and Fe3+/ Fe2+
H1+/H2 and Ni2+/Ni
Cr/Cr3+ and Ti2+/Ti

Include Eo (V) for each please!

Answer 1

Your provided link to table of standard reduction potentials is not accessible to me. It popped up a message of "We're sorry. You do not have access tothis page."
To solve this kind of problem, always remember to balance the charge. When you have the standard reduction potentials given, life becomes quite easy. Youfirst find the two relevant equations of half reaction. Time each equation with appropriate integer parameters to make the coefficient of the charge (e-)the same. Then you reverse the direction of the reaction with lower potential (also change the sign of that potential at the same time), and adding the tworesulted equations together (also adding the potentials at the same time). Finally cancel the identical charges from both sides of the reactionequation.

a. H1+/H2 and Cu2+/Cu E° =
2H+ + Cu = H2 + Cu(2+)

b. Cr/Cr3+ and Ag1+/Ag E° =
3Ag+ + Cr = 3Ag + Cr(3+)

These are the net ionic equations. You may add spectators to each equation. For example:
a) H2SO4 + Cu = H2 + CuSO4
b) 3AgNO3 + Cr = 3Ag + Cr(NO3)3

Answer 2

a)
Oxidation : 3Cu ---> 3Cu2+ + 6e -0.15 *3 = 0.45 (E oxidation)
Reduction 2Cr3+ + 6e ---> 2Cr -2*.74 = 1.48 (E reduction)
Ecell = -(.45+1.48) = -2.93 V

b)
Oxidation 3Hg ---> 3Hg2+ + 6e -.79*2 = -1.58
Reduction 2Fe3+ +2e ----> 2Fe -.44*3 = -1.32
Ecell = -3.80 V

c)