Question

When Hg2+ concentration is 6.35x10^-4 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 1.499V. What is the Cr3+ concentration?

3H^2+ (aq) + 2Cr(s)= 3Hg(l) + 2Cr^3+(aq)

Answer 1

10.5 × 10^5 M

Step-by-step explanation:

E°cell = E°cathode - E°anode

E°cell = 0.85 - (-0.74) = 1.59 V

From Nernst's equation;

Ecell = E°cell - 0.0592/n log Q

1.499 = 1.59 - 0.0592/6 log [Cr^3+]/6.35x10^-4

1.499 - 1.59 = - 0.0592/6 log [Cr^3+]/6.35x10^-4

-0.091 = -0.00987 log [Cr^3+]/6.35x10^-4

-0.091/ -0.00987 = log [Cr^3+]/6.35x10^-4

9.22 = log [Cr^3+]/6.35x10^-4

Antilog (9.22) = [Cr^3+]/6.35x10^-4

1.66 × 10^9 = [Cr^3+]/6.35x10^-4

[Cr^3+] = 1.66 × 10^9 × 6.35x10^-4

[Cr^3+] = 10.5 × 10^5 M