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atroni [7]
3 years ago
15

For the reaction, 2cr2+ + cl2(g) ---> 2cr3+ + 2cl- e cell (standard conditions) = 1.78v calculate ecell (standard conditions)

for the related reaction cr3+ + cl- ----> cr2+ + 1/2cl2(g)
A. 0.89 v
B. - 0.89 v
C. 1.78 v
D. - 1.78 v
E. Non of these
Chemistry
1 answer:
PolarNik [594]3 years ago
3 0

Answer:

-1.78 V

Explanation:

There are several rules required to calculate the cell potential:

  • given standard cell potential, we may reverse the equation: the products of a given reaction become our reactants, while reactants become our products in the reversed equation. For a reversed equation, we change the sign of the cell potential to the opposite sign;
  • if we multiply the whole equation by some number, this doesn't influence the cell potential value. It only produces a different expression in the equilibrium constant.

That said, notice that the initial reaction with respect to the final reaction is:

  • reversed: chromium(III) cation and chloride anion become our reactants as opposed to the products in the initial reaction, so we change the sign of the cell potential to a negative value of -1.78 V;
  • each coefficient is multiplied by a fraction of \frac{1}{2}. It doesn't influence the value of the cell potential.

Thus, we have a cell of E = -1.78 V.

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Initial              0.0375       0        0
Change         -x               +x      +x
Equilibrium    0.0375-x     x        x

We calculate [H+] from Ka:     
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Approximating that x is negligible compared to 0.0375 simplifies the equation to         
     3.0x10^-8 = (x)(x)/0.0375     
     3.0x10^-8 = x2/0.0375     
     x2 = (3.0x10^-8)(0.0375) = 1.125x10^-9     
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in which 0.0000335 is indeed negligible compared to 0.0375.

We can now calculate pH:     
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