Using the relationship given in part 1, calculate Ecell if [Cu2+] = 1.453 M and [Cr3+] = 0.00176 M, assuming that the temperature remains at 25°C. Give your answer to two places after the decimal.

1.05 V

Explanation:

Now recall Nernst equation;

Ecell= E°cell - 0.0592/n log [Red]/[Ox]

Now we look at the values of E°cathode and E°anode

E°cathode= 0.34 V

E°anode = -0.74 V

E°cell= 0.34-(-0.74)

E°cell= 1.08 V

[Red] = 1.453 M

[Ox] = 0.00176 M

n= 6

Ecell= E°cell - 0.0592/n log [Red]/[Ox]

Ecell= 1.08 - 0.0592/6 × log[1.453]/[0.00176]

Ecell= 1.08 - 0.0592/6 × 2.917

Ecell= 1.08 - 0.02878

Ecell= 1.05 V

i think it is d. e because the next one is igneous.