I already understand parts a & b, but can someone explain part c) to me? Conside
ID: 1061989 • Letter: I
Question
I already understand parts a & b, but can someone explain part c) to me?
Consider the following overall cell reaction: (3 points) Cr_2O_7^2- + 14H^+ + 6I^-1 rightarrow 2Cr^3+ + 3I_2 + 7H_2O write the cell reactions at the cathode and anode. What is the E^0 for the cell? What is the emf of the cell (Ecell) when [Cr_2O_7]^2- = 1.5 M, [H^+] = 1.0 M, [I^-] = 1.0M, and [Cr^3+] = 1.7 times 10^-5M ? Cr_2O_7^2- + 14H^+ + 6e^-1 rightarrow 2Cr^3+ +7HzO E degree = + 1.33V 2I^-1 rightarrow I_2 + 2e^-1 E degree = +0.54 VExplanation / Answer
a) the cell reaction at cathode [Reduction]
Cr2O7-2+ 14 H+ + 6e ---> 2 Cr+3 + 7H2O E0 =1.33
The cell reaction at anode [oxidation]
2I - --> I2 + 2e E0 = 0.54
b) E0cell = E0cathode - E0anode = 1.33 - 0. 54 = 0.79 V
c) Now here we will use Nernst equation to calculate the Ecell which is given as
Ecell = E0cell- [2.303 RT / nF] / n log Q
F= 96485
T = 298 K
R = 8.314
Ecell = E0cell- 0.0592 / n log Q
n = 6 [the change in electrons]
Q = [Cr+3]2 / [Cr2O7-2] X [H+]14 X [I-]6
Putting values
Q = [1.7 X 10^-5]^2 / [1.5 X (1)^14 (1)^6
Q = 1.13 X 10^-5
log Q = -4.94
Ecell = 0.79 - 0.0592/ 6 (-4.94)
Ecell = 0.79 + 0.049 = 0.839 V
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