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Inessa [10]
2 years ago
14

Using Equation (10), calculate [Ag+] in the cell, where it is in equilibrium with 1 M Cl- ion. (Ecell in Equation (10) is the ne

gative of the measured value if the polarity is not the same as the standard cell.) Take [Cu2+] to be 1 M. Show your calculations. Ecell=0.4249v, Ecell=-0.00191. M E =E- 0.0592/2 • log ([Cu2+]/[Ag+1?) (10) cell
Chemistry
1 answer:
mixer [17]2 years ago
5 0

Answer:

7.16x10⁻⁸M = [Ag+]

Explanation:

Using the equation:

E(Cell) =E⁰ - 0.0592/2 • log ([Cu2+]/[Ag+]²)

<em>Where E</em>⁰<em>= 0.4249V</em>

<em>E(Cell) = -(-0.0019V) -Measured value-</em>

<em>[Cu2+] = 1M</em>

<em />

Replacing:

0.0019V = 0.4249V - 0.0592/2 • log (1M/[Ag+]²)

-0.423V = - 0.0296 • log (1M/[Ag+]²)

14.29 = log (1M/[Ag+]²)

1.95x10¹⁴ = 1M / [Ag+]²

[Ag+]² = 5.12x10⁻¹⁵M

7.16x10⁻⁸M = [Ag+]

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